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Lemma 14.5.\n\n\\[ \n\\Gamma \\left( {x, y, z}\\right) = \\frac{{\\Delta }_{x + y + z}!{\\Delta }_{x - 1}!{\\Delta }_{y - 1}!{\\Delta }_{z - 1}!}{{\\Delta }_{y + z - 1}!{\\Delta }_{z + x - 1}!{\\Delta }_{x + y - 1}!}. \n\\]
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Proof. Consider the equations depicted in Figure 14.4; as usual a symbol beside a line is a count of the number of parallel arcs that it represents. The first equality follows from the defining relation of Figure 13.6 for \\( {f}^{\\left( y + z - 1\\right) } \\) (together with \\( {f}^{\\left( z\\right) }{e}_{z - 1} = 0 \\) ), and the second line follows by iterating the first line. Next, the defining relation for \\( {f}^{\\left( y + z\\right) } \\) followed by a double application of Figure 14.4 produces the identity of Figure 14.5.\n\nNow apply this last identity to Figure 14.3, using the formulae of Figures 13.4 and 13.5. The following recurrence relation results:\n\n\\[ \n\\Gamma \\left( {x, y, z}\\right) = \n\\]\n\n\\[ \n\\Gamma \\left( {x, y, z - 1}\\right) {\\Delta }_{x + z}/{\\Delta }_{x + z - 1} - \\Gamma \\left( {x + 1, y - 1, z - 1}\\right) {\\left( {\\Delta }_{y - 1}\\right) }^{2}/\\left( {{\\Delta }_{y + z - 1}{\\Delta }_{y + z - 2}}\\right) . \n\\]\n\nThis is ready for a verification of the given formula by induction on \\( z \\). That formula is clearly true when \\( z = 0 \\), and inserting it into this recurrence relation reduces the proof to a demonstration of the equality\n\n\\[ \n{\\Delta }_{x + y + z}{\\Delta }_{z - 1} = {\\Delta }_{x + z}{\\Delta }_{y + z - 1} - {\\Delta }_{y - 1}{\\Delta }_{x} \n\\]\n\nThe truth of this can however easily be checked either directly from the formula for \\( {\\Delta }_{n} \\) or using a double induction on\n\n\\[ \n{\\Delta }_{x + y} = {\\Delta }_{x}{\\Delta }_{y} - {\\Delta }_{x - 1}{\\Delta }_{y - 1} \n\\]
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Yes
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Lemma 14.7. Let \( \left( {a, b, c}\right) \) be admissible and let \( A \) be a primitive \( 4{r}^{\text{th }} \) root of unity. Then \( {\tau }_{a, b, c}^{ * } \) is non-zero if and only if \( a + b + c \leq 2\left( {r - 2}\right) \) .
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Proof. \( \mathcal{S}{D}^{\prime } \) has a base consisting of all diagrams in \( {D}^{\prime } \) with no crossing. For all but one of these diagrams there is an arc from a point of one of the three specified subsets (for example, that with \( a \) points) to another point of the same subset. As usual (using \( {f}^{\left( a\right) }{e}_{i} = 0 \) ), \( {\tau }_{a, b, c}^{ * } \) annihilates such an element. There remains to consider the base element of \( \mathcal{S}{D}^{\prime } \) that consists of \( z \) arcs from the first boundary subset to the second such subset, \( x \) from the second to the third and \( y \) from the third to the first. Of course, \( {T}_{a, b, c}^{ * } \) maps this element to \( \Gamma \left( {x, y, z}\right) \) . It follows from Lemma 14.5 that as \( x + y + z \) increases, this is non-zero until \( {\Delta }_{x + y + z}! = 0 \) and that this occurs when \( x + y + z = r - 1 \) .
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Yes
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Lemma 14.9. Suppose \( A \) is not a root of unity. \( A \) base for \( {Q}_{a, b, c, d} \) is the set of elements as in Figure 14.10 (the boundary of the disc is not shown), where \( j \) takes all values such that \( \left( {a, b, j}\right) \) and \( \left( {c, d, j}\right) \) are both admissible.
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Proof. Note that the proposed base elements each consist of two triads glued together; there is an \( {f}^{\left( j\right) } \) on the central line. Certainly \( {Q}_{a, b, c, d} \) is spanned by all elements of the form shown in Figure 14.11, where the lines all represent multiple parallel arcs, for, as usual, any other diagrams interact with the idempotents to give zero. Without loss of generality, it is assumed that \( b + d \geq a + c \), and it is clear that the diagonal line represents \( \frac{1}{2}\{ b + d - a - c\} \) parallel arcs. The number of arcs represented by the other lines can vary. Suppose there are \( j \) arcs crossing the vertical dotted line. In the Temperley-Lieb algebra \( T{L}_{j} \), recall that \( \mathbf{1} - {f}^{\left( j\right) } \) is in the ideal generated by the \( {e}_{i} \) . Thus a diagram with \( j \) arcs crossing the dotted line can be replaced with a linear sum of diagrams, one with \( j \) arcs containing an \( {f}^{\left( j\right) } \) and others that cross the vertical line fewer than \( j \) times (coming from the \( {e}_{i} \) ). Thus, by induction on the number of arcs crossing the vertical line, it is seen that the given elements span the space.
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Yes
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Lemma 14.10. Suppose \( A \) is a primitive \( 4{r}^{\text{th }} \) root of unity. A base for \( {Q}_{a, b, c, d}^{ * } \) (this being \( {Q}_{a, b, c, d} \) regarded as maps of diagrams outside the disc) is the set of elements as in Figure 14.10 where \( j \) takes all values such that \( \left( {a, b, j}\right) \) and \( \left( {c, d, j}\right) \) are both \( r \) -admissible.
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Proof. The proof that the given elements span is the same as in Lemma 14.9 with a small modification. Now, \( {f}^{\left( n\right) } \) does not exist for \( n \geq r \) . However, \( {f}^{\left( r - 1\right) } \) is the zero map of outsides. Thus working in this dual context, any diagram as in the above proof, with at least \( \left( {r - 1}\right) \) arcs crossing the dotted vertical line, can be replaced by a sum of diagrams with fewer such arcs. Further, any triad encountered that is not \( r \) -admissible may be discarded, since it represents the zero map. The proof of independence is essentially the same as before (though the map used now goes from and to the dual spaces).
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Yes
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Lemma 14.11. In \( \mathcal{S}\left( {{S}^{1} \times I}\right) ,{S}_{a}\left( \alpha \right) {S}_{b}\left( \alpha \right) = \mathop{\sum }\limits_{c}{S}_{c}\left( \alpha \right) \) where the summation is over all \( c \) such that \( \left( {a, b, c}\right) \) is admissible. If \( \bar{A} \) is a primitive \( 4{r}^{\text{th }} \) root of unity regarding both sides of the equation as maps of outsides (of immersed annuli as in Chapter 13), \( {S}_{a}\left( \alpha \right) {S}_{b}\left( \alpha \right) = \mathop{\sum }\limits_{c}{S}_{c}\left( \alpha \right) \), where now the sum is over all \( c \) such that \( \left( {a, b, c}\right) \) is \( r \) -admissible.
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Proof. In fact the first part of this lemma is almost immediate. This is because it is a result on Chebyshev polynomials that \( {S}_{a}\left( x\right) {S}_{b}\left( x\right) = \mathop{\sum }\limits_{c}{S}_{c}\left( x\right) \), the sum being over all \( c \) such that \( \left( {a, b, c}\right) \) is admissible. This follows by induction on \( b \) . However, another proof is shown in Figure 14.16, where the result of Figure 14.15 is first applied at the top of the diagram and then the result of Figure 14.8 is applied at the bottom. The advantage of this alternative proof is that it also works in the \( r \) -admissible case as well.
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Yes
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Theorem 14.12. Let \( {F}_{g} \) be the closed orientable surface of genus \( g \) and let \( A \) be a primitive \( 4{r}^{\text{th }} \) root of unity, \( r \geq 3 \) . Then \( {\mathcal{I}}_{A}\left( {{S}^{1} \times {F}_{g}}\right) \) is an integer. It is \( r - 1 \) when \( g = 1 \) . Otherwise it is the number of ways of labelling the \( 3\left( {g - 1}\right) \) edges of the graph of Figure 14.18 with integers \( {a}_{i},0 \leq {a}_{i} \leq r - 2 \), so that the three labels at any node form an \( r \) -admissible triple.
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Proof. The 3-manifold \( {S}^{1} \times {F}_{g} \) is obtained by surgery on a link that consists of \( g \) copies of the Borromean rings summed together on one component, each component having the zero framing. (Proving this is an interesting exercise.) A diagram \( D \) for such a link is obtained by taking \( g \) annuli, each containing a link as on the left of Figure 14.17, threading an unknotted closed curve through these annuli and then taking the resultant diagram of \( {2g} + 1 \) components. Then \( < \omega ,\omega ,\ldots ,\omega { > }_{D} = < \omega ,{\beta }^{g}{ > }_{H} \), where \( H \) is just the two-crossing diagram of the simple Hopf link of two curves. Thus, as the signature of the linking matrix of\n\n\n\nFigure 14.18\n\nthis link is zero (and using \( \left\langle {{\mu \omega }{ > }_{U} = {\mu }^{-1}}\right\rangle \) ,\n\n\[{\mathcal{I}}_{A}\left( {{S}^{1} \times {F}_{g}}\right) = {\mu }^{{2g} + 2} < \omega ,{\beta }^{g}{ > }_{H}.\n\]\n\nNow \( \beta = \mathop{\sum }\limits_{{a = 0}}^{{r - 2}}{\mu }^{-2}{\left( {S}_{a}\left( \alpha \right) \right) }^{2} \), so \( {\beta }^{g} \) can be expressed as a sum of the \( {S}_{n}\left( \alpha \right) \) ’s by Lemma 14.11. Then, by Lemma 13.9, \( \langle \omega ,{\beta }^{g}{\rangle }_{H} \) is \( {\mu }^{-2 - {2g}}N \), where \( N \) is the number of times \( {S}_{0}\left( \alpha \right) \) appears in the expansion (by Lemma 14.11) for \( {\left( \mathop{\sum }\limits_{{a = 0}}^{{r - 2}}{\left( {S}_{a}\left( \alpha \right) \right) }^{2}\right) }^{g} \) as a sum of the \( {S}_{n}\left( \alpha \right) \) ’s. This \( N \) is the number of \( r \) -admissible labellings of the edges of Figure 14.18.
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No
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Theorem 14.13. If \( p \) and \( q \) are coprime positive integers, then the Jones polynomial of the \( \left( {p, q}\right) \) -torus knot is\n\n\[ \n{t}^{\left( {p - 1}\right) \left( {q - 1}\right) /2}{\left( 1 - {t}^{2}\right) }^{-1}\left( {1 - {t}^{p + 1} - {t}^{q + 1} + {t}^{p + q}}\right) .\n\]
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Proof. Consider the diagram of Figure 14.23, which shows \( p \) arcs traversing a rectangle. Suppose \( q \) copies of this are placed side by side and the result is closed up by joining the \( p \) points on the left to those on the right, using \( p \) crossing-free arcs encircling an annulus \( {S}^{1} \times I \) to form a diagram \( T\left( {p, q}\right) \) in that annulus. It is desired to evaluate this diagram in the skein of the annulus in terms of the base elements \( \left\{ {{S}_{n}\left( \alpha \right) }\right\} \) . Then, placing the annulus in the plane will at once give a value for the Jones polynomial of the \( \left( {p, q}\right) \) -torus knot. For some fixed \( k \) (which will here later be taken to be 1), consider \( p \) arcs side by side in a diagram, each labelled with an \( {f}^{\left( k\right) } \) so that as usual each arc represents \( k \) parallel arcs with the idempotent inserted. Applying the identity of Figure 14.15 ( \( p - 1 \) ) times shows that this is of the form of Figure 14.24, where the coefficient \( \Lambda \left( {{i}_{1},{i}_{2},\ldots ,{i}_{p - 2}, a}\right) \) is the quotient of a product of \( \Delta \) ’s by a product of \( \theta \) ’s and the summation is over all \( \left( {{i}_{1},{i}_{2},\ldots ,{i}_{p - 2}, a}\right) \) that produce an admissible triple at each vertex of the diagram.\n\n\n\nFigure 14.23\n\n\n\n\n\nFigure 14.24\n\nThe diagram of Figure 14.25 is, of course, a multiple of \( {f}^{\left( a\right) } \) ; let it be denoted, without any summation convention, by\n\n\[ \n{\left\{ \Lambda \left( {i}_{1},{i}_{2},\ldots ,{i}_{p - 2}, a\right) \Lambda \left( {j}_{1},{j}_{2},\ldots ,{j}_{p - 2}, a\right) \right\} }^{-1/2}M{\left( a\right) }_{\mathbf{j}}^{\mathbf{i}}{f}^{\left( a\right) }.\n\]\n\nThe \( \left\{ {M{\left( a\right) }_{\mathrm{i}}^{\mathrm{i}}}\right\} \) will be regarded as a matrix \( M\left( a\right) \) with rows and columns indexed by i and \( \mathbf{j} \), each representing a multi-suffix \( \left( {{i}_{1},{i}_{2},\ldots ,{i}_{p - 2}}\right) \) or \( \left( {{j}_{1},{j}_{2},\ldots ,{j}_{p - 2}}\right) \) .\n\n\n\nFigure 14.25\n\nSuppose \( T{\left( p, q\right) }^{\left( k\right) } \) is the diagram \( T\left( {p, q}\right) \) decorated by \( {S}_{k}\left( \alpha \right) \) . Suppose that in \( T{\left( p, q\right) }^{\left( k\right) } \), between consecutive occurrences of (the \( k \) -weighted) Figure 14.23, the \( p \) parallel strings are \
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Yes
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Lemma 15.1. Suppose that \( p \) and \( q \) are two arcs in \( {\mathbb{R}}^{2} \) meeting only at their end points \( A \) and \( B \), and let \( R \) be the compact region bounded by \( p \cup q \) . Suppose that \( {t}_{1},{t}_{2},\ldots {t}_{n} \) are arcs in \( R \), each meeting \( p \cup q \) at just its end points, one in \( p \) and one in \( q \) . Suppose that every \( {t}_{i} \cap {t}_{j} \) is at most one point, that intersections of arcs are transverse and that there are no triple points. The graph, with vertices all intersections of these arcs and edges comprising \( p \cup q \cup \mathop{\bigcup }\limits_{i}{t}_{i} \), separates \( R \) into a collection of \( v \) -gons; amongst these \( v \) -gons there is a 3 -gon with an edge in \( p \) and a 3-gon with an edge in \( q \) .
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Proof. Proceed by induction on the number \( n \) of arcs. The result is trivial if \( n = 1 \), so assume \( n > 1 \) . Amongst the end points of the \( {t}_{i} \) that lie on \( p \), let \( X \) be the nearest to \( A \) . If then \( X \) is an end of \( {t}_{j} \), let \( {B}^{\prime } \) be the other end of \( {t}_{j} \) on \( q \) . If possible, from \( \left\{ {{t}_{i} : i \neq j}\right\} \) select a \( {t}_{k} \) with \( {t}_{k} \cap {t}_{j} = {X}^{\prime },{t}_{k} \cap q = {A}^{\prime } \), such that \( {t}_{k} \) has no point of intersection with a \( {t}_{i} \) between \( {A}^{\prime } \) and \( {X}^{\prime } \) . Select such a \( {t}_{k} \) with \( {X}^{\prime } \) as near as possible to \( {B}^{\prime } \), see Figure 15.1. If there is no such \( {t}_{k} \), select \( p \) instead, taking \( {A}^{\prime } = A \) and \( {X}^{\prime } = X \) . Now let \( {p}^{\prime } \) be an arc starting at \( {A}^{\prime } \), proceeding along\n\n\n\nFigure 15.1\n\n\( {t}_{k} \) (or \( p \) if there is no \( {t}_{k} \) ) to \( {X}^{\prime } \) and then along \( {t}_{j} \) to \( {B}^{\prime } \) (it helps not to think of a corner at \( {X}^{\prime } \) ). Let \( {q}^{\prime } \) be the sub-arc of \( q \) from \( {A}^{\prime } \) to \( {B}^{\prime } \) and let \( {R}^{\prime } \) be the region bounded by \( {p}^{\prime } \cup {q}^{\prime } \) . If no \( {t}_{i} \) meets the interior of \( {R}^{\prime } \), then \( {R}^{\prime } \) is a 3 -gon with an edge in \( q \) . Otherwise \( {R}^{\prime } \) meets \( \left\{ {{t}_{i} : i \neq j, i \neq k}\right\} \) in fewer than \( n \) arcs and so, by induction, there is a 3-gon in \( {R}^{\prime } \) with an edge in \( {q}^{\prime } \) . The choice made for \( {t}_{k} \) ensures that \( {A}^{\prime } \) is not a vertex of this 3-gon (which is important, as \( {X}^{\prime } \) is not a vertex of a \( v \) -gon of \( {R}^{\prime } \) ), and so it is one of the original 3-gons having an edge contained in \( q \) . Similarly, there is a 3-gon with an edge in \( p \) .
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Yes
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Theorem 15.5. There exists a function \[ \Lambda : \left\{ {\text{ Unoriented links diagrams in }{S}^{2}}\right\} \rightarrow \mathbb{Z}\left\lbrack {{a}^{\pm 1},{z}^{\pm 1}}\right\rbrack \] that is defined uniquely by the following: (i) \( \Lambda \left( U\right) = 1 \), where \( U \) is the zero-crossing diagram of the unknot; (ii) \( \Lambda \left( D\right) \) is unchanged by Reidemeister moves of Types II and III on the diagram \( D \) ; (iii) \( \Lambda \left( { \curvearrowright \sim }\right) = {a\Lambda }\left( \curvearrowright \right) \) ; (iv) If \( {D}_{ + },{D}_{ - },{D}_{0} \) and \( {D}_{\infty } \) are four diagrams exactly the same except near a point where they are as shown in Figure 15.6, then \[ \left( {\star \star }\right) \;\Lambda \left( {D}_{ + }\right) + \Lambda \left( {D}_{ - }\right) = z\left( {\Lambda \left( {D}_{0}\right) + \Lambda \left( {D}_{\infty }\right) }\right) . \]
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Proof. Note that, when considering a crossing in an unoriented diagram, it has no claim to be termed \( {D}_{ + } \) rather than \( {D}_{ - } \) in the above notation. However, this never matters, since \( {D}_{ + } \) and \( {D}_{ - } \) feature symmetrically in the formula \( \left( {\star \star }\right) \) ; the treatment of \( {D}_{0} \) and \( {D}_{\infty } \) is likewise symmetric. Observe that the equation \( \left( {\star \star }\right) \) determines uniquely any one of \( \Lambda \left( {D}_{ + }\right) ,\Lambda \left( {D}_{ - }\right) ,\Lambda \left( {D}_{0}\right) \) and \( \Lambda \left( {D}_{\infty }\right) \) from knowledge of the other three. Observe also that a solution to \( \left( {\star \star }\right) \) is \( \left( {\Lambda \left( {D}_{ + }\right) ,\Lambda \left( {D}_{ - }\right) ,\Lambda \left( {D}_{0}\right) ,\Lambda \left( {D}_{\infty }\right) }\right) = \left( {{ax},{a}^{-1}x, x,{\delta x}}\right) \), where \( x \) is arbitrary and \( \delta = \left( {a + {a}^{-1}}\right) {z}^{-1} - 1 \)
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Yes
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Proposition 16.6. For an oriented link \( L \) , \[ V\left( L\right) = F\left( L\right) \text{ when }\left( {a, z}\right) = \left( {-{t}^{-3/4},\;\left( {{t}^{-1/4} + {t}^{1/4}}\right) }\right) , \] \[ {\left( V\left( L\right) \right) }^{2} = {\left( -1\right) }^{\# L - 1}F\left( L\right) \text{ when }t = - {q}^{-2},\;\left( {a, z}\right) = \left( {{q}^{3},{q}^{-1} + q}\right) . \]
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Proof. Underlying the Jones polynomial is the Kauffman bracket. Reverting to the notation for that, given in Definition 3.1, \[ \langle > < \rangle = A\langle > < \rangle + {A}^{-1}\langle < \rangle \] \[ \langle > < \rangle = {A}^{-1}\langle > < \rangle + A\langle > < \rangle . \] Adding these equations gives \[ \langle > < \rangle + \langle > < \rangle = \left( {A + {A}^{-1}}\right) \left( {\langle > < \rangle +\langle < \rangle }\right) . \] However, \( \langle \sim \rangle = - {A}^{3}\langle \sim \rangle \), and the Kauffman bracket is invariant under regular isotopy (Reidemeister Type II and III moves). However, these are the defining rules for the polynomial \( \Lambda \left( D\right) \) of Theorem 15.5 with the variables changed to \( \left( {a, z}\right) = \left( {-{A}^{3},\left( {A + {A}^{-1}}\right) }\right) \) . The substitution \( t = {A}^{-4} \) gives the first result. Subtracting the square of one of the above two equations from the square of the other gives \[ {\left( > < \right) }^{2} - \langle > < {\rangle }^{2} = \left( {{A}^{2} - {A}^{-2}}\right) \left( {\langle > < {\rangle }^{2}-\langle < {\rangle }^{2}}\right) . \] Of course \( \langle \sim \sim {\rangle }^{2} = {A}^{6}\langle \sim {\rangle }^{2} \), and so the square of the Kauffman bracket is an instance of the \( {\Lambda }^{ \star }\left( D\right) \) polynomial, defined at the very end of Chapter 15, that satisfies \[ {\Lambda }^{ \star }\left( {D}_{ + }\right) - {\Lambda }^{ \star }\left( {D}_{ - }\right) = \omega \left( {{\Lambda }^{ \star }\left( {D}_{0}\right) - {\Lambda }^{ \star }\left( {D}_{\infty }\right) }\right) . \] Now, translating the notation by \( \omega = \left( {{A}^{2} - {A}^{-2}}\right) ,\alpha = {A}^{6}, t = - {q}^{-2} = {A}^{-4} \) and (from Chapter 15) \( \left( {a, z}\right) = \left( {{i\alpha }, - {i\omega }}\right) \) the second result follows.
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Yes
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Proposition 16.9. Suppose \( L \) is an oriented link with \( \# L \) components. Then \( (1 - \) \( \# L) \) is the lowest power both of \( m \) in \( P\left( L\right) \) and of \( z \) in \( F\left( L\right) \), and
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\[ {\left\lbrack {z}^{\# L - 1}F\left( L\right) \right\rbrack }_{\left( {a, z}\right) = \left( {l,0}\right) } = {\left\lbrack {\left( -m\right) }^{\# L - 1}P\left( L\right) \right\rbrack }_{m = 0}. \]
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No
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Theorem 16.11. If an oriented link \( L \) is the closure of \( \xi \in {B}_{n} \), let \( T\left( L\right) \) be defined to be \( T\left( \xi \right) \) . This is a well-defined link invariant.
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Proof. Because \( T \) is essentially a trace function, if \( \xi ,\eta \in {B}_{n} \) then \( T\left( {{\eta }^{-1}{\xi \eta }}\right) = \n\n\( T\left( \xi \right) \) . Using the properties of \( \mu \), it is easy to show that \( T\left( {\xi {\sigma }_{n}}\right) = T\left( {\xi {\sigma }_{n}^{-1}}\right) = \n\n\( T\left( \xi \right) \) . The result then follows from Proposition 16.10.
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No
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Theorem 2.1 Suppose that \( f \) is an integrable function on the circle with \( \widehat{f}\left( n\right) = 0 \) for all \( n \in \mathbb{Z} \) . Then \( f\left( {\theta }_{0}\right) = 0 \) whenever \( f \) is continuous at the point \( {\theta }_{0} \) .
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Proof. We suppose first that \( f \) is real-valued, and argue by contradiction. Assume, without loss of generality, that \( f \) is defined on \( \left\lbrack {-\pi ,\pi }\right\rbrack \), that \( {\theta }_{0} = 0 \), and \( f\left( 0\right) > 0 \) . The idea now is to construct a family of trigonometric polynomials \( \left\{ {p}_{k}\right\} \) that \
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No
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Corollary 2.3 Suppose that \( f \) is a continuous function on the circle and that the Fourier series of \( f \) is absolutely convergent, \( \mathop{\sum }\limits_{{n = - \infty }}^{\infty }\left| {\widehat{f}\left( n\right) }\right| < \infty \) . Then, the Fourier series converges uniformly to \( f \), that is,\n\n\[ \mathop{\lim }\limits_{{N \rightarrow \infty }}{S}_{N}\left( f\right) \left( \theta \right) = f\left( \theta \right) \;\text{ uniformly in }\theta . \]
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Proof. Recall that if a sequence of continuous functions converges uniformly, then the limit is also continuous. Now observe that the assumption \( \sum \left| {\widehat{f}\left( n\right) }\right| < \infty \) implies that the partial sums of the Fourier\n\nseries of \( f \) converge absolutely and uniformly, and therefore the function \( g \) defined by\n\n\[ g\left( \theta \right) = \mathop{\sum }\limits_{{n = - \infty }}^{\infty }\widehat{f}\left( n\right) {e}^{in\theta } = \mathop{\lim }\limits_{{N \rightarrow \infty }}\mathop{\sum }\limits_{{n = - N}}^{N}\widehat{f}\left( n\right) {e}^{in\theta } \]\n\nis continuous on the circle. Moreover, the Fourier coefficients of \( g \) are precisely \( \widehat{f}\left( n\right) \) since we can interchange the infinite sum with the integral (a consequence of the uniform convergence of the series). Therefore, the previous corollary applied to the function \( f - g \) yields \( f = g \), as desired.
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Yes
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Corollary 2.4 Suppose that \( f \) is a twice continuously differentiable function on the circle. Then\n\n\[ \widehat{f}\left( n\right) = O\left( {1/{\left| n\right| }^{2}}\right) \;\text{ as }\left| n\right| \rightarrow \infty ,\] \n\nso that the Fourier series of \( f \) converges absolutely and uniformly to \( f \) .
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Proof. The estimate on the Fourier coefficients is proved by integrating by parts twice for \( n \neq 0 \) . We obtain\n\n\[ {2\pi }\widehat{f}\left( n\right) = {\int }_{0}^{2\pi }f\left( \theta \right) {e}^{-{in\theta }}{d\theta } \]\n\n\[ = {\left\lbrack f\left( \theta \right) \cdot \frac{-{e}^{-{in\theta }}}{in}\right\rbrack }_{0}^{2\pi } + \frac{1}{in}{\int }_{0}^{2\pi }{f}^{\prime }\left( \theta \right) {e}^{-{in\theta }}{d\theta } \]\n\n\[ = \frac{1}{in}{\int }_{0}^{2\pi }{f}^{\prime }\left( \theta \right) {e}^{-{in\theta }}{d\theta } \]\n\n\[ = \frac{1}{in}{\left\lbrack {f}^{\prime }\left( \theta \right) \cdot \frac{-{e}^{-{in\theta }}}{in}\right\rbrack }_{0}^{2\pi } + \frac{1}{{\left( in\right) }^{2}}{\int }_{0}^{2\pi }{f}^{\prime \prime }\left( \theta \right) {e}^{-{in\theta }}{d\theta } \]\n\n\[ = \frac{-1}{{n}^{2}}{\int }_{0}^{2\pi }{f}^{\prime \prime }\left( \theta \right) {e}^{-{in\theta }}{d\theta } \]\n\nThe quantities in brackets vanish since \( f \) and \( {f}^{\prime } \) are periodic. Therefore\n\n\[ {2\pi }{\left| n\right| }^{2}\left| {\widehat{f}\left( n\right) }\right| \leq \left| {{\int }_{0}^{2\pi }{f}^{\prime \prime }\left( \theta \right) {e}^{-{in\theta }}{d\theta }}\right| \leq {\int }_{0}^{2\pi }\left| {{f}^{\prime \prime }\left( \theta \right) }\right| {d\theta } \leq C, \]\n\nwhere the constant \( C \) is independent of \( n \) . (We can take \( C = {2\pi B} \) where \( B \) is a bound for \( {f}^{\prime \prime } \) .) Since \( \sum 1/{n}^{2} \) converges, the proof of the corollary is complete.
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Yes
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Proposition 3.1 Suppose that \( f, g \), and \( h \) are \( {2\pi } \) -periodic integrable functions. Then:\n\n(i) \( f * \left( {g + h}\right) = \left( {f * g}\right) + \left( {f * h}\right) \) .
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Proof. Properties (i) and (ii) follow at once from the linearity of the integral.
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No
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Lemma 3.2 Suppose \( f \) is integrable on the circle and bounded by \( B \) . Then there exists a sequence \( {\left\{ {f}_{k}\right\} }_{k = 1}^{\infty } \) of continuous functions on the circle so that\n\n\[ \mathop{\sup }\limits_{{x \in \left\lbrack {-\pi ,\pi }\right\rbrack }}\left| {{f}_{k}\left( x\right) }\right| \leq B\;\text{ for all }k = 1,2,\ldots ,\]\n\nand\n\n\[ {\int }_{-\pi }^{\pi }\left| {f\left( x\right) - {f}_{k}\left( x\right) }\right| {dx} \rightarrow 0\;\text{ as }k \rightarrow \infty . \]
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Using this result, we may complete the proof of the proposition as follows. Apply Lemma 3.2 to \( f \) and \( g \) to obtain sequences \( \left\{ {f}_{k}\right\} \) and \( \left\{ {g}_{k}\right\} \) of approximating continuous functions. Then\n\n\[ f * g - {f}_{k} * {g}_{k} = \left( {f - {f}_{k}}\right) * g + {f}_{k} * \left( {g - {g}_{k}}\right) .\n\nBy the properties of the sequence \( \left\{ {f}_{k}\right\} \) ,\n\n\[ \left| {\left( {f - {f}_{k}}\right) * g\left( x\right) }\right| \leq \frac{1}{2\pi }{\int }_{-\pi }^{\pi }\left| {f\left( {x - y}\right) - {f}_{k}\left( {x - y}\right) }\right| \left| {g\left( y\right) }\right| {dy}\n\n\[ \leq \frac{1}{2\pi }\mathop{\sup }\limits_{y}\left| {g\left( y\right) }\right| {\int }_{-\pi }^{\pi }\left| {f\left( y\right) - {f}_{k}\left( y\right) }\right| {dy}\n\n\[ \rightarrow 0\;\text{ as }k \rightarrow \infty \text{. }\n\nHence \( \left( {f - {f}_{k}}\right) * g \rightarrow 0 \) uniformly in \( x \) . Similarly, \( {f}_{k} * \left( {g - {g}_{k}}\right) \rightarrow 0 \) uniformly, and therefore \( {f}_{k} * {g}_{k} \) tends uniformly to \( f * g \) . Since each \( {f}_{k} * {g}_{k} \) is continuous, it follows that \( f * g \) is also continuous, and we have (v).
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No
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Theorem 4.1 Let \( {\left\{ {K}_{n}\right\} }_{n = 1}^{\infty } \) be a family of good kernels, and \( f \) an integrable function on the circle. Then\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {f * {K}_{n}}\right) \left( x\right) = f\left( x\right) \]\n\nwhenever \( f \) is continuous at \( x \) . If \( f \) is continuous everywhere, then the above limit is uniform.
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Proof of Theorem 4.1. If \( \epsilon > 0 \) and \( f \) is continuous at \( x \), choose \( \delta \) so that \( \left| y\right| < \delta \) implies \( \left| {f\left( {x - y}\right) - f\left( x\right) }\right| < \epsilon \) . Then, by the first property of good kernels, we can write\n\n\[ \left( {f * {K}_{n}}\right) \left( x\right) - f\left( x\right) = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }{K}_{n}\left( y\right) f\left( {x - y}\right) {dy} - f\left( x\right) \]\n\n\[ = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }{K}_{n}\left( y\right) \left\lbrack {f\left( {x - y}\right) - f\left( x\right) }\right\rbrack {dy}. \]\n\nHence,\n\n\[ \left| {\left( {f * {K}_{n}}\right) \left( x\right) - f\left( x\right) }\right| = \left| {\frac{1}{2\pi }{\int }_{-\pi }^{\pi }{K}_{n}\left( y\right) \left\lbrack {f\left( {x - y}\right) - f\left( x\right) }\right\rbrack {dy}}\right| \]\n\n\[ \leq \frac{1}{2\pi }{\int }_{\left| y\right| < \delta }\left| {{K}_{n}\left( y\right) }\right| \left| {f\left( {x - y}\right) - f\left( x\right) }\right| {dy} \]\n\n\[ + \frac{1}{2\pi }{\int }_{\delta \leq \left| y\right| \leq \pi }\left| {{K}_{n}\left( y\right) }\right| \left| {f\left( {x - y}\right) - f\left( x\right) }\right| {dy} \]\n\n\[ \leq \frac{\epsilon }{2\pi }{\int }_{-\pi }^{\pi }\left| {{K}_{n}\left( y\right) }\right| {dy} + \frac{2B}{2\pi }{\int }_{\delta \leq \left| y\right| \leq \pi }\left| {{K}_{n}\left( y\right) }\right| {dy} \]\n\nwhere \( B \) is a bound for \( f \) . The first term is bounded by \( {\epsilon M}/{2\pi } \) because of the second property of good kernels. By the third property we see that for all large \( n \), the second term will be less than \( \epsilon \) . Therefore, for some constant \( C > 0 \) and all large \( n \) we have\n\n\[ \left| {\left( {f * {K}_{n}}\right) \left( x\right) - f\left( x\right) }\right| \leq {C\epsilon } \]\n\nthereby proving the first assertion in the theorem. If \( f \) is continuous everywhere, then it is uniformly continuous, and \( \delta \) can be chosen independent of \( x \) . This provides the desired conclusion that \( f * {K}_{n} \rightarrow f \) uniformly.
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Yes
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Lemma 5.1 We have\n\n\[ \n{F}_{N}\left( x\right) = \frac{1}{N}\frac{{\sin }^{2}\left( {{Nx}/2}\right) }{{\sin }^{2}\left( {x/2}\right) }\n\]\n\nand the Fejér kernel is a good kernel.
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The proof of the formula for \( {F}_{N} \) (a simple application of trigonometric identities) is outlined in Exercise 15. To prove the rest of the lemma, note that \( {F}_{N} \) is positive and \( \frac{1}{2\pi }{\int }_{-\pi }^{\pi }{F}_{N}\left( x\right) {dx} = 1 \), in view of the fact that a similar identity holds for the Dirichlet kernels \( {D}_{n} \) . However, \( {\sin }^{2}\left( {x/2}\right) \geq \) \( {c}_{\delta } > 0 \), if \( \delta \leq \left| x\right| \leq \pi \), hence \( {F}_{N}\left( x\right) \leq 1/\left( {N{c}_{\delta }}\right) \), from which it follows that\n\n\[ \n{\int }_{\delta \leq \left| x\right| \leq \pi }\left| {{F}_{N}\left( x\right) }\right| {dx} \rightarrow 0\;\text{ as }N \rightarrow \infty .\n\]
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No
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Corollary 5.3 If \( f \) is integrable on the circle and \( \widehat{f}\left( n\right) = 0 \) for all \( n \) , then \( f = 0 \) at all points of continuity of \( f \) .
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The proof is immediate since all the partial sums are 0 , hence all the Cesàro means are 0 .
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No
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Corollary 5.4 Continuous functions on the circle can be uniformly approximated by trigonometric polynomials.
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This means that if \( f \) is continuous on \( \left\lbrack {-\pi ,\pi }\right\rbrack \) with \( f\left( {-\pi }\right) = f\left( \pi \right) \) and \( \epsilon > 0 \), then there exists a trigonometric polynomial \( P \) such that\n\n\[ \left| {f\left( x\right) - P\left( x\right) }\right| < \epsilon \;\text{ for all } - \pi \leq x \leq \pi . \]\n\nThis follows immediately from the theorem since the partial sums, hence the Cesàro means, are trigonometric polynomials.
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Yes
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Lemma 5.5 If \( 0 \leq r < 1 \), then\n\n\[ \n{P}_{r}\left( \theta \right) = \frac{1 - {r}^{2}}{1 - {2r}\cos \theta + {r}^{2}}.\n\]
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Proof. The identity \( {P}_{r}\left( \theta \right) = \frac{1 - {r}^{2}}{1 - {2r}\cos \theta + {r}^{2}} \) has already been derived in Section 1.1. Note that\n\n\[ \n1 - {2r}\cos \theta + {r}^{2} = {\left( 1 - r\right) }^{2} + {2r}\left( {1 - \cos \theta }\right) .\n\]\n\nHence if \( 1/2 \leq r \leq 1 \) and \( \delta \leq \left| \theta \right| \leq \pi \), then\n\n\[ \n1 - {2r}\cos \theta + {r}^{2} \geq {c}_{\delta } > 0.\n\]\n\nThus \( {P}_{r}\left( \theta \right) \leq \left( {1 - {r}^{2}}\right) /{c}_{\delta } \) when \( \delta \leq \left| \theta \right| \leq \pi \), and the third property of good kernels is verified. Clearly \( {P}_{r}\left( \theta \right) \geq 0 \), and integrating the expression (4) term by term (which is justified by the absolute convergence of the series) yields\n\n\[ \n\frac{1}{2\pi }{\int }_{-\pi }^{\pi }{P}_{r}\left( \theta \right) {d\theta } = 1\n\]\n\nthereby concluding the proof that \( {P}_{r} \) is a good kernel.
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Yes
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Lemma 1.2 (Best approximation) If \( f \) is integrable on the circle with Fourier coefficients \( {a}_{n} \), then\n\n\[ \begin{Vmatrix}{f - {S}_{N}\left( f\right) }\end{Vmatrix} \leq \begin{Vmatrix}{f - \mathop{\sum }\limits_{{\left| n\right| \leq N}}{c}_{n}{e}_{n}}\end{Vmatrix} \]\n\nfor any complex numbers \( {c}_{n} \) . Moreover, equality holds precisely when \( {c}_{n} = {a}_{n} \) for all \( \left| n\right| \leq N \) .
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Proof. This follows immediately by applying the Pythagorean theorem to\n\n\[ f - \mathop{\sum }\limits_{{\left| n\right| \leq N}}{c}_{n}{e}_{n} = f - {S}_{N}\left( f\right) + \mathop{\sum }\limits_{{\left| n\right| \leq N}}{b}_{n}{e}_{n} \]\n\nwhere \( {b}_{n} = {a}_{n} - {c}_{n} \) .
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Yes
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Lemma 1.5 Suppose \( F \) and \( G \) are integrable on the circle with\n\n\[ F \sim \sum {a}_{n}{e}^{in\theta }\;\text{ and }\;G \sim \sum {b}_{n}{e}^{in\theta }.\]\n\nThen\n\n\[ \frac{1}{2\pi }{\int }_{0}^{2\pi }F\left( \theta \right) \overline{G\left( \theta \right) }{d\theta } = \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{a}_{n}\overline{{b}_{n}}.\]
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Proof. The proof follows from Parseval's identity and the fact that\n\n\[ \left( {F, G}\right) = \frac{1}{4}\left\lbrack {\parallel F + G{\parallel }^{2} - \parallel F - G{\parallel }^{2} + i\left( {\parallel F + {iG}{\parallel }^{2} - \parallel F - {iG}{\parallel }^{2}}\right) }\right\rbrack \]\n\nwhich holds in every Hermitian inner product space. The verification of this fact is left to the reader.
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No
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Theorem 2.1 Let \( f \) be an integrable function on the circle which is differentiable at a point \( {\theta }_{0} \) . Then \( {S}_{N}\left( f\right) \left( {\theta }_{0}\right) \rightarrow f\left( {\theta }_{0}\right) \) as \( N \) tends to infinity.
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Proof. Define\n\n\[ F\left( t\right) = \left\{ \begin{array}{ll} \frac{f\left( {{\theta }_{0} - t}\right) - f\left( {\theta }_{0}\right) }{t} & \text{ if }t \neq 0\text{ and }\left| t\right| < \pi \\ - {f}^{\prime }\left( {\theta }_{0}\right) & \text{ if }t = 0. \end{array}\right. \]\n\nFirst, \( F \) is bounded near 0 since \( f \) is differentiable there. Second, for all small \( \delta \) the function \( F \) is integrable on \( \left\lbrack {-\pi , - \delta }\right\rbrack \cup \left\lbrack {\delta ,\pi }\right\rbrack \) because \( f \) has this property and \( \left| t\right| > \delta \) there. As a consequence of Proposition 1.4 in the appendix, the function \( F \) is integrable on all of \( \left\lbrack {-\pi ,\pi }\right\rbrack \) . We know that \( {S}_{N}\left( f\right) \left( {\theta }_{0}\right) = \left( {f * {D}_{N}}\right) \left( {\theta }_{0}\right) \), where \( {D}_{N} \) is the Dirichlet kernel. Since \( \frac{1}{2\pi }\int {D}_{N} = 1 \), we find that\n\n\[ {S}_{N}\left( f\right) \left( {\theta }_{0}\right) - f\left( {\theta }_{0}\right) = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }f\left( {{\theta }_{0} - t}\right) {D}_{N}\left( t\right) {dt} - f\left( {\theta }_{0}\right) \]\n\n\[ = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }\left\lbrack {f\left( {{\theta }_{0} - t}\right) - f\left( {\theta }_{0}\right) }\right\rbrack {D}_{N}\left( t\right) {dt} \]\n\n\[ = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }F\left( t\right) t{D}_{N}\left( t\right) {dt} \]\n\nWe recall that\n\n\[ t{D}_{N}\left( t\right) = \frac{t}{\sin \left( {t/2}\right) }\sin \left( {\left( {N + 1/2}\right) t}\right) ,\]\n\nwhere the quotient \( \frac{t}{\sin \left( {t/2}\right) } \) is continuous in the interval \( \left\lbrack {-\pi ,\pi }\right\rbrack \) . Since we can write\n\n\[ \sin \left( {\left( {N + 1/2}\right) t}\right) = \sin \left( {Nt}\right) \cos \left( {t/2}\right) + \cos \left( {Nt}\right) \sin \left( {t/2}\right) ,\]\n\nwe can apply the Riemann-Lebesgue lemma to the Riemann integrable functions \( F\left( t\right) t\cos \left( {t/2}\right) /\sin \left( {t/2}\right) \) and \( F\left( t\right) t \) to finish the proof of the theorem.
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Yes
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Theorem 2.2 Suppose \( f \) and \( g \) are two integrable functions defined on the circle, and for some \( {\theta }_{0} \) there exists an open interval \( I \) containing \( {\theta }_{0} \) such that\n\n\[ f\left( \theta \right) = g\left( \theta \right) \;\text{ for all }\theta \in I. \]\n\nThen \( {S}_{N}\left( f\right) \left( {\theta }_{0}\right) - {S}_{N}\left( g\right) \left( {\theta }_{0}\right) \rightarrow 0 \) as \( N \) tends to infinity.
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Proof. The function \( f - g \) is 0 in \( I \), so it is differentiable at \( {\theta }_{0} \), and we may apply the previous theorem to conclude the proof.
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No
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Lemma 2.3 Suppose that the Abel means \( {A}_{r} = \mathop{\sum }\limits_{{n = 1}}^{\infty }{r}^{n}{c}_{n} \) of the series \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{c}_{n} \) are bounded as \( r \) tends to 1 (with \( r < 1 \) ). If \( {c}_{n} = O\left( {1/n}\right) \), then the partial sums \( {S}_{N} = \mathop{\sum }\limits_{{n = 1}}^{N}{c}_{n} \) are bounded.
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Proof. Let \( r = 1 - 1/N \) and choose \( M \) so that \( n\left| {c}_{n}\right| \leq M \) . We estimate the difference\n\n\[ \n{S}_{N} - {A}_{r} = \mathop{\sum }\limits_{{n = 1}}^{N}\left( {{c}_{n} - {r}^{n}{c}_{n}}\right) - \mathop{\sum }\limits_{{n = N + 1}}^{\infty }{r}^{n}{c}_{n} \n\]\n\nas follows:\n\n\[ \n\left| {{S}_{N} - {A}_{r}}\right| \leq \mathop{\sum }\limits_{{n = 1}}^{N}\left| {c}_{n}\right| \left( {1 - {r}^{n}}\right) + \mathop{\sum }\limits_{{n = N + 1}}^{\infty }{r}^{n}\left| {c}_{n}\right| \n\]\n\n\[ \n\leq M\mathop{\sum }\limits_{{n = 1}}^{N}\left( {1 - r}\right) + \frac{M}{N}\mathop{\sum }\limits_{{n = N + 1}}^{\infty }{r}^{n} \n\]\n\n\[ \n\leq {MN}\left( {1 - r}\right) + \frac{M}{N}\frac{1}{1 - r} \n\]\n\n\[ \n= {2M} \n\]\n\nwhere we have used the simple observation that\n\n\[ \n1 - {r}^{n} = \left( {1 - r}\right) \left( {1 + r + \cdots + {r}^{n - 1}}\right) \leq n\left( {1 - r}\right) . \n\]\n\nSo we see that if \( M \) satisfies both \( \left| {A}_{r}\right| \leq M \) and \( n\left| {c}_{n}\right| \leq M \), then \( \left| {S}_{N}\right| \leq \) \( {3M} \) .
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Yes
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Lemma 2.4\n\n\[ \n{S}_{M}\left( {P}_{N}\right) = \left\{ \begin{array}{ll} {P}_{N} & \text{ if }M \geq {3N} \\ {\widetilde{P}}_{N} & \text{ if }M = {2N} \\ 0 & \text{ if }M < N \end{array}\right. \n\]
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This is clear from what has been said above and from Figure 3.
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No
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Lemma 2.2 If \( f \) is continuous and periodic of period 1, and \( \gamma \) is irrational, then\n\n\[ \n\frac{1}{N}\mathop{\sum }\limits_{{n = 1}}^{N}f\left( {n\gamma }\right) \rightarrow {\int }_{0}^{1}f\left( x\right) {dx}\;\text{ as }N \rightarrow \infty .\n\]
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The proof of the lemma is divided into three steps.\n\nStep 1. We first check the validity of the limit in the case when \( f \) is one of the exponentials \( 1,{e}^{2\pi ix},\ldots ,{e}^{2\pi ikx},\ldots \) . If \( f = 1 \), the limit\nsurely holds. If \( f = {e}^{2\pi ikx} \) with \( k \neq 0 \), then the integral is 0 . Since \( \gamma \) is irrational, we have \( {e}^{2\pi ik\gamma } \neq 1 \), therefore\n\n\[ \n\frac{1}{N}\mathop{\sum }\limits_{{n = 1}}^{N}f\left( {n\gamma }\right) = \frac{{e}^{2\pi ik\gamma }}{N}\frac{1 - {e}^{2\pi ikN\gamma }}{1 - {e}^{2\pi ik\gamma }}\n\]\n\nwhich goes to 0 as \( N \rightarrow \infty \) .\n\nStep 2. It is clear that if \( f \) and \( g \) satisfy the lemma, then so does \( {Af} + {Bg} \) for any \( A, B \in \mathbb{C} \) . Therefore, the first step implies that the lemma is true for all trigonometric polynomials.\n\nStep 3. Let \( \epsilon > 0 \) . If \( f \) is any continuous periodic function of period 1, choose a trigonometric polynomial \( P \) so that \( \mathop{\sup }\limits_{{x \in \mathbb{R}}}\left| {f\left( x\right) - P\left( x\right) }\right| < \epsilon /3 \) (this is possible by Corollary 5.4 in Chapter 2). Then, by step 1, for all large \( N \) we have\n\n\[ \n\left| {\frac{1}{N}\mathop{\sum }\limits_{{n = 1}}^{N}P\left( {n\gamma }\right) - {\int }_{0}^{1}P\left( x\right) {dx}}\right| < \epsilon /3\n\]\n\nTherefore\n\n\[ \n\left| {\frac{1}{N}\mathop{\sum }\limits_{{n = 1}}^{N}f\left( {n\gamma }\right) - {\int }_{0}^{1}f\left( x\right) {dx}}\right| \leq \frac{1}{N}\mathop{\sum }\limits_{{n = 1}}^{N}\left| {f\left( {n\gamma }\right) - P\left( {n\gamma }\right) }\right| +\n\]\n\n\[ \n+ \left| {\frac{1}{N}\mathop{\sum }\limits_{{n = 1}}^{N}P\left( {n\gamma }\right) - {\int }_{0}^{1}P\left( x\right) {dx}}\right| +\n\]\n\n\[ \n+ {\int }_{0}^{1}\left| {P\left( x\right) - f\left( x\right) }\right| {dx}\n\]\n\n\[ \n< \epsilon\n\]\n\nand the lemma is proved.
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Yes
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Corollary 2.3 The conclusion of Lemma 2.2 holds for every function \( f \) which is Riemann integrable in \( \left\lbrack {0,1}\right\rbrack \), and periodic of period 1 .
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Proof. Assume \( f \) is real-valued, and consider a partition of the interval \( \left\lbrack {0,1}\right\rbrack \), say \( 0 = {x}_{0} < {x}_{1} < \cdots < {x}_{N} = 1 \) . Next, define \( {f}_{U}\left( x\right) = \) \( \mathop{\sup }\limits_{{{x}_{j - 1} \leq y \leq {x}_{j}}}f\left( y\right) \) if \( x \in \left\lbrack {{x}_{j - 1},{x}_{j}}\right) \) and \( {f}_{L}\left( x\right) = \mathop{\inf }\limits_{{{x}_{j - 1} \leq y \leq {x}_{j}}}f\left( y\right) \) for \( x \in \) \( \left( {{x}_{j - 1},{x}_{j}}\right) \) . Then clearly \( {f}_{L} \leq f \leq {f}_{U} \) and\n\n\[ \n{\int }_{0}^{1}{f}_{L}\left( x\right) {dx} \leq {\int }_{0}^{1}f\left( x\right) {dx} \leq {\int }_{0}^{1}{f}_{U}\left( x\right) {dx}.\n\]\n\nMoreover, by making the partition sufficiently fine we can guarantee that for a given \( \epsilon > 0 \) ,\n\n\[ \n{\int }_{0}^{1}{f}_{U}\left( x\right) {dx} - {\int }_{0}^{1}{f}_{L}\left( x\right) {dx} \leq \epsilon \n\]\n\nHowever,\n\n\[ \n\frac{1}{N}\mathop{\sum }\limits_{{n = 1}}^{N}{f}_{L}\left( {n\gamma }\right) \rightarrow {\int }_{0}^{1}{f}_{L}\left( x\right) {dx} \n\]\n\nby the theorem, because each \( {f}_{L} \) is a finite linear combination of characteristic functions of intervals; similarly we have\n\n\[ \n\frac{1}{N}\mathop{\sum }\limits_{{n = 1}}^{N}{f}_{U}\left( {n\gamma }\right) \rightarrow {\int }_{0}^{1}{f}_{U}\left( x\right) {dx}.\n\]\n\nFrom these two assertions we can conclude the proof of the corollary by using the previous approximation argument.
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Yes
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Lemma 3.3 If \( {2N} = {2}^{n} \), then\n\n\[{\bigtriangleup }_{2N}\left( f\right) - {\bigtriangleup }_{N}\left( f\right) = {2}^{-{n\alpha }}{e}^{i{2}^{n}x}.\]
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This follows from our previous observation (6) because \( {\bigtriangleup }_{2N}\left( f\right) = \) \( {S}_{2N}\left( f\right) \) and \( {\bigtriangleup }_{N}\left( f\right) = {S}_{N}\left( f\right) \) .
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No
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Proposition 1.1 The integral of a function of moderate decrease defined by (5) satisfies the following properties:\n\n(i) Linearity: if \( f, g \in \mathcal{M}\left( \mathbb{R}\right) \) and \( a, b \in \mathbb{C} \), then\n\n\[ \n{\int }_{-\infty }^{\infty }\left( {{af}\left( x\right) + {bg}\left( x\right) }\right) {dx} = a{\int }_{-\infty }^{\infty }f\left( x\right) {dx} + b{\int }_{-\infty }^{\infty }g\left( x\right) {dx}. \n\]
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We say a few words about the proof. Property (i) is immediate.
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No
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Proposition 1.2 If \( f \in \mathcal{S}\left( \mathbb{R}\right) \) then:\n\n(i) \( f\left( {x + h}\right) \rightarrow \widehat{f}\left( \xi \right) {e}^{2\pi ih\xi } \) whenever \( h \in \mathbb{R} \) .\n\n(ii) \( f\left( x\right) {e}^{-{2\pi ixh}} \rightarrow \widehat{f}\left( {\xi + h}\right) \) whenever \( h \in \mathbb{R} \) .\n\n(iii) \( f\left( {\delta x}\right) \rightarrow {\delta }^{-1}\widehat{f}\left( {{\delta }^{-1}\xi }\right) \) whenever \( \delta > 0 \) .\n\n(iv) \( {f}^{\prime }\left( x\right) \rightarrow {2\pi i\xi }\widehat{f}\left( \xi \right) \) .\n\n(v) \( - {2\pi ixf}\left( x\right) \rightarrow \frac{d}{d\xi }\widehat{f}\left( \xi \right) \) .
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Proof. Property (i) is an immediate consequence of the translation invariance of the integral, and property (ii) follows from the definition. Also, the third property of Proposition 1.1 establishes (iii).\n\nIntegrating by parts gives\n\n\[{\int }_{-N}^{N}{f}^{\prime }\left( x\right) {e}^{-{2\pi ix\xi }}{dx} = {\left\lbrack f\left( x\right) {e}^{-{2\pi ix\xi }}\right\rbrack }_{-N}^{N} + {2\pi i\xi }{\int }_{-N}^{N}f\left( x\right) {e}^{-{2\pi ix\xi }}{dx},\]\n\nso letting \( N \) tend to infinity gives (iv).\n\nFinally, to prove property (v), we must show that \( \widehat{f} \) is differentiable and find its derivative. Let \( \epsilon > 0 \) and consider\n\n\[ \frac{\widehat{f}\left( {\xi + h}\right) - \widehat{f}\left( \xi \right) }{h} - \left( {-\widehat{2\pi ix}f}\right) \left( \xi \right) = \]\n\n\[ {\int }_{-\infty }^{\infty }f\left( x\right) {e}^{-{2\pi ix\xi }}\left\lbrack {\frac{{e}^{-{2\pi ixh}} - 1}{h} + {2\pi ix}}\right\rbrack {dx}. \]\n\nSince \( f\left( x\right) \) and \( {xf}\left( x\right) \) are of rapid decrease, there exists an integer \( N \) so that \( {\int }_{\left| x\right| \geq N}\left| {f\left( x\right) }\right| {dx} \leq \epsilon \) and \( {\int }_{\left| x\right| \geq N}\left| x\right| \left| {f\left( x\right) }\right| {dx} \leq \epsilon \) . Moreover, for \( \left| x\right| \leq N \), there exists \( {h}_{0} \) so that \( \left| h\right| < {h}_{0} \) implies\n\n\[ \left| {\frac{{e}^{-{2\pi ixh}} - 1}{h} + {2\pi ix}}\right| \leq \frac{\epsilon }{N} \]\n\nHence for \( \left| h\right| < {h}_{0} \) we have\n\n\[ \left| {\frac{\widehat{f}\left( {\xi + h}\right) - \widehat{f}\left( \xi \right) }{h} - \left( {\widehat{-{2\pi ix}}f}\right) \left( \xi \right) }\right| \]\n\n\[ \leq {\int }_{-N}^{N}\left| {f\left( x\right) {e}^{-{2\pi ix\xi }}\left\lbrack {\frac{{e}^{-{2\pi ixh}} - 1}{h} + {2\pi ix}}\right\rbrack }\right| {dx} + {C\epsilon } \]\n\n\[ \leq {C}^{\prime }\epsilon \]
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Yes
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Theorem 1.3 If \( f \in \mathcal{S}\left( \mathbb{R}\right) \), then \( \widehat{f} \in \mathcal{S}\left( \mathbb{R}\right) \) .
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The proof is an easy application of the fact that the Fourier transform interchanges differentiation and multiplication. In fact, note that if \( f \in \) \( \mathcal{S}\left( \mathbb{R}\right) \), its Fourier transform \( \widehat{f} \) is bounded; then also, for each pair of non-negative integers \( \ell \) and \( k \), the expression\n\n\[ \n{\xi }^{k}{\left( \frac{d}{d\xi }\right) }^{\ell }\widehat{f}\left( \xi \right)\n\]\n\nis bounded, since by the last proposition, it is the Fourier transform of\n\n\[ \n\frac{1}{{\left( 2\pi i\right) }^{k}}{\left( \frac{d}{dx}\right) }^{k}\left\lbrack {{\left( -2\pi ix\right) }^{\ell }f\left( x\right) }\right\rbrack .\n\]
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Yes
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Theorem 1.4 If \( f\left( x\right) = {e}^{-\pi {x}^{2}} \), then \( \widehat{f}\left( \xi \right) = f\left( \xi \right) \) .
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Proof. Define\n\n\[ F\left( \xi \right) = \widehat{f}\left( \xi \right) = {\int }_{-\infty }^{\infty }{e}^{-\pi {x}^{2}}{e}^{-{2\pi ix\xi }}{dx} \]\n\nand observe that \( F\left( 0\right) = 1 \), by our previous calculation. By property (v) in Proposition 1.2, and the fact that \( {f}^{\prime }\left( x\right) = - {2\pi xf}\left( x\right) \), we obtain\n\n\[ {F}^{\prime }\left( \xi \right) = {\int }_{-\infty }^{\infty }f\left( x\right) \left( {-{2\pi ix}}\right) {e}^{-{2\pi ix\xi }}{dx} = i{\int }_{-\infty }^{\infty }{f}^{\prime }\left( x\right) {e}^{-{2\pi ix\xi }}{dx}. \]\n\nBy (iv) of the same proposition, we find that\n\n\[ {F}^{\prime }\left( \xi \right) = i\left( {2\pi i\xi }\right) \widehat{f}\left( \xi \right) = - {2\pi \xi F}\left( \xi \right) . \]\n\nIf we define \( G\left( \xi \right) = F\left( \xi \right) {e}^{\pi {\xi }^{2}} \), then from what we have seen above, it follows that \( {G}^{\prime }\left( \xi \right) = 0 \), hence \( G \) is constant. Since \( F\left( 0\right) = 1 \), we conclude that \( G \) is identically equal to 1, therefore \( F\left( \xi \right) = {e}^{-\pi {\xi }^{2}} \), as was to be shown.
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Yes
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Theorem 1.6 The collection \( {\left\{ {K}_{\delta }\right\} }_{\delta > 0} \) is a family of good kernels as \( \delta \rightarrow 0 \) .
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Proof. First, we claim that \( f \) is uniformly continuous on \( \mathbb{R} \) . Indeed, given \( \epsilon > 0 \) there exists \( R > 0 \) so that \( \left| {f\left( x\right) }\right| < \epsilon /4 \) whenever \( \left| x\right| \geq R \) . Moreover, \( f \) is continuous, hence uniformly continuous on the compact interval \( \left\lbrack {-R, R}\right\rbrack \), and together with the previous observation, we can find \( \eta > 0 \) so that \( \left| {f\left( x\right) - f\left( y\right) }\right| < \epsilon \) whenever \( \left| {x - y}\right| < \eta \) . Now we argue as usual. Using the first property of good kernels, we can write\n\n\[ \left( {f * {K}_{\delta }}\right) \left( x\right) - f\left( x\right) = {\int }_{-\infty }^{\infty }{K}_{\delta }\left( t\right) \left\lbrack {f\left( {x - t}\right) - f\left( x\right) }\right\rbrack {dt}, \]\n\nand since \( {K}_{\delta } \geq 0 \), we find\n\n\[ \left| {\left( {f * {K}_{\delta }}\right) \left( x\right) - f\left( x\right) }\right| \leq {\int }_{\left| t\right| > \eta } + {\int }_{\left| t\right| \leq \eta }{K}_{\delta }\left( t\right) \left| {f\left( {x - t}\right) - f\left( x\right) }\right| {dt}. \]\n\nThe first integral is small by the third property of good kernels, and the fact that \( f \) is bounded, while the second integral is also small since \( f \) is uniformly continuous and \( \int {K}_{\delta } = 1 \) . This concludes the proof of the corollary.
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Yes
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Corollary 1.7 If \( f \in \mathcal{S}\left( \mathbb{R}\right) \), then\n\n\[ \left( {f * {K}_{\delta }}\right) \left( x\right) \rightarrow f\left( x\right) \;\text{ uniformly in }x\text{ as }\delta \rightarrow 0. \]
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Proof. First, we claim that \( f \) is uniformly continuous on \( \mathbb{R} \) . Indeed, given \( \epsilon > 0 \) there exists \( R > 0 \) so that \( \left| {f\left( x\right) }\right| < \epsilon /4 \) whenever \( \left| x\right| \geq R \) . Moreover, \( f \) is continuous, hence uniformly continuous on the compact interval \( \left\lbrack {-R, R}\right\rbrack \), and together with the previous observation, we can find \( \eta > 0 \) so that \( \left| {f\left( x\right) - f\left( y\right) }\right| < \epsilon \) whenever \( \left| {x - y}\right| < \eta \) . Now we argue as usual. Using the first property of good kernels, we can write\n\n\[ \left( {f * {K}_{\delta }}\right) \left( x\right) - f\left( x\right) = {\int }_{-\infty }^{\infty }{K}_{\delta }\left( t\right) \left\lbrack {f\left( {x - t}\right) - f\left( x\right) }\right\rbrack {dt}, \]\n\nand since \( {K}_{\delta } \geq 0 \), we find\n\n\[ \left| {\left( {f * {K}_{\delta }}\right) \left( x\right) - f\left( x\right) }\right| \leq {\int }_{\left| t\right| > \eta } + {\int }_{\left| t\right| \leq \eta }{K}_{\delta }\left( t\right) \left| {f\left( {x - t}\right) - f\left( x\right) }\right| {dt}. \]\n\nThe first integral is small by the third property of good kernels, and the fact that \( f \) is bounded, while the second integral is also small since \( f \) is uniformly continuous and \( \int {K}_{\delta } = 1 \) . This concludes the proof of the corollary.
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Yes
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Proposition 1.8 If \( f, g \in \mathcal{S}\left( \mathbb{R}\right) \), then\n\n\[{\int }_{-\infty }^{\infty }f\left( x\right) \widehat{g}\left( x\right) {dx} = {\int }_{-\infty }^{\infty }\widehat{f}\left( y\right) g\left( y\right) {dy}.\]
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To prove the proposition, we need to digress briefly to discuss the interchange of the order of integration for double integrals. Suppose \( F\left( {x, y}\right) \) is a continuous function in the plane \( \left( {x, y}\right) \in {\mathbb{R}}^{2} \) . We will assume the following decay condition on \( F \) :\n\n\[ \left| {F\left( {x, y}\right) }\right| \leq A/\left( {1 + {x}^{2}}\right) \left( {1 + {y}^{2}}\right) . \]\n\nThen, we can state that for each \( x \) the function \( F\left( {x, y}\right) \) is of moderate decrease in \( y \), and similarly for each fixed \( y \) the function \( F\left( {x, y}\right) \) is of moderate decrease in \( x \) . Moreover, the function \( {F}_{1}\left( x\right) = {\int }_{-\infty }^{\infty }F\left( {x, y}\right) {dy} \) is continuous and of moderate decrease; similarly for the function \( {F}_{2}\left( y\right) = \) \( {\int }_{-\infty }^{\infty }F\left( {x, y}\right) {dx} \) . Finally\n\n\[ {\int }_{-\infty }^{\infty }{F}_{1}\left( x\right) {dx} = {\int }_{-\infty }^{\infty }{F}_{2}\left( y\right) {dy}. \]\n\nThe proof of these facts may be found in the appendix.\n\nWe now apply this to \( F\left( {x, y}\right) = f\left( x\right) g\left( y\right) {e}^{-{2\pi ixy}} \) . Then \( {F}_{1}\left( x\right) = \) \( f\left( x\right) \widehat{g}\left( x\right) \), and \( {F}_{2}\left( y\right) = \widehat{f}\left( y\right) g\left( y\right) \) so\n\n\[ {\int }_{-\infty }^{\infty }f\left( x\right) \widehat{g}\left( x\right) {dx} = {\int }_{-\infty }^{\infty }\widehat{f}\left( y\right) g\left( y\right) {dy} \]\n\nwhich is the assertion of the proposition.
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Yes
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Theorem 1.9 (Fourier inversion) If \( f \in \mathcal{S}\left( \mathbb{R}\right) \), then\n\n\[ f\left( x\right) = {\int }_{-\infty }^{\infty }\widehat{f}\left( \xi \right) {e}^{2\pi ix\xi }{d\xi } \]
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Proof. We first claim that\n\n\[ f\left( 0\right) = {\int }_{-\infty }^{\infty }\widehat{f}\left( \xi \right) {d\xi } \]\n\nLet \( {G}_{\delta }\left( x\right) = {e}^{-{\pi \delta }{x}^{2}} \) so that \( \widehat{{G}_{\delta }}\left( \xi \right) = {K}_{\delta }\left( \xi \right) \). By the multiplication formula we get\n\n\[ {\int }_{-\infty }^{\infty }f\left( x\right) {K}_{\delta }\left( x\right) {dx} = {\int }_{-\infty }^{\infty }\widehat{f}\left( \xi \right) {G}_{\delta }\left( \xi \right) {d\xi }. \]\n\nSince \( {K}_{\delta } \) is a good kernel, the first integral goes to \( f\left( 0\right) \) as \( \delta \) tends to 0. Since the second integral clearly converges to \( {\int }_{-\infty }^{\infty }\widehat{f}\left( \xi \right) {d\xi } \) as \( \delta \) tends to 0, our claim is proved. In general, let \( F\left( y\right) = f\left( {y + x}\right) \) so that\n\n\[ f\left( x\right) = F\left( 0\right) = {\int }_{-\infty }^{\infty }\widehat{F}\left( \xi \right) {d\xi } = {\int }_{-\infty }^{\infty }\widehat{f}\left( \xi \right) {e}^{2\pi ix\xi }{d\xi }. \]
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Yes
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Proposition 1.11 If \( f, g \in \mathcal{S}\left( \mathbb{R}\right) \) then:\n\n(i) \( f * g \in \mathcal{S}\left( \mathbb{R}\right) \) .\n\n(ii) \( f * g = g * f \) .\n\n(iii) \( \widehat{\left( f * g\right) }\left( \xi \right) = \widehat{f}\left( \xi \right) \widehat{g}\left( \xi \right) \) .
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Proof. To prove that \( f * g \) is rapidly decreasing, observe first that for any \( \ell \geq 0 \) we have \( \mathop{\sup }\limits_{x}{\left| x\right| }^{\ell }\left| {g\left( {x - y}\right) }\right| \leq {A}_{\ell }{\left( 1 + \left| y\right| \right) }^{\ell } \), because \( g \) is rapidly decreasing (to check this assertion, consider separately the two cases \( \left. {\left| x\right| \leq 2\left| y\right| \text{and}\left| x\right| \geq 2\left| y\right| }\right) \) . From this, we see that\n\n\[ \mathop{\sup }\limits_{x}\left| {{x}^{\ell }\left( {f * g}\right) \left( x\right) }\right| \leq {A}_{\ell }{\int }_{-\infty }^{\infty }\left| {f\left( y\right) }\right| {\left( 1 + \left| y\right| \right) }^{\ell }{dy} \]\n\nso that \( {x}^{\ell }\left( {f * g}\right) \left( x\right) \) is a bounded function for every \( \ell \geq 0 \) . These estimates carry over to the derivatives of \( f * g \), thereby proving that \( f * g \in \mathcal{S}\left( \mathbb{R}\right) \) because\n\n\[ {\left( \frac{d}{dx}\right) }^{k}\left( {f * g}\right) \left( x\right) = \left( {f * {\left( \frac{d}{dx}\right) }^{k}g}\right) \left( x\right) \;\text{ for }k = 1,2,\ldots \]\n\nThis identity is proved first for \( k = 1 \) by differentiating under the integral defining \( f * g \) . The interchange of differentiation and integration is justified in this case by the rapid decrease of \( {dg}/{dx} \) . The identity then follows for every \( k \) by iteration.\n\nFor fixed \( x \), the change of variables \( x - y = u \) shows that\n\n\[ \left( {f * g}\right) \left( x\right) = {\int }_{-\infty }^{\infty }f\left( {x - u}\right) g\left( u\right) {du} = \left( {g * f}\right) \left( x\right) . \]\n\nThis change of variables is a composition of two changes, \( y \mapsto - y \) and \( y \mapsto y - h \) (with \( h = x \) ). For the first one we use the observation that \( {\int }_{-\infty }^{\infty }F\left( x\right) {dx} = {\int }_{-\infty }^{\infty }F\left( {-x}\right) {dx} \) for any Schwartz function \( F \), and for the second, we apply (ii) of Proposition 1.1\n\nFinally, consider \( F\left( {x, y}\right) = f\left( y\right) g\left( {x - y}\right) {e}^{-{2\pi ix\xi }} \) . Since \( f \) and \( g \) are rapidly decreasing, considering separately the two cases \( \left| x\right| \leq 2\left| y\right| \) and \( \left| x\right| \geq 2\left| y\right| \), we see that the discussion of the change of order of integration after Proposition 1.8 applies to \( F \) . In this case \( {F}_{1}\left( x\right) = \left( {f * g}\right) \left( x\right) {e}^{-{2\pi ix\xi }} \) , and \( {F}_{2}\left( y\right) = f\left( y\right) {e}^{-{2\pi iy\xi }}\widehat{g}\left( \xi \right) \) . Thus \( {\int }_{-\infty }^{\infty }{F}_{1}\left( x\right) {dx} = {\int }_{-\infty }^{\infty }{F}_{2}\left( y\right) {dy} \), which implies (iii). The proposition is therefore proved.
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Yes
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Theorem 1.12 (Plancherel) If \( f \in \mathcal{S}\left( \mathbb{R}\right) \) then \( \parallel \widehat{f}\parallel = \parallel f\parallel \) .
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Proof. If \( f \in \mathcal{S}\left( \mathbb{R}\right) \) define \( {f}^{b}\left( x\right) = \overline{f\left( {-x}\right) } \) . Then \( \widehat{{f}^{b}}\left( \xi \right) = \overline{\widehat{f}\left( \xi \right) } \) . Now let \( h = f * {f}^{b} \) . Clearly, we have\n\n\[ \n\widehat{h}\left( \xi \right) = {\left| \widehat{f}\left( \xi \right) \right| }^{2}\;\text{ and }\;h\left( 0\right) = {\int }_{-\infty }^{\infty }{\left| f\left( x\right) \right| }^{2}{dx}.\n\]\n\nThe theorem now follows from the inversion formula applied with \( x = 0 \) , that is,\n\n\[ \n{\int }_{-\infty }^{\infty }\widehat{h}\left( \xi \right) {d\xi } = h\left( 0\right)\n\]
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Yes
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Corollary 2.2 \( u\left( {\cdot, t}\right) \) belongs to \( \mathcal{S}\left( \mathbb{R}\right) \) uniformly in \( t \), in the sense that for any \( T > 0 \)\n\n(9)\n\n\[ \mathop{\sup }\limits_{\substack{{x \in \mathbb{R}} \\ {0 < t < T} }}{\left| x\right| }^{k}\left| {\frac{{\partial }^{\ell }}{\partial {x}^{\ell }}u\left( {x, t}\right) }\right| < \infty \;\text{ for each }k,\ell \geq 0. \]
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Proof. This result is a consequence of the following estimate:\n\n\[ \left| {u\left( {x, t}\right) }\right| \leq {\int }_{\left| y\right| \leq \left| x\right| /2}\left| {f\left( {x - y}\right) }\right| {\mathcal{H}}_{t}\left( y\right) {dy} + {\int }_{\left| y\right| \geq \left| x\right| /2}\left| {f\left( {x - y}\right) }\right| {\mathcal{H}}_{t}\left( y\right) {dy} \]\n\n\[ \leq \frac{{C}_{N}}{{\left( 1 + \left| x\right| \right) }^{N}} + \frac{C}{\sqrt{t}}{e}^{-c{x}^{2}/t}. \]\n\nIndeed, since \( f \) is rapidly decreasing, we have \( \left| {f\left( {x - y}\right) }\right| \leq {C}_{N}/{\left( 1 + \left| x\right| \right) }^{N} \) when \( \left| y\right| \leq \left| x\right| /2 \) . Also, if \( \left| y\right| \geq \left| x\right| /2 \) then \( {\mathcal{H}}_{t}\left( y\right) \leq C{t}^{-1/2}{e}^{-c{x}^{2}/t} \), and we obtain the above inequality. Consequently, we see that \( u\left( {x, t}\right) \) is rapidly decreasing uniformly for \( 0 < t < T \) .\n\nThe same argument can be applied to the derivatives of \( u \) in the \( x \) variable since we may differentiate under the integral sign and apply the above estimate with \( f \) replaced by \( {f}^{\prime } \), and so on.
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Yes
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Theorem 2.3 Suppose \( u\left( {x, t}\right) \) satisfies the following conditions:\n\n(i) \( u \) is continuous on the closure of the upper half-plane.\n\n(ii) \( u \) satisfies the heat equation for \( t > 0 \) .\n\n(iii) \( u \) satisfies the boundary condition \( u\left( {x,0}\right) = 0 \) .\n\n(iv) \( u\left( {\cdot, t}\right) \in \mathcal{S}\left( \mathbb{R}\right) \) uniformly in \( t \), as in (9).\n\nThen, we conclude that \( u = 0 \) .
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Proof. We define the energy at time \( t \) of the solution \( u\left( {x, t}\right) \) by\n\n\[ E\left( t\right) = {\int }_{\mathbb{R}}{\left| u\left( x, t\right) \right| }^{2}{dx} \]\n\nClearly \( E\left( t\right) \geq 0 \) . Since \( E\left( 0\right) = 0 \) it suffices to show that \( E \) is a decreasing function, and this is achieved by proving that \( {dE}/{dt} \leq 0 \) . The assumptions on \( u \) allow us to differentiate \( E\left( t\right) \) under the integral sign\n\n\[ \frac{dE}{dt} = {\int }_{\mathbb{R}}\left\lbrack {{\partial }_{t}u\left( {x, t}\right) \bar{u}\left( {x, t}\right) + u\left( {x, t}\right) {\partial }_{t}\bar{u}\left( {x, t}\right) }\right\rbrack {dx}. \]\n\nBut \( u \) satisfies the heat equation, therefore \( {\partial }_{t}u = {\partial }_{x}^{2}u \) and \( {\partial }_{t}\bar{u} = {\partial }_{x}^{2}\bar{u} \), so that after an integration by parts, where we use the fact that \( u \) and its \( x \) derivatives decrease rapidly as \( \left| x\right| \rightarrow \infty \), we find\n\n\[ \frac{dE}{dt} = {\int }_{\mathbb{R}}\left\lbrack {{\partial }_{x}^{2}u\left( {x, t}\right) \bar{u}\left( {x, t}\right) + u\left( {x, t}\right) {\partial }_{x}^{2}\bar{u}\left( {x, t}\right) }\right\rbrack {dx} \]\n\n\[ = - {\int }_{\mathbb{R}}\left\lbrack {{\partial }_{x}u\left( {x, t}\right) {\partial }_{x}\bar{u}\left( {x, t}\right) + {\partial }_{x}u\left( {x, t}\right) {\partial }_{x}\bar{u}\left( {x, t}\right) }\right\rbrack {dx} \]\n\n\[ = - 2{\int }_{\mathbb{R}}{\left| {\partial }_{x}u\left( x, t\right) \right| }^{2}{dx} \]\n\n\[ \leq 0 \]\n\nas claimed. Thus \( E\left( t\right) = 0 \) for all \( t \), hence \( u = 0 \) .
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Yes
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Lemma 2.4 The following two identities hold:\n\n\[ \n{\int }_{-\infty }^{\infty }{e}^{-{2\pi }\left| \xi \right| y}{e}^{2\pi i\xi x}{d\xi } = {\mathcal{P}}_{y}\left( x\right) \n\]\n\n\[ \n{\int }_{-\infty }^{\infty }{\mathcal{P}}_{y}\left( x\right) {e}^{-{2\pi ix\xi }}{dx} = {e}^{-{2\pi }\left| \xi \right| y}. \n\]
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Proof. The first formula is fairly straightforward since we can split the integral from \( - \infty \) to 0 and 0 to \( \infty \) . Then, since \( y > 0 \) we have\n\n\[ \n{\int }_{0}^{\infty }{e}^{-{2\pi \xi y}}{e}^{2\pi i\xi x}{d\xi } = {\int }_{0}^{\infty }{e}^{{2\pi i}\left( {x + {iy}}\right) \xi }{d\xi } = {\left\lbrack \frac{{e}^{{2\pi i}\left( {x + {iy}}\right) \xi }}{{2\pi i}\left( {x + {iy}}\right) }\right\rbrack }_{0}^{\infty } = \n\]\n\n\[ \n- \frac{1}{{2\pi i}\left( {x + {iy}}\right) } \n\]\n\nand similarly,\n\n\[ \n{\int }_{-\infty }^{0}{e}^{2\pi \xi y}{e}^{2\pi i\xi x}{d\xi } = \frac{1}{{2\pi i}\left( {x - {iy}}\right) }. \n\]\n\nTherefore\n\n\[ \n{\int }_{-\infty }^{\infty }{e}^{-{2\pi }\left| \xi \right| y}{e}^{2\pi i\xi x}{d\xi } = \frac{1}{{2\pi i}\left( {x - {iy}}\right) } - \frac{1}{{2\pi i}\left( {x + {iy}}\right) } = \frac{y}{\pi \left( {{x}^{2} + {y}^{2}}\right) }. \n\]\n\nThe second formula is now a consequence of the Fourier inversion theorem applied in the case when \( f \) and \( \widehat{f} \) are of moderate decrease.
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Yes
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Lemma 2.5 The Poisson kernel is a good kernel on \( \mathbb{R} \) as \( y \rightarrow 0 \) .
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Proof. Setting \( \xi = 0 \) in the second formula of the lemma shows that \( {\int }_{-\infty }^{\infty }{\mathcal{P}}_{y}\left( x\right) {dx} = 1 \), and clearly \( {\mathcal{P}}_{y}\left( x\right) \geq 0 \), so it remains to check the last property of good kernels. Given a fixed \( \delta > 0 \), we may change variables \( u = x/y \) so that\n\n\[{\int }_{\delta }^{\infty }\frac{y}{{x}^{2} + {y}^{2}}{dx} = {\int }_{\delta /y}^{\infty }\frac{du}{1 + {u}^{2}} = {\left\lbrack \arctan u\right\rbrack }_{\delta /y}^{\infty } = \pi /2 - \arctan \left( {\delta /y}\right) ,\]\n\nand this quantity goes to 0 as \( y \rightarrow 0 \) . Since \( {\mathcal{P}}_{y}\left( x\right) \) is an even function, the proof is complete.
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Yes
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Theorem 2.6 Given \( f \in \mathcal{S}\left( \mathbb{R}\right) \), let \( u\left( {x, y}\right) = \left( {f * {\mathcal{P}}_{y}}\right) \left( x\right) \) . Then:\n\n(i) \( u\left( {x, y}\right) \) is \( {C}^{2} \) in \( {\mathbb{R}}_{ + }^{2} \) and \( \bigtriangleup u = 0 \) .\n\n(ii) \( u\left( {x, y}\right) \rightarrow f\left( x\right) \) uniformly as \( y \rightarrow 0 \) .\n\n(iii) \( {\int }_{-\infty }^{\infty }{\left| u\left( x, y\right) - f\left( x\right) \right| }^{2}{dx} \rightarrow 0 \) as \( y \rightarrow 0 \) .\n\n(iv) If \( u\left( {x,0}\right) = f\left( x\right) \), then \( u \) is continuous on the closure \( \overline{{\mathbb{R}}_{ + }^{2}} \) of the upper half-plane, and vanishes at infinity in the sense that\n\n\[ u\left( {x, y}\right) \rightarrow 0\;\text{ as }\left| x\right| + y \rightarrow \infty . \]
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Proof. The proofs of parts (i), (ii), and (iii) are similar to the case of the heat equation, and so are left to the reader. Part (iv) is a consequence of two easy estimates whenever \( f \) is of moderate decrease. First, we have\n\n\[ \left| {\left( {f * {\mathcal{P}}_{y}}\right) \left( x\right) }\right| \leq C\left( {\frac{1}{\left( 1 + {x}^{2}\right) } + \frac{y}{{x}^{2} + {y}^{2}}}\right) \]\n\nwhich is proved (as in the case of the heat equation) by splitting the integral \( {\int }_{-\infty }^{\infty }f\left( {x - t}\right) {\mathcal{P}}_{y}\left( t\right) {dt} \) into the part where \( \left| t\right| \leq \left| x\right| /2 \) and the part where \( \left| t\right| \geq \left| x\right| /2 \) . Also, we have \( \left| {\left( {f * {\mathcal{P}}_{y}}\right) \left( x\right) }\right| \leq C/y \), since \( \mathop{\sup }\limits_{x}{\mathcal{P}}_{y}\left( x\right) \leq \) \( c/y \) .\n\nUsing the first estimate when \( \left| x\right| \geq \left| y\right| \) and the second when \( \left| x\right| \leq \left| y\right| \) gives the desired decrease at infinity.
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No
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Theorem 2.7 Suppose \( u \) is continuous on the closure of the upper half-plane \( \overline{{\mathbb{R}}_{ + }^{2}} \), satisfies \( \bigtriangleup u = 0 \) for \( \left( {x, y}\right) \in {\mathbb{R}}_{ + }^{2}, u\left( {x,0}\right) = 0 \), and \( u\left( {x, y}\right) \) vanishes at infinity. Then \( u = 0 \) .
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To prove Theorem 2.7 we argue by contradiction. Considering separately the real and imaginary parts of \( u \), we may suppose that \( u \) itself is real-valued, and is somewhere strictly positive, say \( u\left( {{x}_{0},{y}_{0}}\right) > 0 \) for some \( {x}_{0} \in \mathbb{R} \) and \( {y}_{0} > 0 \) . We shall see that this leads to a contradiction. First, since \( u \) vanishes at infinity, we can find a large semi-disc of radius \( R,{D}_{R}^{ + } = \left\{ {\left( {x, y}\right) : {x}^{2} + {y}^{2} \leq R,\;y \geq 0}\right\} \) outside of which \( u\left( {x, y}\right) \leq \) \( \frac{1}{2}u\left( {{x}_{0},{y}_{0}}\right) \) . Next, since \( u \) is continuous in \( {D}_{R}^{ + } \), it attains its maximum \( M \) there, so there exists a point \( \left( {{x}_{1},{y}_{1}}\right) \in {D}_{R}^{ + } \) with \( u\left( {{x}_{1},{y}_{1}}\right) = M \), while \( u\left( {x, y}\right) \leq M \) in the semi-disc; also, since \( u\left( {x, y}\right) \leq \frac{1}{2}u\left( {{x}_{0},{y}_{0}}\right) \leq M/2 \) outside of the semi-disc, we have \( u\left( {x, y}\right) \leq M \) throughout the entire upper half-plane. Now the mean-value property for harmonic functions implies\n\n\[ u\left( {{x}_{1},{y}_{1}}\right) = \frac{1}{2\pi }{\int }_{0}^{2\pi }u\left( {{x}_{1} + \rho \cos \theta ,{y}_{1}
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Yes
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Lemma 2.8 (Mean-value property) Suppose \( \Omega \) is an open set in \( {\mathbb{R}}^{2} \) and let \( u \) be a function of class \( {C}^{2} \) with \( \bigtriangleup u = 0 \) in \( \Omega \) . If the closure of the disc centered at \( \left( {x, y}\right) \) and of radius \( R \) is contained in \( \Omega \), then\n\n\[ u\left( {x, y}\right) = \frac{1}{2\pi }{\int }_{0}^{2\pi }u\left( {x + r\cos \theta, y + r\sin \theta }\right) {d\theta } \]\n\nfor all \( 0 \leq r \leq R \) .
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Proof. Let \( U\left( {r,\theta }\right) = u\left( {x + r\cos \theta, y + r\sin \theta }\right) \) . Expressing the Laplacian in polar coordinates, the equation \( \bigtriangleup u = 0 \) then implies\n\n\[ 0 = \frac{{\partial }^{2}U}{\partial {\theta }^{2}} + r\frac{\partial }{\partial r}\left( {r\frac{\partial U}{\partial r}}\right) . \]\n\nIf we define \( F\left( r\right) = \frac{1}{2\pi }{\int }_{0}^{2\pi }U\left( {r,\theta }\right) {d\theta } \), the above gives\n\n\[ r\frac{\partial }{\partial r}\left( {r\frac{\partial F}{\partial r}}\right) = \frac{1}{2\pi }{\int }_{0}^{2\pi } - \frac{{\partial }^{2}U}{\partial {\theta }^{2}}\left( {r,\theta }\right) {d\theta }. \]\n\nThe integral of \( {\partial }^{2}U/\partial {\theta }^{2} \) over the circle vanishes since \( \partial U/\partial \theta \) is periodic, hence \( r\frac{\partial }{\partial r}\left( {r\frac{\partial F}{\partial r}}\right) = 0 \), and consequently \( r\partial F/\partial r \) must be constant. Evaluating this expression at \( r = 0 \) we find that \( \partial F/\partial r = 0 \) . Thus \( F \) is constant, but since \( F\left( 0\right) = u\left( {x, y}\right) \), we finally find that \( F\left( r\right) = u\left( {x, y}\right) \) for all \( 0 \leq r \leq R \), which is the mean-value property.\n\nFinally, note that the argument above is implicit in the proof of Theorem 5.7, Chapter 2.
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Yes
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Theorem 3.1 (Poisson summation formula) If \( f \in \mathcal{S}\left( \mathbb{R}\right) \), then\n\n\[ \mathop{\sum }\limits_{{n = - \infty }}^{\infty }f\left( {x + n}\right) = \mathop{\sum }\limits_{{n = - \infty }}^{\infty }\widehat{f}\left( n\right) {e}^{2\pi inx}. \]\n\nIn particular, setting \( x = 0 \) we have\n\n\[ \mathop{\sum }\limits_{{n = - \infty }}^{\infty }f\left( n\right) = \mathop{\sum }\limits_{{n = - \infty }}^{\infty }\widehat{f}\left( n\right) \]\n\nIn other words, the Fourier coefficients of the periodization of \( f \) are given precisely by the values of the Fourier transform of \( f \) on the integers.
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Proof. To check the first formula it suffices, by Theorem 2.1 in Chapter 2, to show that both sides (which are continuous) have the same Fourier coefficients (viewed as functions on the circle). Clearly, the \( {m}^{\text{th }} \) Fourier coefficient of the right-hand side is \( \widehat{f}\left( m\right) \) . For the left-hand side we have\n\n\[ {\int }_{0}^{1}\left( {\mathop{\sum }\limits_{{n = - \infty }}^{\infty }f\left( {x + n}\right) }\right) {e}^{-{2\pi imx}}{dx} = \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{\int }_{0}^{1}f\left( {x + n}\right) {e}^{-{2\pi imx}}{dx} \]\n\n\[ = \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{\int }_{n}^{n + 1}f\left( y\right) {e}^{-{2\pi imy}}{dy} \]\n\n\[ = {\int }_{-\infty }^{\infty }f\left( y\right) {e}^{-{2\pi imy}}{dy} \]\n\n\[ = \widehat{f}\left( m\right) \]\n\nwhere the interchange of the sum and integral is permissible since \( f \) is rapidly decreasing. This completes the proof of the theorem.
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Yes
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Theorem 3.2 \( {s}^{-1/2}\vartheta \left( {1/s}\right) = \vartheta \left( s\right) \) whenever \( s > 0 \) .
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The proof of this identity consists of a simple application of the Poisson summation formula to the pair\n\n\[ f\left( x\right) = {e}^{-{\pi s}{x}^{2}}\;\text{ and }\;\widehat{f}\left( \xi \right) = {s}^{-1/2}{e}^{-\pi {\xi }^{2}/s}. \]
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Yes
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Theorem 3.3 The heat kernel on the circle is the periodization of the heat kernel on the real line:
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\[ {H}_{t}\left( x\right) = \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{\mathcal{H}}_{t}\left( {x + n}\right) \]
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Yes
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Corollary 3.4 The kernel \( {H}_{t}\left( x\right) \) is a good kernel for \( t \rightarrow 0 \) .
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Proof. We already observed that \( {\int }_{\left| x\right| \leq 1/2}{H}_{t}\left( x\right) {dx} = 1 \) . Now note that \( {H}_{t} \geq 0 \), which is immediate from the above formula since \( {\mathcal{H}}_{t} \geq 0 \) . Finally, we claim that when \( \left| x\right| \leq 1/2 \) ,\n\n\[ \n{H}_{t}\left( x\right) = {\mathcal{H}}_{t}\left( x\right) + {\mathcal{E}}_{t}\left( x\right) \n\]\n\nwhere the error satisfies \( \left| {{\mathcal{E}}_{t}\left( x\right) }\right| \leq {c}_{1}{e}^{-{c}_{2}/t} \) with \( {c}_{1},{c}_{2} > 0 \) and \( 0 < t \leq 1 \) . To see this, note again that the formula in the theorem gives\n\n\[ \n{H}_{t}\left( x\right) = {\mathcal{H}}_{t}\left( x\right) + \mathop{\sum }\limits_{{\left| n\right| \geq 1}}{\mathcal{H}}_{t}\left( {x + n}\right) \n\]\n\ntherefore, since \( \left| x\right| \leq 1/2 \) ,\n\n\[ \n{\mathcal{E}}_{t}\left( x\right) = \frac{1}{\sqrt{4\pi t}}\mathop{\sum }\limits_{{\left| n\right| \geq 1}}{e}^{-{\left( x + n\right) }^{2}/{4t}} \leq C{t}^{-1/2}\mathop{\sum }\limits_{{n \geq 1}}{e}^{-c{n}^{2}/t}. \n\]\n\nNote that \( {n}^{2}/t \geq {n}^{2} \) and \( {n}^{2}/t \geq 1/t \) whenever \( 0 < t \leq 1 \), so \( {e}^{-c{n}^{2}/t} \leq \) \( {e}^{-\frac{c}{2}{n}^{2}}{e}^{-\frac{c}{2}\frac{1}{t}} \) . Hence\n\n\[ \n\left| {{\mathcal{E}}_{t}\left( x\right) }\right| \leq C{t}^{-1/2}{e}^{-\frac{c}{2}\frac{1}{t}}\mathop{\sum }\limits_{{n \geq 1}}{e}^{-\frac{c}{2}{n}^{2}} \leq {c}_{1}{e}^{-{c}_{2}/t}. \n\]\n\nThe proof of the claim is complete, and as a result \( {\int }_{\left| x\right| \leq 1/2}\left| {{\mathcal{E}}_{t}\left( x\right) }\right| {dx} \rightarrow 0 \) as \( t \rightarrow 0 \) . It is now clear that \( {H}_{t} \) satisfies\n\n\[ \n{\int }_{\eta < \left| x\right| \leq 1/2}\left| {{H}_{t}\left( x\right) }\right| {dx} \rightarrow 0\;\text{ as }t \rightarrow 0, \n\]\n\nbecause \( {\mathcal{H}}_{t} \) does.
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Yes
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Theorem 3.5 \( {P}_{r}\left( {2\pi x}\right) = \mathop{\sum }\limits_{{n \in \mathbb{Z}}}{\mathcal{P}}_{y}\left( {x + n}\right) \) where \( r = {e}^{-{2\pi y}} \) .
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This is again an immediate corollary of the Poisson summation formula applied to \( f\left( x\right) = {\mathcal{P}}_{y}\left( x\right) \) and \( \widehat{f}\left( \xi \right) = {e}^{-{2\pi }\left| \xi \right| y} \) . Of course, here we use the Poisson summation formula under the assumptions that \( f \) and \( \widehat{f} \) are of moderate decrease.
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Yes
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Theorem 4.1 Suppose \( \psi \) is a function in \( \mathcal{S}\left( \mathbb{R}\right) \) which satisfies the normalizing condition \( {\int }_{-\infty }^{\infty }{\left| \psi \left( x\right) \right| }^{2}{dx} = 1 \) . Then\n\n\[ \left( {{\int }_{-\infty }^{\infty }{x}^{2}{\left| \psi \left( x\right) \right| }^{2}{dx}}\right) \left( {{\int }_{-\infty }^{\infty }{\xi }^{2}{\left| \widehat{\psi }\left( \xi \right) \right| }^{2}{d\xi }}\right) \geq \frac{1}{{16}{\pi }^{2}} \]\n\nand equality holds if and only if \( \psi \left( x\right) = A{e}^{-B{x}^{2}} \) where \( B > 0 \) and \( {\left| A\right| }^{2} = \) \( \sqrt{{2B}/\pi } \) .
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Proof. The second inequality actually follows from the first by replacing \( \psi \left( x\right) \) by \( {e}^{-{2\pi ix}{\xi }_{0}}\psi \left( {x + {x}_{0}}\right) \) and changing variables. To prove the first inequality, we argue as follows. Beginning with our normalizing assumption \( \int {\left| \psi \right| }^{2} = 1 \), and recalling that \( \psi \) and \( {\psi }^{\prime } \) are rapidly decreasing, an integration by parts gives\n\n\[ 1 = {\int }_{-\infty }^{\infty }{\left| \psi \left( x\right) \right| }^{2}{dx} \]\n\n\[ = - {\int }_{-\infty }^{\infty }x\frac{d}{dx}{\left| \psi \left( x\right) \right| }^{2}{dx} \]\n\n\[ = - {\int }_{-\infty }^{\infty }\left( {x{\psi }^{\prime }\left( x\right) \overline{\psi \left( x\right) } + x\overline{{\psi }^{\prime }\left( x\right) }\psi \left( x\right) }\right) {dx}. \]\n\nThe last identity follows because \( {\left| \psi \right| }^{2} = \psi \bar{\psi } \). Therefore\n\n\[ 1 \leq 2{\int }_{-\infty }^{\infty }\left| x\right| \left| {\psi \left( x\right) }\right| \left| {{\psi }^{\prime }\left( x\right) }\right| {dx} \]\n\n\[ \leq 2{\left( {\int }_{-\infty }^{\infty }{x}^{2}{\left| \psi \left( x\right) \right| }^{2}dx\right) }^{1/2}{\left( {\int }_{-\infty }^{\infty }{\left| {\psi }^{\prime }\left( x\right) \right| }^{2}dx\right) }^{1/2}, \]\n\nwhere we have used the Cauchy-Schwarz inequality. The identity\n\n\[ {\int }_{-\infty }^{\infty }{\left| {\psi }^{\prime }\left( x\right) \right| }^{2}{dx} = 4{\pi }^{2}{\int }_{-\infty }^{\infty }{\xi }^{2}{\left| \widehat{\psi }\left( \xi \right) \right| }^{2}{d\xi } \]\n\nwhich holds because of the properties of the Fourier transform and the Plancherel formula, concludes the proof of the inequality in the theorem.\n\nIf equality holds, then we must also have equality where we applied the Cauchy-Schwarz inequality, and as a result we find that \( {\psi }^{\prime }\left( x\right) = {\beta x\psi }\left( x\right) \) for some constant \( \beta \). The solutions to this equation are \( \psi \left( x\right) = A{e}^{\beta {x}^{2}/2} \), where \( A \) is constant. Since we want \( \psi \) to be a Schwartz function, we must take \( \beta = - {2B} < 0 \), and since we impose the condition \( {\int }_{-\infty }^{\infty }{\left| \psi \left( x\right) \right| }^{2}{dx} = 1 \) we find that \( {\left| A\right| }^{2} = \sqrt{{2B}/\pi } \), as was to be shown.
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Yes
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Proposition 2.1 Let \( f \in \mathcal{S}\left( {\mathbb{R}}^{d}\right) \). (i) \( f\left( {x + h}\right) \rightarrow \widehat{f}\left( \xi \right) {e}^{{2\pi i\xi } \cdot h} \) whenever \( h \in {\mathbb{R}}^{d} \). (ii) \( f\left( x\right) {e}^{-{2\pi ixh}} \rightarrow \widehat{f}\left( {\xi + h}\right) \) whenever \( h \in {\mathbb{R}}^{d} \). (iii) \( f\left( {\delta x}\right) \rightarrow {\delta }^{-d}\widehat{f}\left( {{\delta }^{-1}\xi }\right) \) whenever \( \delta > 0 \). (iv) \( {\left( \frac{\partial }{\partial x}\right) }^{\alpha }f\left( x\right) \rightarrow {\left( 2\pi i\xi \right) }^{\alpha }\widehat{f}\left( \xi \right) \). (v) \( {\left( -2\pi ix\right) }^{\alpha }f\left( x\right) \rightarrow {\left( \frac{\partial }{\partial \xi }\right) }^{\alpha }\widehat{f}\left( \xi \right) \). (vi) \( f\left( {Rx}\right) \rightarrow \widehat{f}\left( {R\xi }\right) \) whenever \( R \) is a rotation.
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The first five properties are proved in the same way as in the one-dimensional case. To verify the last property, simply change variables \( y = {Rx} \) in the integral. Then, recall that \( \left| {\det \left( R\right) }\right| = 1 \), and \( {R}^{-1}y \cdot \xi = y \cdot {R\xi } \), because \( R \) is a rotation.
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Yes
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Corollary 2.3 The Fourier transform of a radial function is radial.
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This follows at once from property (vi) in the last proposition. Indeed, the condition \( f\left( {Rx}\right) = f\left( x\right) \) for all \( R \) implies that \( \widehat{f}\left( {R\xi }\right) = \widehat{f}\left( \xi \right) \) for all \( R \), thus \( \widehat{f} \) is radial whenever \( f \) is.
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Yes
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A solution of the Cauchy problem for the wave equation is\n\n\[ u\left( {x, t}\right) = {\int }_{{\mathbb{R}}^{d}}\left\lbrack {\widehat{f}\left( \xi \right) \cos \left( {{2\pi }\left| \xi \right| t}\right) + \widehat{g}\left( \xi \right) \frac{\sin \left( {{2\pi }\left| \xi \right| t}\right) }{{2\pi }\left| \xi \right| }}\right\rbrack {e}^{{2\pi ix} \cdot \xi }{d\xi }.\]
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Proof. We first verify that \( u \) solves the wave equation. This is straightforward once we note that we can differentiate in \( x \) and \( t \) under the integral sign (because \( f \) and \( g \) are both Schwartz functions) and therefore \( u \) is at least \( {C}^{2} \). On the one hand we differentiate the exponential with respect to the \( x \) variables to get\n\n\[ \bigtriangleup u\left( {x, t}\right) = {\int }_{{\mathbb{R}}^{d}}\left\lbrack {\widehat{f}\left( \xi \right) \cos \left( {{2\pi }\left| \xi \right| t}\right) + \widehat{g}\left( \xi \right) \frac{\sin \left( {{2\pi }\left| \xi \right| t}\right) }{{2\pi }\left| \xi \right| }}\right\rbrack \left( {-4{\pi }^{2}{\left| \xi \right| }^{2}}\right) {e}^{{2\pi ix} \cdot \xi }{d\xi } \]\n\nwhile on the other hand we differentiate the terms in brackets with respect to \( t \) twice to get\n\n\[ \frac{{\partial }^{2}u}{\partial {t}^{2}}\left( {x, t}\right) = \]\n\n\[ {\int }_{{\mathbb{R}}^{d}}\left\lbrack {-4{\pi }^{2}{\left| \xi \right| }^{2}\widehat{f}\left( \xi \right) \cos \left( {{2\pi }\left| \xi \right| t}\right) - 4{\pi }^{2}{\left| \xi \right| }^{2}\widehat{g}\left( \xi \right) \frac{\sin \left( {{2\pi }\left| \xi \right| t}\right) }{{2\pi }\left| \xi \right| }}\right\rbrack {e}^{{2\pi ix} \cdot \xi }{d\xi }. \]\n\nThis shows that \( u \) solves equation (2). Setting \( t = 0 \) we get\n\n\[ u\left( {x,0}\right) = {\int }_{{\mathbb{R}}^{d}}\widehat{f}\left( \xi \right) {e}^{{2\pi ix} \cdot \xi }{d\xi } = f\left( x\right) \]\n\nby the Fourier inversion theorem. Finally, differentiating once with respect to \( t \), setting \( t = 0 \), and using the Fourier inversion shows that\n\n\[ \frac{\partial u}{\partial t}\left( {x,0}\right) = g\left( x\right) \]\n\nThus \( u \) also verifies the initial conditions, and the proof of the theorem is complete.
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Yes
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Theorem 3.2 If \( u \) is the solution of the wave equation given by formula (3), then \( E\left( t\right) \) is conserved, that is,\n\n\[ E\left( t\right) = E\left( 0\right) ,\;\text{ for all }t \in \mathbb{R}. \]
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The proof requires the following lemma.\n\nLemma 3.3 Suppose a and \( b \) are complex numbers and \( \alpha \) is real. Then\n\n\[ {\left| a\cos \alpha + b\sin \alpha \right| }^{2} + {\left| -a\sin \alpha + b\cos \alpha \right| }^{2} = {\left| a\right| }^{2} + {\left| b\right| }^{2}. \]\n\nThis follows directly because \( {e}_{1} = \left( {\cos \alpha ,\sin \alpha }\right) \) and \( {e}_{2} = \left( {-\sin \alpha ,\cos \alpha }\right) \) are a pair of orthonormal vectors, hence with \( Z = \left( {a, b}\right) \in {\mathbb{C}}^{2} \), we have\n\n\[ {\left| Z\right| }^{2} = {\left| Z \cdot {e}_{1}\right| }^{2} + {\left| Z \cdot {e}_{2}\right| }^{2} \]\n\nwhere \( \cdot \) represents the inner product in \( {\mathbb{C}}^{2} \).\n\nNow by Plancherel's theorem,\n\n\[ {\int }_{{\mathbb{R}}^{d}}{\left| \frac{\partial u}{\partial t}\right| }^{2}{dx} = {\int }_{{\mathbb{R}}^{d}}\left| {-{2\pi }}\right| \xi \left| {\widehat{f}\left( \xi \right) \sin \left( {{2\pi }\left| \xi \right| t}\right) + \widehat{g}\left( \xi \right) \cos \left( {{2\pi }\left| \xi \right| t}\right) }\right| {}^{2}{d\xi }. \]\n\nSimilarly,\n\n\[ {\int }_{{\mathbb{R}}^{d}}\mathop{\sum }\limits_{{j = 1}}^{d}{\left| \frac{\partial u}{\partial {x}_{j}}\right| }^{2}{dx} = {\int }_{{\mathbb{R}}^{d}}\left| {2\pi }\right| \xi \left| {\widehat{f}\left( \xi \right) \cos \left( {{2\pi }\left| \xi \right| t}\right) + \widehat{g}\left( \xi \right) \sin \left( {{2\pi }\left| \xi \right| t}\right) }\right| {}^{2}{d\xi }. \]\n\nWe now apply the lemma with\n\n\[ a = {2\pi }\left| \xi \right| \widehat{f}\left( \xi \right) ,\;b = \widehat{g}\left( \xi \right) \;\text{ and }\;\alpha = {2\pi }\left| \xi \right| t. \]\n\nThe result is that\n\n\[ E\left( t\right) = {\int }_{{\mathbb{R}}^{d}}{\left| \frac{\partial u}{\partial t}\right| }^{2} + {\left| \frac{\partial u}{\partial {x}_{1}}\right| }^{2} + \cdots + {\left| \frac{\partial u}{\partial {x}_{d}}\right| }^{2}{dx} \]\n\n\[ = {\int }_{{\mathbb{R}}^{d}}\left( {4{\pi }^{2}{\left| \xi \right| }^{2}{\left| \widehat{f}\left( \xi \right) \right| }^{2} + {\left| \widehat{g}\left( \xi \right) \right| }^{2}}\right) {d\xi } \]\n\nwhich is clearly independent of \( t \) . Thus Theorem 3.2 is proved.
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Yes
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Lemma 3.3 Suppose a and \( b \) are complex numbers and \( \alpha \) is real. Then\n\n\[ \n{\left| a\cos \alpha + b\sin \alpha \right| }^{2} + {\left| -a\sin \alpha + b\cos \alpha \right| }^{2} = {\left| a\right| }^{2} + {\left| b\right| }^{2}.\n\]
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This follows directly because \( {e}_{1} = \left( {\cos \alpha ,\sin \alpha }\right) \) and \( {e}_{2} = \left( {-\sin \alpha ,\cos \alpha }\right) \) are a pair of orthonormal vectors, hence with \( Z = \left( {a, b}\right) \in {\mathbb{C}}^{2} \), we have\n\n\[ \n{\left| Z\right| }^{2} = {\left| Z \cdot {e}_{1}\right| }^{2} + {\left| Z \cdot {e}_{2}\right| }^{2}\n\]\n\nwhere \( \cdot \) represents the inner product in \( {\mathbb{C}}^{2} \) .
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Yes
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Lemma 3.4 If \( f \in \mathcal{S}\left( {\mathbb{R}}^{3}\right) \) and \( t \) is fixed, then \( {M}_{t}\left( f\right) \in \mathcal{S}\left( {\mathbb{R}}^{3}\right) \) . Moreover, \( {M}_{t}\left( f\right) \) is indefinitely differentiable in \( t \), and each \( t \) -derivative also belongs to \( \mathcal{S}\left( {\mathbb{R}}^{3}\right) \) .
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Proof. Let \( F\left( x\right) = {M}_{t}\left( f\right) \left( x\right) \) . To show that \( F \) is rapidly decreasing, start with the inequality \( \left| {f\left( x\right) }\right| \leq {A}_{N}/\left( {1 + {\left| x\right| }^{N}}\right) \) which holds for every fixed \( N \geq 0 \) . As a simple consequence, whenever \( t \) is fixed, we have\n\n\[ \left| {f\left( {x - {\gamma t}}\right) }\right| \leq {A}_{N}^{\prime }/\left( {1 + {\left| x\right| }^{N}}\right) \;\text{ for all }\gamma \in {S}^{2}. \]\n\nTo see this consider separately the cases when \( \left| x\right| \leq 2\left| t\right| \), and \( \left| x\right| > 2\left| t\right| \) . Therefore, by integration\n\n\[ \left| {F\left( x\right) }\right| \leq {A}_{N}^{\prime }/\left( {1 + {\left| x\right| }^{N}}\right) \]\n\nand since this holds for every \( N \), the function \( F \) is rapidly decreasing. One next observes that \( F \) is indefinitely differentiable, and\n\n(6)\n\n\[ {\left( \frac{\partial }{\partial x}\right) }^{\alpha }F\left( x\right) = {M}_{t}\left( {f}^{\left( \alpha \right) }\right) \left( x\right) \]\n\nwhere \( {f}^{\left( \alpha \right) }\left( x\right) = {\left( \partial /\partial x\right) }^{\alpha }f \) . It suffices to prove this when \( {\left( \partial /\partial x\right) }^{\alpha } = \) \( \partial /\partial {x}_{k} \), and then proceed by induction to get the general case. Furthermore, it is enough to take \( k = 1 \) . Now\n\n\[ \frac{F\left( {{x}_{1} + h,{x}_{2},{x}_{3}}\right) - F\left( {{x}_{1},{x}_{2},{x}_{3}}\right) }{h} = \frac{1}{4\pi }{\int }_{{S}^{2}}{g}_{h}\left( \gamma \right) {d\sigma }\left( \gamma \right) \]\n\nwhere\n\n\[ {g}_{h}\left( \gamma \right) = \frac{f\left( {x + {e}_{1}h - {\gamma t}}\right) - f\left( {x - {\gamma t}}\right) }{h}, \]\n\nand \( {e}_{1} = \left( {1,0,0}\right) \) . Now, it suffices to observe that \( {g}_{h} \rightarrow \frac{\partial }{\partial {x}_{1}}f\left( {x - {\gamma t}}\right) \) as \( h \rightarrow 0 \) uniformly in \( \gamma \) . As a result, we find that (6) holds, and by the first argument, it follows that \( {\left( \frac{\partial }{\partial x}\right) }^{\alpha }F\left( x\right) \) is also rapidly decreasing, hence \( F \in \mathcal{S} \) . The same argument applies to each \( t \) -derivative of \( {M}_{t}\left( f\right) \) .
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Yes
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Lemma 3.5 \( \frac{1}{4\pi }{\int }_{{S}^{2}}{e}^{-{2\pi i\xi } \cdot \gamma }{d\sigma }\left( \gamma \right) = \frac{\sin \left( {{2\pi }\left| \xi \right| }\right) }{{2\pi }\left| \xi \right| } \)
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Proof. Note that the integral on the left is radial in \( \xi \) . Indeed, if \( R \) is a rotation then\n\n\[ \n{\int }_{{S}^{2}}{e}^{-{2\pi iR}\left( \xi \right) \cdot \gamma }{d\sigma }\left( \gamma \right) = {\int }_{{S}^{2}}{e}^{-{2\pi i\xi } \cdot {R}^{-1}\left( \gamma \right) }{d\sigma }\left( \gamma \right) = {\int }_{{S}^{2}}{e}^{-{2\pi i\xi } \cdot \gamma }{d\sigma }\left( \gamma \right) \n\]\n\nbecause we may change variables \( \gamma \rightarrow {R}^{-1}\left( \gamma \right) \) . (For this, see formula (4) in the appendix.) So if \( \left| \xi \right| = \rho \), it suffices to prove the lemma with\n\n\( \xi = \left( {0,0,\rho }\right) \) . If \( \rho = 0 \), the lemma is obvious. If \( \rho > 0 \), we choose spherical coordinates to find that the left-hand side is equal to\n\n\[ \n\frac{1}{4\pi }{\int }_{0}^{2\pi }{\int }_{0}^{\pi }{e}^{-{2\pi i\rho }\cos \theta }\sin {\theta d\theta d\varphi } \n\]\n\nThe change of variables \( u = - \cos \theta \) gives\n\n\[ \n\frac{1}{4\pi }{\int }_{0}^{2\pi }{\int }_{0}^{\pi }{e}^{-{2\pi i\rho }\cos \theta }\sin {\theta d\theta d\varphi } = \frac{1}{2}{\int }_{0}^{\pi }{e}^{-{2\pi i\rho }\cos \theta }\sin {\theta d\theta } \n\]\n\n\[ \n= \frac{1}{2}{\int }_{-1}^{1}{e}^{2\pi i\rho u}{du} \n\]\n\n\[ \n= \frac{1}{4\pi i\rho }{\left\lbrack {e}^{2\pi i\rho u}\right\rbrack }_{-1}^{1} \n\]\n\n\[ \n= \frac{\sin \left( {2\pi \rho }\right) }{2\pi \rho } \n\]\n\nand the formula is proved.
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Yes
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Theorem 3.6 The solution when \( d = 3 \) of the Cauchy problem for the wave equation\n\n\[ \bigtriangleup u = \frac{{\partial }^{2}u}{\partial {t}^{2}}\;\text{ subject to }\;u\left( {x,0}\right) = f\left( x\right) \;\text{ and }\;\frac{\partial u}{\partial t}\left( {x,0}\right) = g\left( x\right) \]\n\nis given by\n\n\[ u\left( {x, t}\right) = \frac{\partial }{\partial t}\left( {t{M}_{t}\left( f\right) \left( x\right) }\right) + t{M}_{t}\left( g\right) \left( x\right) . \]
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Proof. Consider first the problem\n\n\[ \bigtriangleup u = \frac{{\partial }^{2}u}{\partial {t}^{2}}\;\text{ subject to }\;u\left( {x,0}\right) = 0\;\text{ and }\;\frac{\partial u}{\partial t}\left( {x,0}\right) = g\left( x\right) . \]\n\nThen by Theorem 3.1, we know that its solution \( {u}_{1} \) is given by\n\n\[ {u}_{1}\left( {x, t}\right) = {\int }_{{\mathbb{R}}^{3}}\left\lbrack {\widehat{g}\left( \xi \right) \frac{\sin \left( {{2\pi }\left| \xi \right| t}\right) }{{2\pi }\left| \xi \right| }}\right\rbrack {e}^{{2\pi ix} \cdot \xi }{d\xi } \]\n\n\[ = t{\int }_{{\mathbb{R}}^{3}}\left\lbrack {\widehat{g}\left( \xi \right) \frac{\sin \left( {{2\pi }\left| \xi \right| t}\right) }{{2\pi }\left| \xi \right| t}}\right\rbrack {e}^{{2\pi ix} \cdot \xi }{d\xi } \]\n\n\[ = t{M}_{t}\left( g\right) \left( x\right) \]\n\nwhere we have used (7) applied to \( g \), and the Fourier inversion formula.\n\nAccording to Theorem 3.1 again, the solution to the problem\n\n\[ \bigtriangleup u = \frac{{\partial }^{2}u}{\partial {t}^{2}}\;\text{ subject to }\;u\left( {x,0}\right) = f\left( x\right) \;\text{ and }\;\frac{\partial u}{\partial t}\left( {x,0}\right) = 0 \]\n\nis given by\n\n\[ {u}_{2}\left( {x, t}\right) = {\int }_{{\mathbb{R}}^{3}}\left\lbrack {\widehat{f}\left( \xi \right) \cos \left( {{2\pi }\left| \xi \right| t}\right) }\right\rbrack {e}^{{2\pi ix} \cdot \xi }{d\xi } \]\n\n\[ = \frac{\partial }{\partial t}\left( {t{\int }_{{\mathbb{R}}^{3}}\left\lbrack {\widehat{f}\left( \xi \right) \frac{\sin \left( {{2\pi }\left| \xi \right| t}\right) }{{2\pi }\left| \xi \right| t}}\right\rbrack {e}^{{2\pi ix} \cdot \xi }{d\xi }}\right) \]\n\n\[ = \frac{\partial }{\partial t}\left( {t{M}_{t}\left( f\right) \left( x\right) }\right) \]\n\nWe may now superpose these two solutions to obtain \( u = {u}_{1} + {u}_{2} \) as the solution of our original problem.
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Yes
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Theorem 3.7 A solution of the Cauchy problem for the wave equation in two dimensions with initial data \( f, g \in \mathcal{S}\left( {\mathbb{R}}^{2}\right) \) is given by\n\n\[ u\left( {x, t}\right) = \frac{\partial }{\partial t}\left( {t{\widetilde{M}}_{t}\left( f\right) \left( x\right) }\right) + t{\widetilde{M}}_{t}\left( g\right) \left( x\right) . \]
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Formally, the identity in the theorem arises as follows. If we start with an initial pair of functions \( f \) and \( g \) in \( \mathcal{S}\left( {\mathbb{R}}^{2}\right) \), we may consider the corresponding functions \( \widetilde{f} \) and \( \widetilde{g} \) on \( {\mathbb{R}}^{3} \) that are merely extensions of \( f \) and \( g \) that are constant in the \( {x}_{3} \) variable, that is,\n\n\[ \widetilde{f}\left( {{x}_{1},{x}_{2},{x}_{3}}\right) = f\left( {{x}_{1},{x}_{2}}\right) \;\text{ and }\;\widetilde{g}\left( {{x}_{1},{x}_{2},{x}_{3}}\right) = g\left( {{x}_{1},{x}_{2}}\right) . \]\n\nNow, if \( \widetilde{u} \) is the solution (given in the previous section) of the 3-dimensional wave equation with initial data \( \widetilde{f} \) and \( \widetilde{g} \), then one can expect that \( \widetilde{u} \) is also constant in \( {x}_{3} \) so that \( \widetilde{u} \) satisfies the 2-dimensional wave equation. A difficulty with this argument is that \( \widetilde{f} \) and \( \widetilde{g} \) are not rapidly decreasing since they are constant in \( {x}_{3} \), so that our previous methods do not apply. However, it is easy to modify the argument so as to obtain a proof of Theorem 3.7.\n\nWe fix \( T > 0 \) and consider a function \( \eta \left( {x}_{3}\right) \) that is in \( \mathcal{S}\left( \mathbb{R}\right) \), such that \( \eta \left( {x}_{3}\right) = 1 \) if \( \left| {x}_{3}\right| \leq {3T} \) . The trick is to truncate \( \widetilde{f} \) and \( \widetilde{g} \) in the \( {x}_{3} \) -variable, and consider instead\n\n\[ {\widetilde{f}}^{b}\left( {{x}_{1},{x}_{2},{x}_{3}}\right) = f\left( {{x}_{1},{x}_{2}}\right) \eta \left( {x}_{3}\right) \;\text{ and }\;{\widetilde{g}}^{b}\left( {{x}_{1},{x}_{2},{x}_{3}}\right) = g\left( {{x}_{1},{x}_{2}}\right) \eta \left( {x}_{3}\right) . \]\n\nNow both \( {\widetilde{f}}^{b} \) and \( {\widetilde{g}}^{b} \) are in \( \mathcal{S}\left( {\mathbb{R}}^{3}\right) \), so Theorem 3.6 provides a solution \( {\widetilde{u}}^{b} \) of the wave equation with initial data \( {\widetilde{f}}^{b} \) and \( {\widetilde{g}}^{b} \) . It is easy to see from the formula that \( {\widetilde{u}}^{b}\left( {x, t}\right) \) is independent of \( {x}_{3} \), whenever \( \left| {x}_{3}\right| \leq T \) and \( \left| t\right| \leq T \) . In particular, if we define \( u\left( {{x}_{1},{x}_{2}, t}\right) = {\widetilde{u}}^{b}\left( {{x}_{1},{x}_{2},0, t}\right) \), then \( u \) satisfies the 2-dimensional wave equation when \( \left| t\right| \leq T \) . Since \( T \) is arbitrary, \( u \) is a solution to our problem, and it remains to see why \( u \) has the desired form.
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Yes
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Proposition 5.1 If \( f \in \mathcal{S}\left( {\mathbb{R}}^{3}\right) \), then for each \( \gamma \) the definition of \( {\int }_{{\mathcal{P}}_{t,\gamma }}f \) is independent of the choice of \( {e}_{1} \) and \( {e}_{2} \) . Moreover\n\n\[{\int }_{-\infty }^{\infty }\left( {{\int }_{{\mathcal{P}}_{t,\gamma }}f}\right) {dt} = {\int }_{{\mathbb{R}}^{3}}f\left( x\right) {dx}\]
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Proof. If \( {e}_{1}^{\prime },{e}_{2}^{\prime } \) is another choice of basis vectors so that \( \gamma ,{e}_{1}^{\prime },{e}_{2}^{\prime } \) is orthonormal, consider the rotation \( R \) in \( {\mathbb{R}}^{2} \) which takes \( {e}_{1} \) to \( {e}_{1}^{\prime } \) and \( {e}_{2} \) to \( {e}_{2}^{\prime } \) . Changing variables \( {u}^{\prime } = R\left( u\right) \) in the integral proves that our definition (12) is independent of the choice of basis.\n\nTo prove the formula, let \( R \) denote the rotation which takes the standard basis of unit vectors \( {}^{4} \) in \( {\mathbb{R}}^{3} \) to \( \gamma ,{e}_{1} \), and \( {e}_{2} \) . Then\n\n\[{\int }_{{\mathbb{R}}^{3}}f\left( x\right) {dx} = {\int }_{{\mathbb{R}}^{3}}f\left( {Rx}\right) {dx}\]\n\n\[= {\int }_{{\mathbb{R}}^{3}}f\left( {{x}_{1}\gamma + {x}_{2}{e}_{1} + {x}_{3}{e}_{2}}\right) d{x}_{1}d{x}_{2}d{x}_{3}\]\n\n\[= {\int }_{-\infty }^{\infty }\left( {{\int }_{{\mathcal{P}}_{t,\gamma }}f}\right) {dt}\]
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Yes
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Lemma 5.2 If \( f \in \mathcal{S}\left( {\mathbb{R}}^{3}\right) \), then \( \mathcal{R}\left( f\right) \left( {t,\gamma }\right) \in \mathcal{S}\left( \mathbb{R}\right) \) for each fixed \( \gamma \) . Moreover, \[ \widehat{\mathcal{R}}\left( f\right) \left( {s,\gamma }\right) = \widehat{f}\left( {s\gamma }\right) \]
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Proof. Since \( f \in \mathcal{S}\left( {\mathbb{R}}^{3}\right) \), for every positive integer \( N \) there is a constant \( {A}_{N} < \infty \) so that \[ {\left( 1 + \left| t\right| \right) }^{N}{\left( 1 + \left| u\right| \right) }^{N}\left| {f\left( {{t\gamma } + u}\right) }\right| \leq {A}_{N} \] if we recall that \( x = {t\gamma } + u \), where \( \gamma \) is orthogonal to \( u \) . Therefore, as soon as \( N \geq 3 \), we find \[ {\left( 1 + \left| t\right| \right) }^{N}\mathcal{R}\left( f\right) \left( {t,\gamma }\right) \leq {A}_{N}{\int }_{{\mathbb{R}}^{2}}\frac{du}{{\left( 1 + \left| u\right| \right) }^{N}} < \infty . \] A similar argument for the derivatives shows that \( \mathcal{R}\left( f\right) \left( {t,\gamma }\right) \in \mathcal{S}\left( \mathbb{R}\right) \) for each fixed \( \gamma \) . To establish the identity, we first note that \[ \widehat{\mathcal{R}}\left( f\right) \left( {s,\gamma }\right) = {\int }_{-\infty }^{\infty }\left( {{\int }_{{\mathcal{P}}_{t,\gamma }}f}\right) {e}^{-{2\pi ist}}{dt} \] \[ = {\int }_{-\infty }^{\infty }{\int }_{{\mathbb{R}}^{2}}f\left( {{t\gamma } + {u}_{1}{e}_{1} + {u}_{2}{e}_{2}}\right) d{u}_{1}d{u}_{2}{e}^{-{2\pi ist}}{dt}. \] However, since \( \gamma \cdot u = 0 \) and \( \left| \gamma \right| = 1 \), we may write \[ {e}^{-{2\pi ist}} = {e}^{-{2\pi is\gamma } \cdot \left( {{t\gamma } + u}\right) }. \] As a result, we find that \[ \widehat{\mathcal{R}}\left( f\right) \left( {s,\gamma }\right) = {\int }_{-\infty }^{\infty }{\int }_{{\mathbb{R}}^{2}}f\left( {{t\gamma } + {u}_{1}{e}_{1} + {u}_{2}{e}_{2}}\right) {e}^{-{2\pi is\gamma } \cdot \left( {{t\gamma } + u}\right) }d{u}_{1}d{u}_{2}{dt} \] \[ = {\int }_{-\infty }^{\infty }{\int }_{{\mathbb{R}}^{2}}f\left( {{t\gamma } + u}\right) {e}^{-{2\pi is\gamma } \cdot \left( {{t\gamma } + u}\right) }{dudt}. \] A final rotation from \( \gamma ,{e}_{1},{e}_{2} \) to the standard basis in \( {\mathbb{R}}^{3} \) proves that \( \widehat{\mathcal{R}}\left( f\right) \left( {s,\gamma }\right) = \widehat{f}\left( {s\gamma }\right) \), as desired.
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Yes
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Corollary 5.3 If \( f, g \in \mathcal{S}\left( {\mathbb{R}}^{3}\right) \) and \( \mathcal{R}\left( f\right) = \mathcal{R}\left( g\right) \), then \( f = g \) .
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The proof of the corollary follows from an application of the lemma to the difference \( f - g \) and use of the Fourier inversion theorem.
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No
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Theorem 5.4 If \( f \in \mathcal{S}\left( {\mathbb{R}}^{3}\right) \), then\n\n\[ \bigtriangleup \left( {{\mathcal{R}}^{ * }\mathcal{R}\left( f\right) }\right) = - 8{\pi }^{2}f \]
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We recall that \( \bigtriangleup = \frac{{\partial }^{2}}{\partial {x}_{1}^{2}} + \frac{{\partial }^{2}}{\partial {x}_{2}^{2}} + \frac{{\partial }^{2}}{\partial {x}_{3}^{2}} \) is the Laplacian.\n\nProof. By our previous lemma, we have\n\n\[ \mathcal{R}\left( f\right) \left( {t,\gamma }\right) = {\int }_{-\infty }^{\infty }\widehat{f}\left( {s\gamma }\right) {e}^{2\pi its}{ds}. \]\n\nTherefore\n\n\[ {\mathcal{R}}^{ * }\mathcal{R}\left( f\right) \left( x\right) = {\int }_{{S}^{2}}{\int }_{-\infty }^{\infty }\widehat{f}\left( {s\gamma }\right) {e}^{{2\pi ix} \cdot {\gamma s}}{dsd\sigma }\left( \gamma \right) , \]\n\nhence\n\n\[ \bigtriangleup \left( {{\mathcal{R}}^{ * }\mathcal{R}\left( f\right) }\right) \left( x\right) = {\int }_{{S}^{2}}{\int }_{-\infty }^{\infty }\widehat{f}\left( {s\gamma }\right) \left( {-4{\pi }^{2}{s}^{2}}\right) {e}^{{2\pi ix} \cdot {\gamma s}}{dsd\sigma }\left( \gamma \right) \]\n\n\[ = - 4{\pi }^{2}{\int }_{{S}^{2}}{\int }_{-\infty }^{\infty }\widehat{f}\left( {s\gamma }\right) {e}^{{2\pi ix} \cdot {\gamma s}}{s}^{2}{dsd\sigma }\left( \gamma \right) \]\n\n\[ = - 4{\pi }^{2}{\int }_{{S}^{2}}{\int }_{-\infty }^{0}\widehat{f}\left( {s\gamma }\right) {e}^{{2\pi ix} \cdot {\gamma s}}{s}^{2}{dsd\sigma }\left( \gamma \right) \]\n\n\[ - 4{\pi }^{2}{\int }_{{S}^{2}}{\int }_{0}^{\infty }\widehat{f}\left( {s\gamma }\right) {e}^{{2\pi ix} \cdot {\gamma s}}{s}^{2}{dsd\sigma }\left( \gamma \right) \]\n\n\[ = - 8{\pi }^{2}{\int }_{{S}^{2}}{\int }_{0}^{\infty }\widehat{f}\left( {s\gamma }\right) {e}^{{2\pi ix} \cdot {\gamma s}}{s}^{2}{dsd\sigma }\left( \gamma \right) \]\n\n\[ = - 8{\pi }^{2}f\left( x\right) \text{.} \]\n\nIn the first line, we have differentiated under the integral sign and used the fact \( \bigtriangleup \left( {e}^{{2\pi ix} \cdot {\gamma s}}\right) = \left( {-4{\pi }^{2}{s}^{2}}\right) {e}^{{2\pi ix} \cdot {\gamma s}} \), since \( \left| \gamma \right| = 1 \). The last step follows from the formula for polar coordinates in \( {\mathbb{R}}^{3} \) and the Fourier inversion theorem.
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Yes
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Lemma 1.1 The family \( \left\{ {{e}_{0},\ldots ,{e}_{N - 1}}\right\} \) is orthogonal. In fact,\n\n\[ \left( {{e}_{m},{e}_{\ell }}\right) = \left\{ \begin{array}{ll} N & \text{ if }m = \ell \\ 0 & \text{ if }m \neq \ell \end{array}\right. \]
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Proof. We have\n\n\[ \left( {{e}_{m},{e}_{\ell }}\right) = \mathop{\sum }\limits_{{k = 0}}^{{N - 1}}{\zeta }^{mk}{\zeta }^{-\ell k} = \mathop{\sum }\limits_{{k = 0}}^{{N - 1}}{\zeta }^{\left( {m - \ell }\right) k}. \]\n\nIf \( m = \ell \), each term in the sum is equal to 1, and the sum equals \( N \) . If \( m \neq \ell \), then \( q = {\zeta }^{m - \ell } \) is not equal to 1, and the usual formula\n\n\[ 1 + q + {q}^{2} + \cdots + {q}^{N - 1} = \frac{1 - {q}^{N}}{1 - q} \]\n\nshows that \( \left( {{e}_{m},{e}_{\ell }}\right) = 0 \), because \( {q}^{N} = 1 \) .
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Yes
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Theorem 1.2 If \( F \) is a function on \( \mathbb{Z}\left( N\right) \), then\n\n\[ F\left( k\right) = \mathop{\sum }\limits_{{n = 0}}^{{N - 1}}{a}_{n}{e}^{{2\pi ink}/N}. \]\n\nMoreover,\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{{N - 1}}{\left| {a}_{n}\right| }^{2} = \frac{1}{N}\mathop{\sum }\limits_{{k = 0}}^{{N - 1}}{\left| F\left( k\right) \right| }^{2} \]
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The proof follows directly from (1) once we observe that\n\n\[ {a}_{n} = \frac{1}{N}\left( {F,{e}_{n}}\right) = \frac{1}{\sqrt{N}}\left( {F,{e}_{n}^{ * }}\right) . \]
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Yes
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Lemma 1.4 If we are given \( {\omega }_{2M} = {e}^{-{2\pi i}/\left( {2M}\right) } \), then\n\n\[ \n\# \left( {2M}\right) \leq 2\# \left( M\right) + {8M}.\n\]
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Proof. The calculation of \( {\omega }_{2M},\ldots ,{\omega }_{2M}^{2M} \) requires no more than \( {2M} \) operations. Note that in particular we get \( {\omega }_{M} = {e}^{-{2\pi i}/M} = {\omega }_{2M}^{2} \). The main idea is that for any given function \( F \) on \( \mathbb{Z}\left( {2M}\right) \), we consider two functions \( {F}_{0} \) and \( {F}_{1} \) on \( \mathbb{Z}\left( M\right) \) defined by\n\n\[ \n{F}_{0}\left( r\right) = F\left( {2r}\right) \;\text{ and }\;{F}_{1}\left( r\right) = F\left( {{2r} + 1}\right) .\n\]\n\nWe assume that it is possible to calculate the Fourier coefficients of \( {F}_{0} \) and \( {F}_{1} \) in no more than \( \# \left( M\right) \) operations each. If we denote the Fourier coefficients corresponding to the groups \( \mathbb{Z}\left( {2M}\right) \) and \( \mathbb{Z}\left( M\right) \) by \( {a}_{k}^{2M} \) and \( {a}_{k}^{M} \), respectively, then we have\n\n\[ \n{a}_{k}^{2M}\left( F\right) = \frac{1}{2}\left( {{a}_{k}^{M}\left( {F}_{0}\right) + {a}_{k}^{M}\left( {F}_{1}\right) {\omega }_{2M}^{k}}\right) .\n\]\n\nTo prove this, we sum over odd and even integers in the definition of the Fourier coefficient \( {a}_{k}^{2M}\left( F\right) \), and find\n\n\[ \n{a}_{k}^{2M}\left( F\right) = \frac{1}{2M}\mathop{\sum }\limits_{{r = 0}}^{{{2M} - 1}}F\left( r\right) {\omega }_{2M}^{kr}\n\]\n\n\[ \n= \frac{1}{2}\left( {\frac{1}{M}\mathop{\sum }\limits_{{\ell = 0}}^{{M - 1}}F\left( {2\ell }\right) {\omega }_{2M}^{k\left( {2\ell }\right) } + \frac{1}{M}\mathop{\sum }\limits_{{m = 0}}^{{M - 1}}F\left( {{2m} + 1}\right) {\omega }_{2M}^{k\left( {{2m} + 1}\right) }}\right)\n\]\n\n\[ \n= \frac{1}{2}\left( {\frac{1}{M}\mathop{\sum }\limits_{{\ell = 0}}^{{M - 1}}{F}_{0}\left( \ell \right) {\omega }_{M}^{k\ell } + \frac{1}{M}\mathop{\sum }\limits_{{m = 0}}^{{M - 1}}{F}_{1}\left( m\right) {\omega }_{M}^{km}{\omega }_{2M}^{k}}\right) ,\n\]\n\nwhich establishes our assertion.\n\nAs a result, knowing \( {a}_{k}^{M}\left( {F}_{0}\right) ,{a}_{k}^{M}\left( {F}_{1}\right) \), and \( {\omega }_{2M}^{k} \), we see that each \( {a}_{k}^{2M}\left( F\right) \) can be computed using no more than three operations (one addition and two multiplications). So\n\n\[ \n\# \left( {2M}\right) \leq {2M} + 2\# \left( M\right) + 3 \times {2M} = 2\# \left( M\right) + {8M},\n\]\n\nand the proof of the lemma is complete.
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Yes
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Lemma 2.1 The set \( \widehat{G} \) is an abelian group under multiplication defined \( {by} \)\n\n\[ \left( {{e}_{1} \cdot {e}_{2}}\right) \left( a\right) = {e}_{1}\left( a\right) {e}_{2}\left( a\right) \;\text{ for all }a \in G. \]
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The proof of this assertion is straightforward if one observes that the trivial character plays the role of the unit. We call \( \widehat{G} \) the dual group of \( G \) .
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No
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Lemma 2.2 Let \( G \) be a finite abelian group, and \( e : G \rightarrow \mathbb{C} - \{ 0\} \) a multiplicative function, namely \( e\left( {a \cdot b}\right) = e\left( a\right) e\left( b\right) \) for all \( a, b \in G \) . Then \( e \) is a character.
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Proof. The group \( G \) being finite, the absolute value of \( e\left( a\right) \) is bounded above and below as \( a \) ranges over \( G \) . Since \( \left| {e\left( {b}^{n}\right) }\right| = {\left| e\left( b\right) \right| }^{n} \), we conclude that \( \left| {e\left( b\right) }\right| = 1 \) for all \( b \in G \) .
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Yes
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Theorem 2.3 The characters of \( G \) form an orthonormal family with respect to the inner product defined above.
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Since \( \left| {e\left( a\right) }\right| = 1 \) for any character, we find that\n\n\[ \left( {e, e}\right) = \frac{1}{\left| G\right| }\mathop{\sum }\limits_{{a \in G}}e\left( a\right) \overline{e\left( a\right) } = \frac{1}{\left| G\right| }\mathop{\sum }\limits_{{a \in G}}{\left| e\left( a\right) \right| }^{2} = 1. \]\n\nIf \( e \neq {e}^{\prime } \) and both are characters, we must prove that \( \left( {e,{e}^{\prime }}\right) = 0 \) ; we isolate the key step in a lemma.\n\nLemma 2.4 If \( e \) is a non-trivial character of the group \( G \), then \( \mathop{\sum }\limits_{{a \in G}}e\left( a\right) = 0. \)\n\nProof. Choose \( b \in G \) such that \( e\left( b\right) \neq 1 \) . Then we have\n\n\[ e\left( b\right) \mathop{\sum }\limits_{{a \in G}}e\left( a\right) = \mathop{\sum }\limits_{{a \in G}}e\left( b\right) e\left( a\right) = \mathop{\sum }\limits_{{a \in G}}e\left( {ab}\right) = \mathop{\sum }\limits_{{a \in G}}e\left( a\right) . \]\n\nThe last equality follows because as \( a \) ranges over the group, \( {ab} \) ranges over \( G \) as well. Therefore \( \mathop{\sum }\limits_{{a \in G}}e\left( a\right) = 0 \) .\n\nWe can now conclude the proof of the theorem. Suppose \( {e}^{\prime } \) is a character distinct from \( e \) . Because \( e{\left( {e}^{\prime }\right) }^{-1} \) is non-trivial, the lemma implies that\n\n\[ \mathop{\sum }\limits_{{a \in G}}e\left( a\right) {\left( {e}^{\prime }\left( a\right) \right) }^{-1} = 0. \]\n\nSince \( {\left( {e}^{\prime }\left( a\right) \right) }^{-1} = \overline{{e}^{\prime }\left( a\right) } \), the theorem is proved.
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Yes
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Lemma 2.4 If \( e \) is a non-trivial character of the group \( G \), then \( \mathop{\sum }\limits_{{a \in G}}e\left( a\right) = 0. \)
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Proof. Choose \( b \in G \) such that \( e\left( b\right) \neq 1 \) . Then we have\n\n\[ e\left( b\right) \mathop{\sum }\limits_{{a \in G}}e\left( a\right) = \mathop{\sum }\limits_{{a \in G}}e\left( b\right) e\left( a\right) = \mathop{\sum }\limits_{{a \in G}}e\left( {ab}\right) = \mathop{\sum }\limits_{{a \in G}}e\left( a\right) .\n\]\n\nThe last equality follows because as \( a \) ranges over the group, \( {ab} \) ranges over \( G \) as well. Therefore \( \mathop{\sum }\limits_{{a \in G}}e\left( a\right) = 0 \) .
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Yes
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Lemma 2.6 Suppose \( \\left\\{ {{T}_{1},\\ldots ,{T}_{k}}\\right\\} \) is a commuting family of unitary transformations on the finite-dimensional inner product space \( V \) ; that is,\n\n\[ \n{T}_{i}{T}_{j} = {T}_{j}{T}_{i}\\;\\text{ for all }i, j.\n\]\n\nThen \( {T}_{1},\\ldots ,{T}_{k} \) are simultaneously diagonalizable. In other words, there exists a basis for \( V \) which consists of eigenvectors for every \( {T}_{i}, i = 1,\\ldots, k \) .
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Proof. We use induction on \( k \) . The case \( k = 1 \) is simply the spectral theorem. Suppose that the lemma is true for any family of \( k - 1 \) commuting unitary transformations. The spectral theorem applied to \( {T}_{k} \) says that \( V \) is the direct sum of its eigenspaces\n\n\[ \nV = {V}_{{\\lambda }_{1}} \\oplus \\cdots \\oplus {V}_{{\\lambda }_{s}}\n\]\n\nwhere \( {V}_{{\\lambda }_{i}} \) denotes the subspace of all eigenvectors with eigenvalue \( {\\lambda }_{i} \) . We claim that each one of the \( {T}_{1},\\ldots ,{T}_{k - 1} \) maps each eigenspace \( {V}_{{\\lambda }_{i}} \) to itself. Indeed, if \( v \\in {V}_{{\\lambda }_{i}} \) and \( 1 \\leq j \\leq k - 1 \), then\n\n\[ \n{T}_{k}{T}_{j}\\left( v\\right) = {T}_{j}{T}_{k}\\left( v\\right) = {T}_{j}\\left( {{\\lambda }_{i}v}\\right) = {\\lambda }_{i}{T}_{j}\\left( v\\right)\n\]\n\nso \( {T}_{j}\\left( v\\right) \\in {V}_{{\\lambda }_{i}} \), and the claim is proved.\n\nSince the restrictions to \( {V}_{{\\lambda }_{i}} \) of \( {T}_{1},\\ldots ,{T}_{k - 1} \) form a family of commuting unitary linear transformations, the induction hypothesis guarantees that these are simultaneously diagonalizable on each subspace \( {V}_{{\\lambda }_{i}} \) . This diagonalization provides us with the desired basis for each \( {V}_{{\\lambda }_{i}} \), and thus for \( V \) .
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Yes
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Theorem 2.8 If \( f \) is a function on \( G \), then \( \parallel f{\parallel }^{2} = \mathop{\sum }\limits_{{e \in \widehat{G}}}{\left| \widehat{f}\left( e\right) \right| }^{2} \) .
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Proof. Since the characters of \( G \) form an orthonormal basis for the vector space \( V \), and \( \left( {f, e}\right) = \widehat{f}\left( e\right) \), we have that\n\n\[ \parallel f{\parallel }^{2} = \left( {f, f}\right) = \mathop{\sum }\limits_{{e \in \widehat{G}}}\left( {f, e}\right) \overline{\widehat{f}\left( e\right) } = \mathop{\sum }\limits_{{e \in \widehat{G}}}{\left| \widehat{f}\left( e\right) \right| }^{2}. \]\n\nThe apparent difference of this statement with that of Theorem 1.2 is due to the different normalizations of the Fourier coefficients that are used.
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Yes
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Theorem 1.1 (Euclid’s algorithm) For any integers \( a \) and \( b \) with \( b > 0 \), there exist unique integers \( q \) and \( r \) with \( 0 \leq r < b \) such that\n\n\[ a = {qb} + r. \]
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Proof. First we prove the existence of \( q \) and \( r \) . Let \( S \) denote the set of all non-negative integers of the form \( a - {qb} \) with \( q \in \mathbb{Z} \) . This set is non-empty and in fact \( S \) contains arbitrarily large positive integers since \( b \neq 0 \) . Let \( r \) denote the smallest element in \( S \), so that\n\n\[ r = a - {qb} \]\n\nfor some integer \( q \) . By construction \( 0 \leq r \), and we claim that \( r < b \) . If not, we may write \( r = b + s \) with \( 0 \leq s < r \), so \( b + s = a - {qb} \), which then implies\n\n\[ s = a - \left( {q + 1}\right) b. \]\n\nHence \( s \in S \) with \( s < r \), and this contradicts the minimality of \( r \) . So \( r < b \), hence \( q \) and \( r \) satisfy the conditions of the theorem.\n\nTo prove uniqueness, suppose we also had \( a = {q}_{1}b + {r}_{1} \) where \( 0 \leq {r}_{1} < b \) . By subtraction we find\n\n\[ \left( {q - {q}_{1}}\right) b = {r}_{1} - r \]\n\nThe left-hand side has absolute value 0 or \( \geq b \), while the right-hand side has absolute value \( < b \) . Hence both sides of the equation must be 0, which gives \( q = {q}_{1} \) and \( r = {r}_{1} \) .
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Yes
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Theorem 1.2 If \( \gcd \left( {a, b}\right) = d \), then there exist integers \( x \) and \( y \) such that\n\n\[ \n{ax} + {by} = d\text{.} \n\]
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Proof. Consider the set \( S \) of all positive integers of the form \( {ax} + {by} \) where \( x, y \in \mathbb{Z} \), and let \( s \) be the smallest element in \( S \) . We claim that \( s = \) \( d \) . By construction, there exist integers \( x \) and \( y \) such that\n\n\[ \n{ax} + {by} = s.\n\]\n\nClearly, any divisor of \( a \) and \( b \) divides \( s \), so we must have \( d \leq s \) . The proof will be complete if we can show that \( s\left| {a\text{and}s}\right| b \) . By Euclid’s algorithm, we can write \( a = {qs} + r \) with \( 0 \leq r < s \) . Multiplying the above by \( q \) we find \( {qax} + {qby} = {qs} \), and therefore\n\n\[ \n{qax} + {qby} = a - r.\n\]\n\nHence \( r = a\left( {1 - {qx}}\right) + b\left( {-{qy}}\right) \) . Since \( s \) was minimal in \( S \) and \( 0 \leq r < s \) , we conclude that \( r = 0 \), therefore \( s \) divides \( a \) . A similar argument shows that \( s \) divides \( b \), hence \( s = d \) as desired.
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Yes
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Corollary 1.3 Two positive integers \( a \) and \( b \) are relatively prime if and only if there exist integers \( x \) and \( y \) such that \( {ax} + {by} = 1 \) .
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Proof. If \( a \) and \( b \) are relatively prime, two integers \( x \) and \( y \) with the desired property exist by Theorem 1.2. Conversely, if \( {ax} + {by} = 1 \) holds and \( d \) is positive and divides both \( a \) and \( b \), then \( d \) divides 1, hence \( d = 1 \) .
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Yes
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Corollary 1.4 If a and \( c \) are relatively prime and \( c \) divides \( {ab} \), then \( c \) divides \( b \) . In particular, if \( p \) is a prime that does not divide \( a \) and \( p \) divides \( {ab} \), then \( p \) divides \( b \) .
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Proof. We can write \( 1 = {ax} + {cy} \), so multiplying by \( b \) we find \( b = \) \( {abx} + {cby} \) . Hence \( c \mid b \) .
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Yes
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Corollary 1.5 If \( p \) is prime and \( p \) divides the product \( {a}_{1}\cdots {a}_{r} \), then \( p \) divides \( {a}_{i} \) for some \( i \) .
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Proof. By the previous corollary, if \( p \) does not divide \( {a}_{1} \), then \( p \) divides \( {a}_{2}\cdots {a}_{r} \), so eventually \( p \mid {a}_{i} \) .
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No
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Theorem 1.6 Every positive integer greater than 1 can be factored uniquely into a product of primes.
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Proof. First, we show that such a factorization is possible. We do so by proving that the set \( S \) of positive integers \( > 1 \) which do not have a factorization into primes is empty. Arguing by contradiction, we assume that \( S \neq \varnothing \) . Let \( n \) be the smallest element of \( S \) . Since \( n \) cannot be a prime, there exist integers \( a > 1 \) and \( b > 1 \) such that \( {ab} = n \) . But then \( a < n \) and \( b < n \), so \( a \notin S \) as well as \( b \notin S \) . Hence both \( a \) and \( b \) have prime factorizations and so does their product \( n \) . This implies \( n \notin S \), therefore \( S \) is empty, as desired.\n\nWe now turn our attention to the uniqueness of the factorization. Suppose that \( n \) has two factorizations into primes\n\n\[ n = {p}_{1}{p}_{2}\cdots {p}_{r} \]\n\n\[ = {q}_{1}{q}_{2}\cdots {q}_{s} \]\n\nSo \( {p}_{1} \) divides \( {q}_{1}{q}_{2}\cdots {q}_{s} \), and we can apply Corollary 1.5 to conclude that \( {p}_{1} \mid {q}_{i} \) for some \( i \) . Since \( {q}_{i} \) is prime, we must have \( {p}_{1} = {q}_{i} \) . Continuing with this argument we find that the two factorizations of \( n \) are equal up to a permutation of the factors.
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Yes
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Theorem 1.7 There are infinitely many primes.
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Proof. Suppose not, and denote by \( {p}_{1},\ldots ,{p}_{n} \) the complete set of primes. Define\n\n\[ N = {p}_{1}{p}_{2}\cdots {p}_{n} + 1 \]\n\nSince \( N \) is larger than any \( {p}_{i} \), the integer \( N \) cannot be prime. Therefore, \( N \) is divisible by a prime that belongs to our list. But this is also an absurdity since every prime divides the product, yet no prime divides 1 . This contradiction concludes the proof.
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Yes
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Lemma 1.8 The exponential and logarithm functions satisfy the following properties:\n\n(i) \( {e}^{\log x} = x \) .\n\n(ii) \( \log \left( {1 + x}\right) = x + E\left( x\right) \) where \( \left| {E\left( x\right) }\right| \leq {x}^{2} \) if \( \left| x\right| < 1/2 \) .\n\n(iii) If \( \log \left( {1 + x}\right) = y \) and \( \left| x\right| < 1/2 \), then \( \left| y\right| \leq 2\left| x\right| \) .
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Proof. Property (i) is standard. To prove property (ii) we use the power series expansion of \( \log \left( {1 + x}\right) \) for \( \left| x\right| < 1 \), that is,\n\n(2)\n\n\[ \log \left( {1 + x}\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n + 1}}{n}{x}^{n}. \]\n\nThen we have\n\n\[ E\left( x\right) = \log \left( {1 + x}\right) - x = - \frac{{x}^{2}}{2} + \frac{{x}^{3}}{3} - \frac{{x}^{4}}{4} + \cdots ,\]\n\nand the triangle inequality implies\n\n\[ \left| {E\left( x\right) }\right| \leq \frac{{x}^{2}}{2}\left( {1 + \left| x\right| + {\left| x\right| }^{2} + \cdots }\right) . \]\n\nTherefore, if \( \left| x\right| \leq 1/2 \) we can sum the geometric series on the right-hand side to find that\n\n\[ \left| {E\left( x\right) }\right| \leq \frac{{x}^{2}}{2}\left( {1 + \frac{1}{2} + \frac{1}{{2}^{2}} + \cdots }\right) \]\n\n\[ \leq \frac{{x}^{2}}{2}\left( \frac{1}{1 - 1/2}\right) \]\n\n\[ \leq {x}^{2}\text{.} \]\n\nThe proof of property (iii) is now immediate; if \( x \neq 0 \) and \( \left| x\right| \leq 1/2 \), then\n\n\[ \left| \frac{\log \left( {1 + x}\right) }{x}\right| \leq 1 + \left| \frac{E\left( x\right) }{x}\right| \]\n\n\[ \leq 1 + \left| x\right| \]\n\n\[ \leq 2\text{,} \]\n\nand if \( x = 0 \) ,(iii) is clearly also true.
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Yes
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Proposition 1.9 If \( {A}_{n} = 1 + {a}_{n} \) and \( \sum \left| {a}_{n}\right| \) converges, then the product \( \mathop{\prod }\limits_{n}{A}_{n} \) converges, and this product vanishes if and only if one of its factors \( {A}_{n} \) vanishes. Also, if \( {a}_{n} \neq 1 \) for all \( n \), then \( \mathop{\prod }\limits_{n}1/\left( {1 - {a}_{n}}\right) \) converges.
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Proof. If \( \sum \left| {a}_{n}\right| \) converges, then for all large \( n \) we must have \( \left| {a}_{n}\right| < \) \( 1/2 \) . Disregarding finitely many terms if necessary, we may assume that this inequality holds for all \( n \) . Then we may write the partial products as follows:\n\n\[ \mathop{\prod }\limits_{{n = 1}}^{N}{A}_{n} = \mathop{\prod }\limits_{{n = 1}}^{N}{e}^{\log \left( {1 + {a}_{n}}\right) } = {e}^{{B}_{N}} \]\n\nwhere \( {B}_{N} = \mathop{\sum }\limits_{{n = 1}}^{N}{b}_{n} \) with \( {b}_{n} = \log \left( {1 + {a}_{n}}\right) \) . By the lemma, we know that \( \left| {b}_{n}\right| \leq 2\left| {a}_{n}\right| \), so that \( {B}_{N} \) converges to a real number, say \( B \) . Since the exponential function is continuous, we conclude that \( {e}^{{B}_{N}} \) converges to \( {e}^{B} \) as \( N \) goes to infinity, proving the first assertion of the proposition. Observe also that if \( 1 + {a}_{n} \neq 0 \) for all \( n \), the product converges to a non-zero limit since it is expressed as \( {e}^{B} \) .\n\nFinally observe that the partial products of \( \mathop{\prod }\limits_{n}1/\left( {1 - {a}_{n}}\right) \) are \( 1/\mathop{\prod }\limits_{{n = 1}}^{N}\left( {1 - {a}_{n}}\right) \), so the same argument as above proves that the product in the denominator converges to a non-zero limit.
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Yes
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Theorem 1.10 For every \( s > 1 \), we have\n\n\[ \zeta \left( s\right) = \mathop{\prod }\limits_{p}\frac{1}{1 - 1/{p}^{s}} \]\n\nwhere the product is taken over all primes.
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Proof. Suppose \( M \) and \( N \) are positive integers with \( M > N \). Observe now that any positive integer \( n \leq N \) can be written uniquely as a product of primes, and that each prime must be less than or equal to \( N \) and repeated less than \( M \) times. Therefore\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{N}\frac{1}{{n}^{s}} \leq \mathop{\prod }\limits_{{p \leq N}}\left( {1 + \frac{1}{{p}^{s}} + \frac{1}{{p}^{2s}} + \cdots + \frac{1}{{p}^{Ms}}}\right) \]\n\n\[ \leq \mathop{\prod }\limits_{{p \leq N}}\left( \frac{1}{1 - {p}^{-s}}\right) \]\n\n\[ \leq \mathop{\prod }\limits_{p}\left( \frac{1}{1 - {p}^{-s}}\right) \]\n\nLetting \( N \) tend to infinity now yields\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{{n}^{s}} \leq \mathop{\prod }\limits_{p}\left( \frac{1}{1 - {p}^{-s}}\right) \]\n\nFor the reverse inequality, we argue as follows. Again, by the fundamental theorem of arithmetic, we find that\n\n\[ \mathop{\prod }\limits_{{p \leq N}}\left( {1 + \frac{1}{{p}^{s}} + \frac{1}{{p}^{2s}} + \cdots + \frac{1}{{p}^{Ms}}}\right) \leq \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{{n}^{s}}. \]\n\nLetting \( M \) tend to infinity gives\n\n\[ \mathop{\prod }\limits_{{p \leq N}}\left( \frac{1}{1 - {p}^{-s}}\right) \leq \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{{n}^{s}} \]\n\nHence\n\n\[ \mathop{\prod }\limits_{p}\left( \frac{1}{1 - {p}^{-s}}\right) \leq \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{{n}^{s}} \]\n\nand the proof of the product formula is complete.
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Yes
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Proposition 1.11 The series\n\n\\[ \n\\mathop{\\sum }\\limits_{p}1/p \n\\]\n\n diverges, when the sum is taken over all primes p.
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Proof. We take logarithms of both sides of the Euler formula. Since \\( \\log x \\) is continuous, we may write the logarithm of the infinite product as the sum of the logarithms. Therefore, we obtain for \\( s > 1 \\)\n\n\\[ \n- \\mathop{\\sum }\\limits_{p}\\log \\left( {1 - 1/{p}^{s}}\\right) = \\log \\zeta \\left( s\\right) \n\\]\n\nSince \\( \\log \\left( {1 + x}\\right) = x + O\\left( {\\left| x\\right| }^{2}\\right) \\) whenever \\( \\left| x\\right| \\leq 1/2 \\), we get\n\n\\[ \n- \\mathop{\\sum }\\limits_{p}\\left\\lbrack {-1/{p}^{s} + O\\left( {1/{p}^{2s}}\\right) }\\right\\rbrack = \\log \\zeta \\left( s\\right) \n\\]\n\nwhich gives\n\n\\[ \n\\mathop{\\sum }\\limits_{p}1/{p}^{s} + O\\left( 1\\right) = \\log \\zeta \\left( s\\right) \n\\]\n\nThe term \\( O\\left( 1\\right) \\) appears because \\( \\mathop{\\sum }\\limits_{p}1/{p}^{2s} \\leq \\mathop{\\sum }\\limits_{{n = 1}}^{\\infty }1/{n}^{2} \\) . Now we let \\( s \\) tend to 1 from above, namely \\( s \\rightarrow {1}^{ + } \\), and note that \\( \\zeta \\left( s\\right) \\rightarrow \\infty \\) since \\( \\mathop{\\sum }\\limits_{{n = 1}}^{\\infty }1/{n}^{s} \\geq \\mathop{\\sum }\\limits_{{n = 1}}^{M}1/{n}^{s} \\), and therefore\n\n\\[ \n\\mathop{\\liminf }\\limits_{{s \\rightarrow {1}^{ + }}}\\mathop{\\sum }\\limits_{{n = 1}}^{\\infty }1/{n}^{s} \\geq \\mathop{\\sum }\\limits_{{n = 1}}^{M}1/n\\;\\text{ for every }M. \n\\]\n\nWe conclude that \\( \\mathop{\\sum }\\limits_{p}1/{p}^{s} \\rightarrow \\infty \\) as \\( s \\rightarrow {1}^{ + } \\), and since \\( 1/p > 1/{p}^{s} \\) for all \\( s > 1 \\), we finally have that\n\n\\[ \n\\mathop{\\sum }\\limits_{p}1/p = \\infty \n\\]
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Yes
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Lemma 2.2 The Dirichlet characters are multiplicative. Moreover,
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\[ {\delta }_{\ell }\left( m\right) = \frac{1}{\varphi \left( q\right) }\mathop{\sum }\limits_{\chi }\overline{\chi \left( \ell \right) }\chi \left( m\right) \] where the sum is over all Dirichlet characters. With the above lemma we have taken our first step towards a proof of the theorem, since this lemma shows that \[ \mathop{\sum }\limits_{{p \equiv \ell }}\frac{1}{{p}^{s}} = \mathop{\sum }\limits_{p}\frac{{\delta }_{\ell }\left( p\right) }{{p}^{s}} \] \[ = \frac{1}{\varphi \left( q\right) }\mathop{\sum }\limits_{\chi }\overline{\chi \left( \ell \right) }\mathop{\sum }\limits_{p}\frac{\chi \left( p\right) }{{p}^{s}}. \] Thus it suffices to understand the behavior of \( \mathop{\sum }\limits_{p}\chi \left( p\right) {p}^{-s} \) as \( s \rightarrow {1}^{ + } \) . In fact, we divide the above sum in two parts depending on whether or not \( \chi \) is trivial. So we have \[ \mathop{\sum }\limits_{{p \equiv \ell }}\frac{1}{{p}^{s}} = \frac{1}{\varphi \left( q\right) }\mathop{\sum }\limits_{p}\frac{{\chi }_{0}\left( p\right) }{{p}^{s}} + \frac{1}{\varphi \left( q\right) }\mathop{\sum }\limits_{{\chi \neq {\chi }_{0}}}\overline{\chi \left( \ell \right) }\mathop{\sum }\limits_{p}\frac{\chi \left( p\right) }{{p}^{s}} \] (4) \[ = \frac{1}{\varphi \left( q\right) }\mathop{\sum }\limits_{{p\text{ not dividing }q}}\frac{1}{{p}^{s}} + \frac{1}{\varphi \left( q\right) }\mathop{\sum }\limits_{{\chi \neq {\chi }_{0}}}\overline{\chi \left( \ell \right) }\mathop{\sum }\limits_{p}\frac{\chi \left( p\right) }{{p}^{s}}. \] Since there are only finitely many primes dividing \( q \), Euler’s theorem (Proposition 1.11) implies that the first sum on the right-hand side diverges when \( s \) tends to 1 . These observations show that Dirichlet’s theorem is a consequence of the following assertion.
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Yes
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Theorem 2.3 If \( \chi \) is a nontrivial Dirichlet character, then the sum\n\n\[ \mathop{\sum }\limits_{p}\frac{\chi \left( p\right) }{{p}^{s}} \]\n\nremains bounded as \( s \rightarrow {1}^{ + } \) .
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The proof of Theorem 2.3 requires the introduction of the \( L \) -functions, to which we now turn.
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No
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Proposition 3.1 The logarithm function \( {\log }_{1} \) satisfies the following properties:\n\n(i) If \( \left| z\right| < 1 \), then\n\n\[ \n{e}^{{\log }_{1}\left( \frac{1}{1 - z}\right) } = \frac{1}{1 - z}.\n\]\n\n(ii) If \( \left| z\right| < 1 \), then\n\n\[ \n{\log }_{1}\left( \frac{1}{1 - z}\right) = z + {E}_{1}\left( z\right)\n\]\n\nwhere the error \( {E}_{1} \) satisfies \( \left| {{E}_{1}\left( z\right) }\right| \leq {\left| z\right| }^{2} \) if \( \left| z\right| < 1/2 \) .\n\n(iii) If \( \left| z\right| < 1/2 \), then\n\n\[ \n\left| {{\log }_{1}\left( \frac{1}{1 - z}\right) }\right| \leq 2\left| z\right|\n\]
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Proof. To establish the first property, let \( z = r{e}^{i\theta } \) with \( 0 \leq r < 1 \) , and observe that it suffices to show that\n\n(5)\n\n\[ \n\left( {1 - r{e}^{i\theta }}\right) {e}^{\mathop{\sum }\limits_{{k = 1}}^{\infty }{\left( r{e}^{i\theta }\right) }^{k}/k} = 1.\n\]\n\nTo do so, we differentiate the left-hand side with respect to \( r \), and this gives\n\n\[ \n\left\lbrack {-{e}^{i\theta } + \left( {1 - r{e}^{i\theta }}\right) {\left( \mathop{\sum }\limits_{{k = 1}}^{\infty }{\left( r{e}^{i\theta }\right) }^{k}/k\right) }^{\prime }}\right\rbrack {e}^{\mathop{\sum }\limits_{{k = 1}}^{\infty }{\left( r{e}^{i\theta }\right) }^{k}/k}.\n\]\n\nThe term in brackets equals\n\n\[ \n- {e}^{i\theta } + \left( {1 - r{e}^{i\theta }}\right) {e}^{i\theta }\left( {\mathop{\sum }\limits_{{k = 1}}^{\infty }{\left( r{e}^{i\theta }\right) }^{k - 1}}\right) = - {e}^{i\theta } + \left( {1 - r{e}^{i\theta }}\right) {e}^{i\theta }\frac{1}{1 - r{e}^{i\theta }} = 0.\n\]\n\nHaving found that the left-hand side of the equation (5) is constant, we set \( r = 0 \) and get the desired result.\n\nThe proofs of the second and third properties are the same as their real counterparts given in Lemma 1.8.
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Yes
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Proposition 3.2 If \( \sum \left| {a}_{n}\right| \) converges, and \( {a}_{n} \neq 1 \) for all \( n \), then\n\n\[ \mathop{\prod }\limits_{{n = 1}}^{\infty }\left( \frac{1}{1 - {a}_{n}}\right) \]\n\nconverges. Moreover, this product is non-zero.
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Proof. For \( n \) large enough, \( \left| {a}_{n}\right| < 1/2 \), so we may assume without loss of generality that this inequality holds for all \( n \geq 1 \) . Then\n\n\[ \mathop{\prod }\limits_{{n = 1}}^{N}\left( \frac{1}{1 - {a}_{n}}\right) = \mathop{\prod }\limits_{{n = 1}}^{N}{e}^{{\log }_{1}\left( \frac{1}{1 - {a}_{n}}\right) } = {e}^{\mathop{\sum }\limits_{{n = 1}}^{N}{\log }_{1}\left( \frac{1}{1 - {a}_{n}}\right) }.\]\n\nBut we know from the previous proposition that\n\n\[ \left| {{\log }_{1}\left( \frac{1}{1 - z}\right) }\right| \leq 2\left| z\right| \]\n\nso the fact that the series \( \sum \left| {a}_{n}\right| \) converges, immediately implies that the limit\n\n\[ \mathop{\lim }\limits_{{N \rightarrow \infty }}\mathop{\sum }\limits_{{n = 1}}^{N}{\log }_{1}\left( \frac{1}{1 - {a}_{n}}\right) = A \]\n\nexists. Since the exponential function is continuous, we conclude that the product converges to \( {e}^{A} \), which is clearly non-zero.
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Yes
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Proposition 3.3 Suppose \( {\chi }_{0} \) is the trivial Dirichlet character,\n\n\[ \n{\chi }_{0}\left( n\right) = \left\{ \begin{array}{ll} 1 & \text{ if }n\text{ and }q\text{ are relatively prime,} \\ 0 & \text{ otherwise,} \end{array}\right.\n\]\n\nand \( q = {p}_{1}^{{a}_{1}}\cdots {p}_{N}^{{a}_{N}} \) is the prime factorization of \( q \) . Then\n\n\[ \nL\left( {s,{\chi }_{0}}\right) = \left( {1 - {p}_{1}^{-s}}\right) \left( {1 - {p}_{2}^{-s}}\right) \cdots \left( {1 - {p}_{N}^{-s}}\right) \zeta \left( s\right) .\n\]\n\nTherefore \( L\left( {s,{\chi }_{0}}\right) \rightarrow \infty \) as \( s \rightarrow {1}^{ + } \) .
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Proof. The identity follows at once on comparing the Dirichlet and Euler product formulas. The final statement holds because \( \zeta \left( s\right) \rightarrow \infty \) as \( s \rightarrow {1}^{ + } \)
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Yes
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Lemma 3.5 If \( \chi \) is a non-trivial Dirichlet character, then\n\n\[ \left| {\mathop{\sum }\limits_{{n = 1}}^{k}\chi \left( n\right) }\right| \leq q,\;\text{ for any }k. \]
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Proof. First, we recall that\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{q}\chi \left( n\right) = 0 \]\n\nIn fact, if \( S \) denotes the sum and \( a \in {\mathbb{Z}}^{ * }\left( q\right) \), then the multiplicative property of the Dirichlet character \( \chi \) gives\n\n\[ \chi \left( a\right) S = \sum \chi \left( a\right) \chi \left( n\right) = \sum \chi \left( {an}\right) = \sum \chi \left( n\right) = S. \]\n\nSince \( \chi \) is non-trivial, \( \chi \left( a\right) \neq 1 \) for some \( a \), hence \( S = 0 \) . We now write \( k = {aq} + b \) with \( 0 \leq b < q \), and note that\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{k}\chi \left( n\right) = \mathop{\sum }\limits_{{n = 1}}^{{aq}}\chi \left( n\right) + \mathop{\sum }\limits_{{{aq} < n \leq {aq} + b}}\chi \left( n\right) = \mathop{\sum }\limits_{{{aq} < n \leq {aq} + b}}\chi \left( n\right) ,\]\n\nand there are no more than \( q \) terms in the last sum. The proof is complete once we recall that \( \left| {\chi \left( n\right) }\right| \leq 1 \) .
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Yes
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Proposition 3.6 If \( s > 1 \), then\n\n\[ \n{e}^{{\log }_{2}L\left( {s,\chi }\right) } = L\left( {s,\chi }\right) \n\]\n\nMoreover\n\n\[ \n{\log }_{2}L\left( {s,\chi }\right) = \mathop{\sum }\limits_{p}{\log }_{1}\left( \frac{1}{1 - \chi \left( p\right) /{p}^{s}}\right) .\n\]
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Proof. Differentiating \( {e}^{-{\log }_{2}L\left( {s,\chi }\right) }L\left( {s,\chi }\right) \) with respect to \( s \) gives\n\n\[ \n- \frac{{L}^{\prime }\left( {s,\chi }\right) }{L\left( {s,\chi }\right) }{e}^{-{\log }_{2}L\left( {s,\chi }\right) }L\left( {s,\chi }\right) + {e}^{-{\log }_{2}L\left( {s,\chi }\right) }{L}^{\prime }\left( {s,\chi }\right) = 0.\n\]\n\nSo \( {e}^{-{\log }_{2}L\left( {s,\chi }\right) }L\left( {s,\chi }\right) \) is constant, and this constant can be seen to be 1 by letting \( s \) tend to infinity. This proves the first conclusion.\n\nTo prove the equality between the logarithms, we fix \( s \) and take the exponential of both sides. The left-hand side becomes \( {e}^{{\log }_{2}L\left( {s,\chi }\right) } = L\left( {s,\chi }\right) \), and the right-hand side becomes\n\n\[ \n{e}^{\mathop{\sum }\limits_{p}{\log }_{1}\left( \frac{1}{1 - \chi \left( p\right) /{p}^{s}}\right) } = \mathop{\prod }\limits_{p}{e}^{{\log }_{1}\left( \frac{1}{1 - \chi \left( p\right) /{p}^{s}}\right) } = \mathop{\prod }\limits_{p}\left( \frac{1}{1 - \chi \left( p\right) /{p}^{s}}\right) = L\left( {s,\chi }\right) ,\n\]\n\nby (i) in Proposition 3.1 and the Dirichlet product formula. Therefore, for each \( s \) there exists an integer \( M\left( s\right) \) so that\n\n\[ \n{\log }_{2}L\left( {s,\chi }\right) - \mathop{\sum }\limits_{p}{\log }_{1}\left( \frac{1}{1 - \chi \left( p\right) /{p}^{s}}\right) = {2\pi iM}\left( s\right) .\n\]\n\nAs the reader may verify, the left-hand side is continuous in \( s \), and this implies the continuity of the function \( M\left( s\right) \). But \( M\left( s\right) \) is integer-valued so we conclude that \( M\left( s\right) \) is constant, and this constant can be seen to be 0 by letting \( s \) go to infinity.
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Yes
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Lemma 3.8 If \( s > 1 \), then\n\n\[ \mathop{\prod }\limits_{\chi }L\left( {s,\chi }\right) \geq 1 \]\n\nwhere the product is taken over all Dirichlet characters. In particular the product is real-valued.
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Proof. We have shown earlier that for \( s > 1 \)\n\n\[ L\left( {s,\chi }\right) = \exp \left( {\mathop{\sum }\limits_{p}{\log }_{1}\left( \frac{1}{1 - \chi \left( p\right) {p}^{-s}}\right) }\right) .\n\nHence,\n\n\[ \mathop{\prod }\limits_{\chi }L\left( {s,\chi }\right) = \exp \left( {\mathop{\sum }\limits_{\chi }\mathop{\sum }\limits_{p}{\log }_{1}\left( \frac{1}{1 - \chi \left( p\right) {p}^{-s}}\right) }\right)\n\n\[ = \exp \left( {\mathop{\sum }\limits_{\chi }\mathop{\sum }\limits_{p}\mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{1}{k}\frac{\chi \left( {p}^{k}\right) }{{p}^{ks}}}\right)\n\n\[ = \exp \left( {\mathop{\sum }\limits_{p}\mathop{\sum }\limits_{{k = 1}}^{\infty }\mathop{\sum }\limits_{\chi }\frac{1}{k}\frac{\chi \left( {p}^{k}\right) }{{p}^{ks}}}\right) .\n\nBecause of Lemma 2.2 (with \( \ell = 1 \) ) we have \( \mathop{\sum }\limits_{\chi }\chi \left( {p}^{k}\right) = \varphi \left( q\right) {\delta }_{1}\left( {p}^{k}\right) \), and hence\n\n\[ \mathop{\prod }\limits_{\chi }L\left( {s,\chi }\right) = \exp \left( {\varphi \left( q\right) \mathop{\sum }\limits_{p}\mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{1}{k}\frac{{\delta }_{1}\left( {p}^{k}\right) }{{p}^{ks}}}\right) \geq 1,\n\nsince the term in the exponential is non-negative.
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Yes
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Proposition 3.10 If \( N \) is a positive integer, then:\n\n(i) \( \mathop{\sum }\limits_{{1 \leq n \leq N}}\frac{1}{n} = {\int }_{1}^{N}\frac{dx}{x} + O\left( 1\right) = \log N + O\left( 1\right) \).\n\n(ii) More precisely, there exists a real number \( \gamma \), called Euler’s constant, so that\n\n\[ \mathop{\sum }\limits_{{1 \leq n \leq N}}\frac{1}{n} = \log N + \gamma + O\left( {1/N}\right) \]
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Proof. It suffices to establish the more refined estimate given in part (ii). Let\n\n\[ {\gamma }_{n} = \frac{1}{n} - {\int }_{n}^{n + 1}\frac{dx}{x} \]\n\nSince \( 1/x \) is decreasing, we clearly have\n\n\[ 0 \leq {\gamma }_{n} \leq \frac{1}{n} - \frac{1}{n + 1} \leq \frac{1}{{n}^{2}} \]\n\nso the series \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{\gamma }_{n} \) converges to a limit which we denote by \( \gamma \) . Moreover, if we estimate \( \sum f\left( n\right) \) by \( \int f\left( x\right) {dx} \), where \( f\left( x\right) = 1/{x}^{2} \), we find\n\n\[ \mathop{\sum }\limits_{{n = N + 1}}^{\infty }{\gamma }_{n} \leq \mathop{\sum }\limits_{{n = N + 1}}^{\infty }\frac{1}{{n}^{2}} \leq {\int }_{N}^{\infty }\frac{dx}{{x}^{2}} = O\left( {1/N}\right) .\n\nTherefore\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{N}\frac{1}{n} - {\int }_{1}^{N}\frac{dx}{x} = \gamma - \mathop{\sum }\limits_{{n = N + 1}}^{\infty }{\gamma }_{n} + {\int }_{N}^{N + 1}\frac{dx}{x}, \]\n\nand this last integral is \( O\left( {1/N}\right) \) as \( N \rightarrow \infty \) .
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Yes
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Proposition 3.11 If \( N \) is a positive integer, then\n\n\[ \mathop{\sum }\limits_{{1 \leq n \leq N}}\frac{1}{{n}^{1/2}} = {\int }_{1}^{N}\frac{dx}{{x}^{1/2}} + {c}^{\prime } + O\left( {1/{N}^{1/2}}\right) \]\n\n\[ = 2{N}^{1/2} + c + O\left( {1/{N}^{1/2}}\right) \text{.} \]
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The proof is essentially a repetition of the proof of the previous proposition, this time using the fact that\n\n\[ \left| {\frac{1}{{n}^{1/2}} - \frac{1}{{\left( n + 1\right) }^{1/2}}}\right| \leq \frac{C}{{n}^{3/2}} \]\n\nThis last inequality follows from the mean-value theorem applied to \( f\left( x\right) = {x}^{-1/2} \), between \( x = n \) and \( x = n + 1 \) .
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Yes
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Theorem 3.12 If \( k \) is a positive integer, then\n\n\[ \frac{1}{N}\mathop{\sum }\limits_{{k = 1}}^{N}d\left( k\right) = \log N + O\left( 1\right) \]\n\nMore precisely,\n\n\[ \frac{1}{N}\mathop{\sum }\limits_{{k = 1}}^{N}d\left( k\right) = \log N + \left( {{2\gamma } - 1}\right) + O\left( {1/{N}^{1/2}}\right) ,\]\n\nwhere \( \gamma \) is Euler’s constant.
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Proof. Let \( {S}_{N} = \mathop{\sum }\limits_{{k = 1}}^{N}d\left( k\right) \) . We observed that summing \( F = 1 \) along hyperbolas gives \( {S}_{N} \) . Summing vertically, we find\n\n\[ {S}_{N} = \mathop{\sum }\limits_{{1 \leq m \leq N}}\mathop{\sum }\limits_{{1 \leq n \leq N/m}}1 \]\n\nBut \( \mathop{\sum }\limits_{{1 \leq n \leq N/m}}1 = \left\lbrack {N/m}\right\rbrack = N/m + O\left( 1\right) \), where \( \left\lbrack x\right\rbrack \) denote the greatest integer \( \leq x \) . Therefore\n\n\[ {S}_{N} = \mathop{\sum }\limits_{{1 \leq m \leq N}}\left( {N/m + O\left( 1\right) }\right) = N\left( {\mathop{\sum }\limits_{{1 \leq m \leq N}}1/m}\right) + O\left( N\right) . \]\n\nHence, by part (i) of Proposition 3.10,\n\n\[ \frac{{S}_{N}}{N} = \log N + O\left( 1\right) \]\n\nwhich gives the first conclusion.
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Yes
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Proposition 3.13 The following statements are true:\n\n(i) \( {S}_{N} \geq c\log N \) for some constant \( c > 0 \) .\n\n(ii) \( {S}_{N} = 2{N}^{1/2}L\left( {1,\chi }\right) + O\left( 1\right) \) .
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It suffices to prove the proposition, since the assumption \( L\left( {1,\chi }\right) = 0 \) would give an immediate contradiction.\n\nWe first sum along hyperbolas. Observe that\n\n\[ \mathop{\sum }\limits_{{{nm} = k}}\frac{\chi \left( n\right) }{{\left( nm\right) }^{1/2}} = \frac{1}{{k}^{1/2}}\mathop{\sum }\limits_{{n \mid k}}\chi \left( n\right) . \]\n\nFor conclusion (i) it will be enough to show the following lemma.\n\nLemma 3.14 \( \mathop{\sum }\limits_{{n \mid k}}\chi \left( n\right) \geq \left\{ \begin{array}{ll} 0 & \text{ for all }k \\ 1 & \text{ if }k = {\ell }^{2}\text{ for some }\ell \in \mathbb{Z}. \end{array}\right. \)\n\nFrom the lemma, we then get\n\n\[ {S}_{N} \geq \mathop{\sum }\limits_{{k = {\ell }^{2},\ell \leq {N}^{1/2}}}\frac{1}{{k}^{1/2}} \geq c\log N \]\n\nwhere the last inequality follows from (i) in Proposition 3.10.\n\nThe proof of the lemma is simple. If \( k \) is a power of a prime, say \( k = {p}^{a} \), then the divisors of \( k \) are \( 1, p,{p}^{2},\ldots ,{p}^{a} \) and\n\n\[ \mathop{\sum }\limits_{{n \mid k}}\chi \left( n\right) = \chi \left( 1\right) + \chi \left( p\right) + \chi \left( {p}^{2}\right) + \cdots + \chi \left( {p}^{a}\right) \]\n\n\[ = 1 + \chi \left( p\right) + \chi {\left( p\right) }^{2} + \cdots + \chi {\left( p\right) }^{a}. \]\n\nSo this sum is equal to\n\n\[ \left\{ \begin{matrix} a + 1 & \text{ if }\chi \left( p\right) = 1, \\ 1 & \text{ if }\chi \left( p\right) = - 1\text{ and }a\text{ is even,} \\ 0 & \text{ if }\chi \left( p\right) = - 1\text{ and }a\text{ is odd,} \\ 1 & \text{ if }\chi \left( p\right) = 0,\text{ that is }p \mid q. \end{matrix}\right. \]\n\nIn general, if \( k = {p}_{1}^{{a}_{1}}\cdots {p}_{N}^{{a}_{N}} \), then any divisor of \( k \) is of the form \( {p}_{1}^{{b}_{1}}\cdots {p}_{N}^{{b}_{N}} \) where \( 0 \leq {b}_{j} \leq {a}_{j} \) for all \( j \) . Therefore, the multiplicative property of \( \chi \) gives\n\n\[ \mathop{\sum }\limits_{{n \mid k}}\chi \left( n\right) = \mathop{\prod }\limits_{{j = 1}}^{N}\left( {\chi \left( 1\right) + \chi \left( {p}_{j}\right) + \chi \left( {p}_{j}^{2}\right) + \cdots + \chi \left( {p}_{j}^{{a}_{j}}\right) }\right) ,\]\n\nand the proof is complete.
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Yes
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Lemma 3.14 \( \mathop{\sum }\limits_{{n \mid k}}\chi \left( n\right) \geq \left\{ \begin{array}{ll} 0 & \text{ for all }k \\ 1 & \text{ if }k = {\ell }^{2}\text{ for some }\ell \in \mathbb{Z}. \end{array}\right. \)
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The proof of the lemma is simple. If \( k \) is a power of a prime, say \( k = {p}^{a} \), then the divisors of \( k \) are \( 1, p,{p}^{2},\ldots ,{p}^{a} \) and\n\n\[ \mathop{\sum }\limits_{{n \mid k}}\chi \left( n\right) = \chi \left( 1\right) + \chi \left( p\right) + \chi \left( {p}^{2}\right) + \cdots + \chi \left( {p}^{a}\right) \]\n\n\[ = 1 + \chi \left( p\right) + \chi {\left( p\right) }^{2} + \cdots + \chi {\left( p\right) }^{a}. \]\n\nSo this sum is equal to\n\n\[ \left\{ \begin{matrix} a + 1 & \text{ if }\chi \left( p\right) = 1, \\ 1 & \text{ if }\chi \left( p\right) = - 1\text{ and }a\text{ is even,} \\ 0 & \text{ if }\chi \left( p\right) = - 1\text{ and }a\text{ is odd,} \\ 1 & \text{ if }\chi \left( p\right) = 0,\text{ that is }p \mid q. \end{matrix}\right. \]\n\nIn general, if \( k = {p}_{1}^{{a}_{1}}\cdots {p}_{N}^{{a}_{N}} \), then any divisor of \( k \) is of the form \( {p}_{1}^{{b}_{1}}\cdots {p}_{N}^{{b}_{N}} \) where \( 0 \leq {b}_{j} \leq {a}_{j} \) for all \( j \) . Therefore, the multiplicative property of \( \chi \) gives\n\n\[ \mathop{\sum }\limits_{{n \mid k}}\chi \left( n\right) = \mathop{\prod }\limits_{{j = 1}}^{N}\left( {\chi \left( 1\right) + \chi \left( {p}_{j}\right) + \chi \left( {p}_{j}^{2}\right) + \cdots + \chi \left( {p}_{j}^{{a}_{j}}\right) }\right) ,\]\n\nand the proof is complete.
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Yes
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