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Prove that a Möbius transformation \( M \) with distinct fixed points \( a \) and \( b \) in \( \mathbb{C} \) either maps each circle in the family \( \mathcal{A} \) of Apollonian circles defined by the point circles \( a \) and \( b \) to itself or it maps no circle in \( \mathcal{A} \) to itself. | Solution By a suitable Möbius transformation \( f \), we can map the point \( a \) to 0 and \( b \) to \( \infty \) . Then \( f \) maps the family \( \mathcal{A} \) of Apollonian circles defined by the point circles \( a \) and \( b \) to the family \( \mathcal{B} \) of concentric circles with centre 0 .\n\nThen \( f \... | Yes |
\( \bigtriangleup {ABC} \) are line segments on \( {S}^{2} \), that is, parts of great circles; and the sum of its angles exceeds \( \pi \) . The sides of \( \bigtriangleup {PQR} \) are line segments in the plane, and the sum of its angles is \( \pi \) . So we cannot simultaneously have geodesics on \( {S}^{2} \) mappi... | To see this, let \( E \) denote the surface of the Earth, a sphere with centre \( O \) and radius \( R \) ; and let \( {P}^{\prime } \in E \) be the centre of a circle \( {C}^{\prime } \) on \( E \) of radius \( r \) (we will suppose that \( r < \frac{1}{2}{\pi R} \), so that \( {C}^{\prime } \) lies in a hemisphere on... | Yes |
In affine geometry, three distinct types of (non-degenerate) conic arise because there are three ways in which a projective conic can meet the ideal Line. In the first figure below, the projective conic and the Line cross at two Points; the corresponding picture in \( {\mathbb{R}}^{2} \) is that of a hyperbola. In the ... | So hyperbolas, parabolas and ellipses are plane conics which correspond to projective conics with two, one and no ideal Points missing, respectively. Since affine transformations correspond to projective transformations which map ideal Points to ideal Points, they cannot change the number of ideal Points on a projectiv... | Yes |
Theorem 7\n\n--- \n\nconic \( C \) . Then \( t\left( \ell \right) \) is a tangent to the conic \( t\left( C\right) \) . | Notice that in this proof of the theorem we did not have to consider tangents whose Point of contact is an ideal Point. But to what does such a tangent correspond in \( {\mathbb{R}}^{2} \) ? The answer depends on how many ideal Points lie on the projective conic.\n\nA projective conic which passes through one ideal Poi... | No |
Theorem 1 A projective reflection is a projective transformation. | Proof Consider the projective reflection \( r \) in the Line \( \ell \) with centre \( F \) and parameter \( k \), and let \( Q = r\left( P\right) \) where \( P \) is a Point of \( {\mathbb{{RP}}}^{2} \) that does not belong to \( \ell \cup \{ F\} \) . Then \( \left( {FRPQ}\right) = k \), where \( R \) is the Point of ... | Yes |
Find the image under \( L \) of the point \( P\left( {r,0}\right) \) in \( \mathcal{D} \) . | Notice first that it is sufficient to consider the image under \( L \) of the \( x \) -axis and the \( d \) -line through \( P \) at right angles to the \( x \) -axis.\n\nThe transformation \( L \) maps the \( x \) -axis to itself (as a line, not pointwise). The \( d \) -line through \( P \) perpendicular to the \( x \... | Yes |
Theorem 1 Let \( t \) be the hyperbolic reflection in the \( d \) -line \( d \) joining the boundary points \( A \) and \( B \) of \( \mathcal{D} \), and let \( F \) be the pole of the (Euclidean) line \( \ell \) through \( A \) and \( B \) . Then \( {Lt}{L}^{-1} \) is the projective reflection \( r \) in \( \ell \) wi... | Proof First we show that \( {Lt}{L}^{-1} \) agrees with \( r \) on \( \mathcal{C} \) . To do this, observe that lines through \( F \) are perpendicular to \( d \), so the hyperbolic reflection \( r \) maps a given point \( C \) of \( \mathcal{C} \) to the other point \( D \) of \( \mathcal{C} \) that lies on \( {FC} \)... | No |
Theorem 1 The geometry on \( {H}_{N} \) is Euclidean geometry, and the group \( {G}_{N} \) acts on \( {H}_{N} \) as Euclidean isometries. | Proof Invert the whole figure in a sphere with centre \( N \) . The point \( N \) is mapped to ’infinity’; \( {S}^{2} \) maps to a plane \( \pi : z = a \), say, and \( {H}_{N} \) maps to a plane parallel to \( \pi \), the plane \( {\pi }^{\prime } : z = b \), say. An element of \( {G}_{N} \) is an inversion in a genera... | Yes |
Theorem 1.1 (Weierstrass’ Theorem for sequences). Suppose that \( {f}_{n}\left( z\right) \) is analytic in a region \( \Omega \) and that the sequence \( \left\{ {{f}_{n}\left( z\right) }\right\} \) converges to a limit function \( f\left( z\right) \) in \( \Omega \), uniformly on every compact subset of \( \Omega \) .... | Proof. The continuity of \( f\left( z\right), z \in \Omega \), is obtained in Section 2.2 in Chapter 2, and so \( {\int }_{\gamma }f\left( z\right) {dz} \) is well defined for any rectifiable arc \( \gamma \subset \Omega \).\n\nBy the assumption for any \( \varepsilon > 0 \) there exists an integer \( N \) such that\n\... | Yes |
Theorem 1.4. If \( f\left( z\right) \) is analytic in the region \( \Omega \), containing \( {z}_{0} \), then the representation\n\n\[ f\left( z\right) = f\left( {z}_{0}\right) + {f}^{\prime }\left( {z}_{0}\right) \left( {z - {z}_{0}}\right) + \frac{{f}^{\prime \prime }\left( {z}_{0}\right) }{2}{\left( z - {z}_{0}\righ... | Proof. Let \( R \) be the largest number so that \( B\left( {{z}_{0}, R}\right) \subset \Omega, z \in B\left( {{z}_{0}, R}\right) \) and let \( \rho \in \left( {\left| {z - {z}_{0}}\right|, R}\right) \) . Then we have\n\n(1.2)\n\n\[ f\left( z\right) = \frac{1}{2\pi i}{\int }_{\left| {\zeta - {z}_{0}}\right| = \rho }\fr... | Yes |
Theorem 1.5. Assume that \( f\left( z\right) \) is analytic in the annulus region \( {R}_{1} < \left| {z - a}\right| < \) \( {R}_{2} \), then \( f \) can be expanded to\n\n(1.5)\n\n\[ f\left( z\right) = \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{a}_{n}{\left( z - a\right) }^{n},{R}_{1} < \left| {z - a}\right| < ... | Proof. We denote by \( A\left( {a,{R}_{1},{R}_{2}}\right) \) the annulus \( {R}_{1} < \left| {z - a}\right| < {R}_{2} \). For any \( z \in \) \( A\left( {a,{R}_{1},{R}_{2}}\right) \) choose \( {r}_{1} > 0 \) and \( {r}_{2} > 0 \) such that\n\n\[ {R}_{1} < {r}_{1} < \left| {z - a}\right| < {r}_{2} < {R}_{2} \]\n\nThen b... | Yes |
The Laurent series of\n\n\\[ f\\left( z\\right) = \\frac{1}{\\left( {z - 1}\\right) \\left( {z - 2}\\right) }.\n\\] | This function can be expanded to be Taylor series in the region \\( \\left| z\\right| < 1 \\), and Laurent series in the regions \\( 1 < \\left| z\\right| < 2 \\) and \\( \\left| {z > 2}\\right| \\) .\n\nWe first write\n\n\\[ f\\left( z\\right) = \\frac{1}{1 - z} - \\frac{1}{2 - z}.\n\\]\n\nIn the region \\( \\left| z\... | Yes |
Lemma 2.2. Let \( f \) be analytic at a. Then a is a zero with order \( h \) of \( f \) iff\n\n\[ f\left( z\right) = {\left( z - a\right) }^{h}g\left( z\right) \]\n\nwhere \( g \) is analytic at \( a \) and \( g\left( a\right) \neq 0 \) . | This follows from the Taylor expansion of \( f \) at \( a \) . If \( a \) is a zero of \( f \) of order \( h \) , then\n\n\[ f\left( z\right) = \frac{{f}^{\left( h\right) }\left( a\right) }{h!}{\left( z - a\right) }^{h} + \frac{{f}^{\left( h + 1\right) }\left( a\right) }{\left( {h + 1}\right) !}{\left( z - a\right) }^{... | Yes |
Theorem 2.4. If \( f\left( z\right) \) and \( g\left( z\right) \) are analytic in \( \Omega \), and if \( f\left( z\right) = g\left( z\right) \) on a set which has an accumulation point in \( \Omega \), then \( f\left( z\right) \) is identically equal to \( g\left( z\right) \) . | The conclusion follows by consideration of the difference \( f\left( z\right) - g\left( z\right) \) . | No |
Theorem 2.5. Let a be an isolated singularity of \( f \) . Then the following conditions are equivalent each other:\n\n(i) a is removable;\n\n(ii) \( \mathop{\lim }\limits_{{z \rightarrow a}}f\left( z\right) \) exists;\n\n(iii) there exists a \( {\delta }_{0} > 0 \) such that \( f \) is bounded in \( {B}^{ * }\left( {a... | Assume \( f \) is analytic in \( {B}^{ * }\left( {a,\delta }\right) \) . It suffices to prove (iv) \( \Rightarrow \) (i). By Laurent’s theorem,\n\n\[ f\left( z\right) = \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{c}_{n}{\left( z - a\right) }^{n},0 < \left| {z - a}\right| < \delta . \]\n\nWe show that \( {c}_{-m} ... | Yes |
Theorem 2.6. Let a be an isolated singularity of \( f \) . Then the following are equivalent:\n\n(i) \( a \) is a pole, say \( \mathop{\lim }\limits_{{z \rightarrow a}}f\left( z\right) = \infty \) .\n\n(ii) \( a \) is a zero of \( g\left( z\right) = 1/f\left( z\right) \) .\n\n(iii) there exists a positive integer \( h ... | If \( a \) is a pole, then \( g\left( z\right) = 1/f\left( z\right) \rightarrow 0 \) as \( z \rightarrow a \) and thus \( a \) is a removable singularity of \( g \) . Thus \( g\left( z\right) \) become analytic at \( a \) if we define \( g\left( a\right) = 0 \), which is what (ii) means. But \( g\left( z\right) \) can ... | Yes |
Theorem 2.7. Let a be an isolated singularity of an analytic function \( f \) . Then\n\n(i) a is removable iff the Laurent development of \( f \) at a has no term of negative power.\n\n(ii) a is a pole iff the Laurent development of \( f \) at a has, but only a finite number of, terms of negative powers.\n\n(iii) a is ... | (i) is in fact proved in the proof of Theorem 2.5. To prove (ii), we first assume \( a \) is a pole of order \( h \) . Then in a neighborhood of \( a \) we have\n\n\[ f\left( z\right) = \frac{g\left( z\right) }{{\left( z - a\right) }^{h}} \]\n\nwhere \( g \) is analytic at \( a \) with \( g\left( a\right) \neq 0 \) . T... | Yes |
Theorem 2.8. An analytic function comes arbitrarily close to any complex value in every neighborhood of an essential singularity. | If the assertion were not true, we could find a complex number \( A \) and an \( {\varepsilon }_{0} > 0 \) such that \( \left| {f\left( z\right) - A}\right| > {\varepsilon }_{0} \) in a punctured neighborhood \( {B}^{ * }\left( {a,\delta }\right) \) of \( a \) (except for \( z = a) \) . Then \( g\left( z\right) = \frac... | Yes |
Theorem 3.1 (Cauchy’s Residue Theorem). Let \( \Omega \) be a region enclosed by a finite number of piecewise differentiable Jordan curves and \( {a}_{j} \in \Omega, j = 1,\ldots, n \) . Then for any analytic function defined on \( \bar{\Omega } \smallsetminus \left\{ {a}_{j}\right\} \) , \[ {\int }_{\partial \Omega }f... | The proof is given by applying Cauchy's theorem to the closed region that is obtained by \( \bar{\Omega } \), omitting the sufficiently small disks \( \left| {z - {a}_{j}}\right| < \delta \) . To make this theorem more useful, we introduce some simple methods to compute residues. | No |
Lemma 3.2. The residue of \( f\left( z\right) \) at an isolated singularity \( a \in \mathbb{C} \) is the unique complex number \( R \) which makes \( f\left( z\right) - R/\left( {z - a}\right) \) the derivative of a single-valued analytic function in an annulus \( 0 < \left| {z - a}\right| < \delta \) . | If \( a \) is a removable singularity, then we have \( {c}_{-n} = 0 \) for all \( n \geq 1 \), and so \( {c}_{-1} = 0 \) . In this case the residue, by definition, equals 0 . | No |
Lemma 3.3. Let \( a \in \mathbb{C} \) be a pole of \( f \) of order \( m \) . Then\n\n\[ \n{\operatorname{Res}}_{z = a}f\left( z\right) = \mathop{\lim }\limits_{{z \rightarrow a}}\frac{{d}^{m - 1}\left\lbrack {{\left( z - a\right) }^{m}f\left( z\right) }\right\rbrack }{d{z}^{m - 1}}. \n\] | Proof. The assumption implies that in a neighborhood of \( a, f \) has a representation\n\n\[ \nf\left( z\right) = \frac{g\left( z\right) }{{\left( z - a\right) }^{m}} \n\]\n\nwhere \( g \) is analytic at \( a \) with \( g\left( a\right) \neq 0 \) . Then\n\n\[ \nf\left( z\right) = \frac{g\left( a\right) + {g}^{\prime }... | Yes |
Corollary 3.4. If \( a \in \mathbb{C} \) is a simple pole of \( f \), then | \[ {\operatorname{Res}}_{z = a}f\left( z\right) = \mathop{\lim }\limits_{{z \rightarrow a}}\left( {z - a}\right) f\left( z\right) . \] | Yes |
Theorem 3.6. Let \( \Omega \) be a region bounded by a finite number of piecewise regular Jordan curves. Let \( f\left( z\right) \) be a meromorphic function in \( \bar{\Omega } \) such that all zero and poles are contained in \( \Omega \) . Then\n\n\[ \frac{1}{2\pi i}{\int }_{\partial \Omega }\frac{{f}^{\prime }\left(... | Proof. Let \( {a}_{j} \) and \( {b}_{k}, j = 1,\ldots, m, k = 1,\ldots, n \), be all zeros and poles, where each \( {a}_{j} \) is repeated as many as its orders, and so does \( {b}_{j} \) . Then by the residue theorem, \( \frac{1}{2\pi i}{\int }_{\partial \Omega }\frac{{f}^{\prime }\left( z\right) }{f\left( z\right) }{... | Yes |
Theorem 3.7. If the functions \( {f}_{n}\left( z\right) \) are analytic and \( \neq 0 \) in a region \( \Omega \), and if \( {f}_{n}\left( z\right) \) converges to \( f\left( z\right) \) uniformly on every compact subset of \( \Omega \), then \( f\left( z\right) \) is either identically zero or never equal to zero in \... | By Weierstrass’s theorem, \( f \) is analytic on \( \Omega \) . If \( f \) has a zero \( {z}_{0} \in \Omega \) but is not identically zero, then \( {z}_{0} \) is an isolated zero of \( f \) and there is a \( \delta > 0 \) so that \( f\left( z\right) \) does not vanish on \( \overline{B\left( {{z}_{0},\delta }\right) } ... | Yes |
Corollary 3.9 (Open Mapping Theorem). A nonconstant analytic function maps open sets onto open sets. | This is merely another way of saying that the image of every sufficiently small disk \( B\left( {{z}_{0},\delta }\right) \) contains a neighborhood \( B\left( {{w}_{0},\rho }\right) \) . In the case \( n = 1 \) there is one-to-one correspondence between the disk \( B\left( {{w}_{0},\rho }\right) \) and an open subset o... | No |
Corollary 3.10 (Inverse Function Theorem). If \( f\left( z\right) \) is analytic at \( {z}_{0} \) with \( {f}^{\prime }\left( {z}_{0}\right) \neq 0 \), it maps a neighborhood of \( {z}_{0} \) conformably and topologically onto a region \( D \) ; and the inverse on \( D \) of \( f \) is also conformal. | From the continuity of the inverse function it follows in the usual way that the inverse function is analytic, and hence the inverse mapping is likewise conformal. Conversely, if the local mapping is one to one, Theorem 3.8 can hold only with \( n = 1 \), and hence \( {f}^{\prime }\left( {z}_{0}\right) \) must be diffe... | Yes |
Theorem 3.11. (The maximum principle.) If \( f\left( z\right) \) is analytic and nonconstant in a region \( \Omega \), then its absolute value \( \left| {f\left( z\right) }\right| \) has no maximum in \( \Omega \) . | The proof is clear. If \( {w}_{0} = f\left( {z}_{0}\right) \) is any value taken in \( \Omega \), there exists a neighborhood \( B\left( {{w}_{0},\rho }\right) \) contained in the image of \( \Omega \). In this neighborhood there are points of modulus \( > \left| {w}_{0}\right| \) and hence \( \left| {f\left( {z}_{0}\r... | No |
Theorem 3.12. If \( f\left( z\right) \) is analytic for \( \left| z\right| < 1 \) and satisfies the conditions \( \left| {f\left( z\right) }\right| \leq \) \( 1, f\left( 0\right) = 0 \), then \( \left| {f\left( z\right) }\right| \leq \left| z\right| \) and \( \left| {{f}^{\prime }\left( 0\right) }\right| \leq 1 \) . If... | We apply the maximum principle to the function \( {f}_{1}\left( z\right) \) which is equal to \( f\left( z\right) /z \) for \( z \neq 0 \) and to \( {f}^{\prime }\left( 0\right) \) for \( z = 0 \) . On the circle \( \left| z\right| = r < 1 \) it is of absolute value \( \leq 1/r \), and hence \( \left| {{f}_{1}\left( z\... | Yes |
Theorem 4.2. The arithmetic mean of a harmonic function over concentric circles \( \left| z\right| = r \) is a linear function of \( \log r \) , | \[ \frac{1}{2\pi }{\int }_{0}^{2\pi }u\left( {r{e}^{i\theta }}\right) {d\theta } = \alpha \log r + \beta ,\] and if \( u \) is harmonic in a disk \( \alpha = 0 \) and the arithmetic mean is constant. In the latter case \( \beta = u\left( 0\right) \), by continuity, and changing to a new origin we find \[ \frac{1}{2\pi ... | Yes |
Theorem 4.3. A nonconstant harmonic function has neither a maximum nor a minimum in its region of definition. Consequently, the maximum and the minimum on a closed bounded set \( E \) are taken on the boundary of \( E \) . | The proof is the same as for the maximum principle of analytic functions and will not be repeated. It applies also to the minimum for the reason that \( - u \) is harmonic together with \( u \) . In the case of analytic functions the corresponding procedure would have been to apply the maximum principle to \( 1/f\left(... | No |
Theorem 4.5. The function \( {P}_{U}\left( z\right) \) is harmonic for \( \left| z\right| < 1 \) and\n\n\[ \mathop{\lim }\limits_{{z \rightarrow {e}^{i{\theta }_{0}}}}{P}_{U}\left( z\right) = U\left( {\theta }_{0}\right) \]\n\nprovided that \( U \) is continuous at \( {\theta }_{0} \) . | We have already remarked that \( {P}_{U} \) is harmonic. To study the boundary behavior, let \( {C}_{1} \) and \( {C}_{2} \) be complementary arcs of the unit circle, and denote by \( {U}_{1} \) the function which coincides with \( U \) on \( {C}_{1} \) and vanishes on \( {C}_{2} \), by \( {U}_{2} \) the corresponding ... | Yes |
Theorem 4.6. Let \( {\Omega }^{ + } \) be the part in the upper half plane of a symmetric region \( \Omega \), and let \( \sigma \) be the part of the real axis in \( \Omega \) . Suppose that \( v\left( x\right) \) is continuous in \( {\Omega }^{ + } + \sigma \), harmonic in \( {\Omega }^{ + } \), and zero on \( \sigma... | For the proof we construct the function \( V\left( z\right) \) which is equal to \( v\left( z\right) \) in \( {\Omega }^{ + },0 \) on \( \sigma \), and equal to \( - v\left( z\right) \) in the mirror image of \( {\Omega }^{ + } \) . We have to show that \( V \) is harmonic on \( \sigma \) . For a point \( {x}_{0} \in \... | Yes |
Theorem 1.1 (Weierstrass’ Theorem for sequences). Suppose that \( {f}_{n}\left( z\right) \) is analytic in a region \( \Omega \) and that the sequence \( \left\{ {{f}_{n}\left( z\right) }\right\} \) converges to a limit function \( f\left( z\right) \) in \( \Omega \), uniformly on every compact subset of \( \Omega \) .... | Proof. The continuity of \( f\left( z\right), z \in \Omega \), is obtained in Section 2.2 in Chapter 2, and so \( {\int }_{\gamma }f\left( z\right) {dz} \) is well defined for any rectifiable arc \( \gamma \subset \Omega \) .\n\nBy the assumption for any \( \varepsilon > 0 \) there exists an integer \( N \) such that\n... | Yes |
Theorem 1.4. If \( f\left( z\right) \) is analytic in the region \( \Omega \), containing \( {z}_{0} \), then the representation\n\n\[ f\left( z\right) = f\left( {z}_{0}\right) + {f}^{\prime }\left( {z}_{0}\right) \left( {z - {z}_{0}}\right) + \frac{{f}^{\prime \prime }\left( {z}_{0}\right) }{2}{\left( z - {z}_{0}\righ... | Proof. Let \( R \) be the largest number so that \( B\left( {{z}_{0}, R}\right) \subset \Omega, z \in B\left( {{z}_{0}, R}\right) \) and let \( \rho \in \left( {\left| {z - {z}_{0}}\right|, R}\right) \) . Then we have\n\n(1.2)\n\n\[ f\left( z\right) = \frac{1}{2\pi i}{\int }_{\left| {\zeta - {z}_{0}}\right| = \rho }\fr... | Yes |
Theorem 1.5. Assume that \( f\left( z\right) \) is analytic in the annulus region \( {R}_{1} < \left| {z - a}\right| < \) \( {R}_{2} \), then \( f \) can be expanded to\n\n\[ f\left( z\right) = \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{a}_{n}{\left( z - a\right) }^{n},{R}_{1} < \left| {z - a}\right| < {R}_{2}, ... | Proof. We denote by \( A\left( {a,{R}_{1},{R}_{2}}\right) \) the annulus \( {R}_{1} < \left| {z - a}\right| < {R}_{2} \). For any \( z \in \) \( A\left( {a,{R}_{1},{R}_{2}}\right) \) choose \( {r}_{1} > 0 \) and \( {r}_{2} > 0 \) such that\n\n\[ {R}_{1} < {r}_{1} < \left| {z - a}\right| < {r}_{2} < {R}_{2} \]\n\nThen b... | Yes |
The Laurent series of\n\n\\[ f\\left( z\\right) = \\frac{1}{\\left( {z - 1}\\right) \\left( {z - 2}\\right) }.\n\\] | This function can be expanded to be Taylor series in the region \\( \\left| z\\right| < 1 \\), and Laurent series in the regions \\( 1 < \\left| z\\right| < 2 \\) and \\( \\left| {z > 2}\\right| \\) .\n\nWe first write\n\n\\[ f\\left( z\\right) = \\frac{1}{1 - z} - \\frac{1}{2 - z}.\n\\]\n\nIn the region \\( \\left| z\... | Yes |
Lemma 2.2. Let \( f \) be analytic at a. Then a is a zero with order \( h \) of \( f \) iff\n\n\[ f\left( z\right) = {\left( z - a\right) }^{h}g\left( z\right) \]\n\nwhere \( g \) is analytic at \( a \) and \( g\left( a\right) \neq 0 \) . | This follows from the Taylor expansion of \( f \) at \( a \) . If \( a \) is a zero of \( f \) of order \( h \) , then\n\n\[ f\left( z\right) = \frac{{f}^{\left( h\right) }\left( a\right) }{h!}{\left( z - a\right) }^{h} + \frac{{f}^{\left( h + 1\right) }\left( a\right) }{\left( {h + 1}\right) !}{\left( z - a\right) }^{... | Yes |
Theorem 2.4. If \( f\left( z\right) \) and \( g\left( z\right) \) are analytic in \( \Omega \), and if \( f\left( z\right) = g\left( z\right) \) on a set which has an accumulation point in \( \Omega \), then \( f\left( z\right) \) is identically equal to \( g\left( z\right) \) . | The conclusion follows by consideration of the difference \( f\left( z\right) - g\left( z\right) \) . | No |
Theorem 2.5. Let a be an isolated singularity of \( f \) . Then the following conditions are equivalent each other:\n\n(i) a is removable;\n\n(ii) \( \mathop{\lim }\limits_{{z \rightarrow a}}f\left( z\right) \) exists;\n\n(iii) there exists a \( {\delta }_{0} > 0 \) such that \( f \) is bounded in \( {B}^{ * }\left( {a... | Assume \( f \) is analytic in \( {B}^{ * }\left( {a,\delta }\right) \) . It suffices to prove (iv) \( \Rightarrow \) (i). By Laurent’s theorem,\n\n\[ f\left( z\right) = \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{c}_{n}{\left( z - a\right) }^{n},0 < \left| {z - a}\right| < \delta . \]\n\nWe show that \( {c}_{-m} ... | Yes |
Theorem 2.6. Let a be an isolated singularity of \( f \) . Then the following are equivalent:\n\n(i) \( a \) is a pole, say \( \mathop{\lim }\limits_{{z \rightarrow a}}f\left( z\right) = \infty \) .\n\n(ii) \( a \) is a zero of \( g\left( z\right) = 1/f\left( z\right) \) .\n\n(iii) there exists a positive integer \( h ... | If \( a \) is a pole, then \( g\left( z\right) = 1/f\left( z\right) \rightarrow 0 \) as \( z \rightarrow a \) and thus \( a \) is a removable singularity of \( g \) . Thus \( g\left( z\right) \) become analytic at \( a \) if we define \( g\left( a\right) = 0 \), which is what (ii) means. But \( g\left( z\right) \) can ... | Yes |
Theorem 2.7. Let a be an isolated singularity of an analytic function \( f \) . Then\n\n(i) a is removable iff the Laurent development of \( f \) at a has no term of negative power.\n\n(ii) a is a pole iff the Laurent development of \( f \) at a has, but only a finite number of, terms of negative powers.\n\n(iii) a is ... | (i) is in fact proved in the proof of Theorem 2.5. To prove (ii), we first assume \( a \) is a pole of order \( h \) . Then in a neighborhood of \( a \) we have\n\n\[ f\left( z\right) = \frac{g\left( z\right) }{{\left( z - a\right) }^{h}} \]\n\nwhere \( g \) is analytic at \( a \) with \( g\left( a\right) \neq 0 \) . T... | Yes |
Theorem 2.8. An analytic function comes arbitrarily close to any complex value in every neighborhood of an essential singularity. | If the assertion were not true, we could find a complex number \( A \) and an \( {\varepsilon }_{0} > 0 \) such that \( \left| {f\left( z\right) - A}\right| > {\varepsilon }_{0} \) in a punctured neighborhood \( {B}^{ * }\left( {a,\delta }\right) \) of \( a \) (except for \( z = a) \) . Then \( g\left( z\right) = \frac... | Yes |
Theorem 3.1 (Cauchy’s Residue Theorem). Let \( \Omega \) be a region enclosed by a finite number of piecewise differentiable Jordan curves and \( {a}_{j} \in \Omega, j = 1,\ldots, n \) . Then for any analytic function defined on \( \bar{\Omega } \smallsetminus \left\{ {a}_{j}\right\} \) , \[ {\int }_{\partial \Omega }f... | The proof is given by applying Cauchy's theorem to the closed region that is obtained by \( \bar{\Omega } \), omitting the sufficiently small disks \( \left| {z - {a}_{j}}\right| < \delta \) . | Yes |
Lemma 3.2. The residue of \( f\left( z\right) \) at an isolated singularity \( a \in \mathbb{C} \) is the unique complex number \( R \) which makes \( f\left( z\right) - R/\left( {z - a}\right) \) the derivative of a single-valued analytic function in an annulus \( 0 < \left| {z - a}\right| < \delta \) . | If \( a \) is a removable singularity, then we have \( {c}_{-n} = 0 \) for all \( n \geq 1 \), and so \( {c}_{-1} = 0 \) . In this case the residue, by definition, equals 0 . | No |
Lemma 3.3. Let \( a \in \mathbb{C} \) be a pole of \( f \) of order \( m \) . Then\n\n\[ \n{\operatorname{Res}}_{z = a}f\left( z\right) = \mathop{\lim }\limits_{{z \rightarrow a}}\frac{{d}^{m - 1}\left\lbrack {{\left( z - a\right) }^{m}f\left( z\right) }\right\rbrack }{d{z}^{m - 1}}. \n\] | Proof. The assumption implies that in a neighborhood of \( a, f \) has a representation\n\n\[ \nf\left( z\right) = \frac{g\left( z\right) }{{\left( z - a\right) }^{m}} \n\]\n\nwhere \( g \) is analytic at \( a \) with \( g\left( a\right) \neq 0 \) . Then\n\n\[ \nf\left( z\right) = \frac{g\left( a\right) + {g}^{\prime }... | Yes |
Corollary 3.4. If \( a \in \mathbb{C} \) is a simple pole of \( f \), then | \[ {\operatorname{Res}}_{z = a}f\left( z\right) = \mathop{\lim }\limits_{{z \rightarrow a}}\left( {z - a}\right) f\left( z\right) . \] | Yes |
Theorem 3.6. Let \( \Omega \) be a region bounded by a finite number of piecewise regular Jordan curves. Let \( f\left( z\right) \) be a meromorphic function in \( \bar{\Omega } \) such that all zero and poles are contained in \( \Omega \) . Then\n\n\[ \frac{1}{2\pi i}{\int }_{\partial \Omega }\frac{{f}^{\prime }\left(... | Proof. Let \( {a}_{j} \) and \( {b}_{k}, j = 1,\ldots, m, k = 1,\ldots, n \), be all zeros and poles, where each \( {a}_{j} \) is repeated as many as its orders, and so does \( {b}_{j} \) . Then by the residue theorem, \( \frac{1}{2\pi i}{\int }_{\partial \Omega }\frac{{f}^{\prime }\left( z\right) }{f\left( z\right) }{... | No |
Theorem 3.7. If the functions \( {f}_{n}\left( z\right) \) are analytic and \( \neq 0 \) in a region \( \Omega \), and if \( {f}_{n}\left( z\right) \) converges to \( f\left( z\right) \) uniformly on every compact subset of \( \Omega \), then \( f\left( z\right) \) is either identically zero or never equal to zero in \... | By Weierstrass’s theorem, \( f \) is analytic on \( \Omega \) . If \( f \) has a zero \( {z}_{0} \in \Omega \) but is not identically zero, then \( {z}_{0} \) is an isolated zero of \( f \) and there is a \( \delta > 0 \) so that \( f\left( z\right) \) does not vanish on \( \overline{B\left( {{z}_{0},\delta }\right) } ... | Yes |
Corollary 3.9 (Open Mapping Theorem). A nonconstant analytic function maps open sets onto open sets. | This is merely another way of saying that the image of every sufficiently small disk \( B\left( {{z}_{0},\delta }\right) \) contains a neighborhood \( B\left( {{w}_{0},\rho }\right) \) . In the case \( n = 1 \) there is one-to-one correspondence between the disk \( B\left( {{w}_{0},\rho }\right) \) and an open subset o... | No |
Corollary 3.10 (Inverse Function Theorem). If \( f\left( z\right) \) is analytic at \( {z}_{0} \) with \( {f}^{\prime }\left( {z}_{0}\right) \neq 0 \), it maps a neighborhood of \( {z}_{0} \) conformably and topologically onto a region \( D \) ; and the inverse on \( D \) of \( f \) is also conformal. | From the continuity of the inverse function it follows in the usual way that the inverse function is analytic, and hence the inverse mapping is likewise conformal. Conversely, if the local mapping is one to one, Theorem 3.8 can hold only with \( n = 1 \), and hence \( {f}^{\prime }\left( {z}_{0}\right) \) must be diffe... | Yes |
Theorem 3.11. (The maximum principle.) If \( f\left( z\right) \) is analytic and nonconstant in a region \( \Omega \), then its absolute value \( \left| {f\left( z\right) }\right| \) has no maximum in \( \Omega \) . | The proof is clear. If \( {w}_{0} = f\left( {z}_{0}\right) \) is any value taken in \( \Omega \), there exists a neighborhood \( B\left( {{w}_{0},\rho }\right) \) contained in the image of \( \Omega \). In this neighborhood there are points of modulus \( > \left| {w}_{0}\right| \) and hence \( \left| {f\left( {z}_{0}\r... | Yes |
Theorem 3.12. If \( f\left( z\right) \) is analytic for \( \left| z\right| < 1 \) and satisfies the conditions \( \left| {f\left( z\right) }\right| \leq \) \( 1, f\left( 0\right) = 0 \), then \( \left| {f\left( z\right) }\right| \leq \left| z\right| \) and \( \left| {{f}^{\prime }\left( 0\right) }\right| \leq 1 \) . If... | We apply the maximum principle to the function \( {f}_{1}\left( z\right) \) which is equal to \( f\left( z\right) /z \) for \( z \neq 0 \) and to \( {f}^{\prime }\left( 0\right) \) for \( z = 0 \) . On the circle \( \left| z\right| = r < 1 \) it is of absolute value \( \leq 1/r \), and hence \( \left| {{f}_{1}\left( z\... | Yes |
Theorem 4.2. The arithmetic mean of a harmonic function over concentric circles \( \left| z\right| = r \) is a linear function of \( \log r \) , | \[ \frac{1}{2\pi }{\int }_{0}^{2\pi }u\left( {r{e}^{i\theta }}\right) {d\theta } = \alpha \log r + \beta \] and if \( u \) is harmonic in a disk \( \alpha = 0 \) and the arithmetic mean is constant. In the latter case \( \beta = u\left( 0\right) \), by continuity, and changing to a new origin we find \[ \frac{1}{2\pi }... | Yes |
Theorem 4.3. A nonconstant harmonic function has neither a maximum nor a minimum in its region of definition. Consequently, the maximum and the minimum on a closed bounded set \( E \) are taken on the boundary of \( E \) . | The proof is the same as for the maximum principle of analytic functions and will not be repeated. It applies also to the minimum for the reason that \( - u \) is harmonic together with \( u \) . In the case of analytic functions the corresponding procedure would have been to apply the maximum principle to \( 1/f\left(... | No |
Theorem 4.5. The function \( {P}_{U}\left( z\right) \) is harmonic for \( \left| z\right| < 1 \) and\n\n\[ \mathop{\lim }\limits_{{z \rightarrow {e}^{i{\theta }_{0}}}}{P}_{U}\left( z\right) = U\left( {\theta }_{0}\right) \]\n\nprovided that \( U \) is continuous at \( {\theta }_{0} \) . | We have already remarked that \( {P}_{U} \) is harmonic. To study the boundary behavior, let \( {C}_{1} \) and \( {C}_{2} \) be complementary arcs of the unit circle, and denote by \( {U}_{1} \) the function which coincides with \( U \) on \( {C}_{1} \) and vanishes on \( {C}_{2} \), by \( {U}_{2} \) the corresponding ... | Yes |
Theorem 4.6. Let \( {\Omega }^{ + } \) be the part in the upper half plane of a symmetric region \( \Omega \), and let \( \sigma \) be the part of the real axis in \( \Omega \) . Suppose that \( v\left( x\right) \) is continuous in \( {\Omega }^{ + } + \sigma \), harmonic in \( {\Omega }^{ + } \), and zero on \( \sigma... | For the proof we construct the function \( V\left( z\right) \) which is equal to \( v\left( z\right) \) in \( {\Omega }^{ + },0 \) on \( \sigma \), and equal to \( - v\left( z\right) \) in the mirror image of \( {\Omega }^{ + } \) . We have to show that \( V \) is harmonic on \( \sigma \) . For a point \( {x}_{0} \in \... | Yes |
Theorem 1.1. Let \( \left\{ {b}_{v}\right\} \) be a sequence of distinct \( 1 \) complex numbers with \( \mathop{\lim }\limits_{{v \rightarrow \infty }}{b}_{v} = \) \( \infty \), and let \( {P}_{v}\left( \zeta \right) \) be polynomials without constant term. Then there are functions which are meromorphic in the whole p... | We may suppose that no \( {b}_{v} \) is zero. The function \( {P}_{v}\left( {1/\left( {z - {b}_{v}}\right) }\right) \) is analytic for \( \left| z\right| \leq \left| {b}_{v}\right| \) and can thus be expanded in a Taylor series about the origin. We choose for \( {p}_{v}\left( z\right) \) a partial sum of this series, e... | Yes |
Theorem 2.1. The infinite product \( \prod \left( {1 + {a}_{n}}\right) \) with \( 1 + {a}_{n} \neq 0 \) converges simultaneously with the series \( \sum \log \left( {1 + {a}_{n}}\right) \) whose terms represent the values of the principal branch of the logarithm. | The question of convergence of a product can thus be reduced to the more familiar question concerning the convergence of a series. It can be further reduced by observing that the series (2.3) converges absolutely at the same time as the simpler series \( \mathop{\sum }\limits_{{n = 1}}^{\infty }\left| {a}_{n}\right| \)... | Yes |
Theorem 3.1. There exists an entire function with arbitrarily prescribed zeros \( {a}_{n} \) provided that, in the case of infinitely many zeros, \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = \infty \) . Every entire function with these and no other zeros can be written in the form\n\n(3.3)\n\n\[ f\left( ... | This theorem is due to Weierstrass. | No |
Corollary 3.2. Every function which is meromorphic in the whole plane is the quotient of two entire functions. | In fact, if \( F\left( z\right) \) is meromorphic in the whole plane, we can find an entire function \( g\left( z\right) \) with the poles of \( F\left( z\right) \) for zeros. The product \( F\left( z\right) g\left( z\right) \) is then an entire function \( f\left( z\right) \), and we obtain \( F\left( z\right) = f\lef... | Yes |
Problem 5.2. Show directly that the function \( J\left( z\right) \) defined by 5.5 is analytic at every \( z \) with \( \operatorname{Rez} > 0 \) . | Proof. Let\n\n\[ f\left( z\right) = \frac{1}{\pi }{\int }_{0}^{\infty }\frac{{y}^{2} - {z}^{2}}{{\left( {y}^{2} + {z}^{2}\right) }^{2}}\log \frac{1}{1 - {e}^{-{2\pi y}}}{dy}. \]\n\nWe show \( {J}^{\prime }\left( z\right) = f\left( z\right) \) . For each \( {z}_{0} \) with \( {\operatorname{Rez}}_{0} \neq 0 \), we have ... | Yes |
Show that \( J\left( z\right) \) is multi-valued analytic in \( \mathbb{C} \smallsetminus \{ 0, - 1, - 2\ldots \} \) with logarithm singularities at \( 0, - 1, - 2,\ldots \), singl-valued analytic in \( \mathbb{C} \smallsetminus (\infty ,0\rbrack \). Show also that \( {e}^{J\left( z\right) } \) is multi-valued analytic... | This can be seen by Stirling's formula\n\n\[ \log \Gamma \left( z\right) = - 1 + \frac{1}{2}\log {2\pi } + z + \left( {z + 1/2}\right) \log z + J\left( z\right) .\n\]\n\nSince \( \Gamma \left( z\right) \) has no zero or pole in \( \mathbb{C} \smallsetminus ( - \infty ,0\rbrack \), we have seen \( J\left( z\right) \) is... | Yes |
Corollary 7.2. An enttire function of fractional order assumes every finite value infinitely many times. | It is clear that \( f \) and \( f - a \) have the same order for any constant \( a \) . Therefore we need only show that \( \mathrm{f} \) has infinitely many zeros. If \( f \) has only a finite number of zeros we can divide by a polynomial and obtain a function of the same order without zeros. By the theorem it must be... | Yes |
Theorem 9.1. For \( \sigma = \operatorname{Res} > 1 \), \[ \frac{1}{\zeta \left( s\right) } = \mathop{\prod }\limits_{{n = 1}}^{\infty }\left( {1 - {p}_{n}^{-s}}\right) \] | According to Theorem 6 the infinite product converges uniformly for \( \sigma = \operatorname{Res} \geq {\sigma }_{0} > 1 \) if the same is true of the series \( \mathop{\sum }\limits_{1}^{\infty }\left| {p}_{n}^{-s}\right| = \mathop{\sum }\limits_{1}^{\infty }{p}_{n}^{-\sigma } \). Since the latter is obtained by omit... | Yes |
Theorem 10.1. For \( \sigma > 1 \) ,\n\n\[ \zeta \left( s\right) = - \frac{\Gamma \left( {1 - s}\right) }{2\pi i}{\int }_{C}\frac{{\left( -z\right) }^{s - 1}}{{e}^{z} - 1}{dz} \]\n\nwhere \( {\left( -z\right) }^{s - 1} \) is defined on the complement of the positive real axis as \( {e}^{\left( {s - 1}\right) \log \left... | The integral is obviously convergent. By Cauchy's theorem its value does not depend on the shape of \( C \) as long as \( C \) does not enclose any multiples of \( {2\pi i} \). In particular, we are free to let \( r \) tend to zero. It is readily seen that the integral over the circle tends to zero with \( r \rightarro... | Yes |
Corollary 10.2. The \( \zeta \) -function can be extended to a mermnorphic function in the whole plane whose only pole is a simple pole at \( s = 1 \) with the residue 1 . | The values \( \zeta \left( {-n}\right) \) at the negative integers and zero can be evaluated explicitly. Recall the expansion (Sec. Ahlfors Chepter 5.1.3, Ex. 4)\n\n\( \left( {10.3}\right) \)\n\n\[ \n\frac{1}{{e}^{z} - 1} = \frac{1}{z} - \frac{1}{2} + \mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{{\left( -1\right) }^{... | Yes |
The function\n\n\[ \xi \left( s\right) = \frac{1}{2}s\left( {1 - s}\right) {\pi }^{-s/2}\Gamma \left( {s/2}\right) \zeta \left( s\right) \]\n\nis entire and satisfies\n\n\[ \xi \left( s\right) = \xi \left( {1 - s}\right) \] | It is evident that \( \xi \left( s\right) \) is entire, for the factor \( 1 - s \) offsets the pole of \( \Gamma \left( s\right) \), and the poles of \( \Gamma \left( {s/2}\right) \) cancel against the trivial zeros of \( \zeta \left( s\right) \) . By use of (11.3) the assertion \( \xi \left( s\right) = \xi \left( {1 -... | Yes |
Lemma 3.1 (i), \( \ell \left( {\widetilde{\alpha }}_{t}\right) \geq \ell \left( \widetilde{\beta }\right) \) | On the other hand, \[ \ell \left( {\widetilde{\alpha }}_{t}\right) = \ell \left( {\alpha }_{t}\right) \leq \ell \left( \beta \right) = \ell \left( \widetilde{\beta }\right) = t \] hence \( \ell \left( {\alpha }_{t}\right) = t \), as was claimed. Since \( t \) is arbitrary, \( M \) is not bounded, which concludes the pr... | No |
Lemma 1.1 If a rectangle is the almost disjoint union of finitely many other rectangles, say \( R = \mathop{\bigcup }\limits_{{k = 1}}^{N}{R}_{k} \), then\n\n\[ \left| R\right| = \mathop{\sum }\limits_{{k = 1}}^{N}\left| {R}_{k}\right| \] | Proof. We consider the grid formed by extending indefinitely the sides of all rectangles \( {R}_{1},\ldots ,{R}_{N} \) . This construction yields finitely many rectangles \( {\widetilde{R}}_{1},\ldots ,{\widetilde{R}}_{M} \), and a partition \( {J}_{1},\ldots ,{J}_{N} \) of the integers between 1 and \( M \), such that... | Yes |
Lemma 1.2 If \( R,{R}_{1},\ldots ,{R}_{N} \) are rectangles, and \( R \subset \mathop{\bigcup }\limits_{{k = 1}}^{N}{R}_{k} \), then\n\n\[ \left| R\right| \leq \mathop{\sum }\limits_{{k = 1}}^{N}\left| {R}_{k}\right| \] | The main idea consists of taking the grid formed by extending all sides of the rectangles \( R,{R}_{1},\ldots ,{R}_{N} \), and noting that the sets corresponding to the \( {J}_{k} \) (in the above proof) need not be disjoint any more. | No |
Theorem 1.3 Every open subset \( \mathcal{O} \) of \( \mathbb{R} \) can be writen uniquely as a countable union of disjoint open intervals. | Proof. For each \( x \in \mathcal{O} \), let \( {I}_{x} \) denote the largest open interval containing \( x \) and contained in \( \mathcal{O} \) . More precisely, since \( \mathcal{O} \) is open, \( x \) is contained in some small (non-trivial) interval, and therefore if\n\n\[ \n{a}_{x} = \inf \{ a < x : \left( {a, x}... | Yes |
Theorem 1.4 Every open subset \( \mathcal{O} \) of \( {\mathbb{R}}^{d}, d \geq 1 \), can be written as a countable union of almost disjoint closed cubes. | Proof. We must construct a countable collection \( \mathcal{Q} \) of closed cubes whose interiors are disjoint, and so that \( \mathcal{O} = \mathop{\bigcup }\limits_{{Q \in \mathcal{Q}}}Q \) . As a first step, consider the grid in \( {\mathbb{R}}^{d} \) formed by taking all closed cubes of side length 1 whose vertices... | Yes |
Property 2 If \( {m}_{ * }\left( E\right) = 0 \), then \( E \) is measurable. In particular, if \( F \) is a subset of a set of exterior measure 0, then \( F \) is measurable. | By Observation 3 of the exterior measure, for every \( \epsilon > 0 \) there exists an open set \( \mathcal{O} \) with \( E \subset \mathcal{O} \) and \( {m}_{ * }\left( \mathcal{O}\right) \leq \epsilon \) . Since \( \left( {\mathcal{O} - E}\right) \subset \mathcal{O} \) , monotonicity implies \( {m}_{ * }\left( {\math... | No |
Property 3 A countable union of measurable sets is measurable. | Suppose \( E = \mathop{\bigcup }\limits_{{j = 1}}^{\infty }{E}_{j} \), where each \( {E}_{j} \) is measurable. Given \( \epsilon > 0 \), we may choose for each \( j \) an open set \( {\mathcal{O}}_{j} \) with \( {E}_{j} \subset {\mathcal{O}}_{j} \) and \( {m}_{ * }\left( {{\mathcal{O}}_{j} - {E}_{j}}\right) \leq \epsil... | Yes |
Property 4 Closed sets are measurable. | First, we observe that it suffices to prove that compact sets are measurable. Indeed, any closed set \( F \) can be written as the union of compact sets, say \( F = \mathop{\bigcup }\limits_{{k = 1}}^{\infty }F \cap {B}_{k} \), where \( {B}_{k} \) denotes the closed ball of radius \( k \) centered at the origin; then P... | Yes |
Lemma 3.1 If \( F \) is closed, \( K \) is compact, and these sets are disjoint, then \( d\left( {F, K}\right) > 0 \) . | Proof. Since \( F \) is closed, for each point \( x \in K \), there exists \( {\delta }_{x} > 0 \) so that \( d\left( {x, F}\right) > 3{\delta }_{x} \) . Since \( \mathop{\bigcup }\limits_{{x \in K}}{B}_{2{\delta }_{x}}\left( x\right) \) covers \( K \), and \( K \) is compact, we may find a subcover, which we denote by... | Yes |
Property 5 The complement of a measurable set is measurable. | If \( E \) is measurable, then for every positive integer \( n \) we may choose an open set \( {\mathcal{O}}_{n} \) with \( E \subset {\mathcal{O}}_{n} \) and \( {m}_{ * }\left( {{\mathcal{O}}_{n} - E}\right) \leq 1/n \) . The complement \( {\mathcal{O}}_{n}^{c} \) is closed, hence measurable, which implies that the un... | Yes |
Property 6 A countable intersection of measurable sets is measurable. | This follows from Properties 3 and 5, since\n\n\[ \mathop{\bigcap }\limits_{{j = 1}}^{\infty }{E}_{j} = {\left( \mathop{\bigcup }\limits_{{j = 1}}^{\infty }{E}_{j}^{c}\right) }^{c} \]\n\n | Yes |
Theorem 3.2 If \( {E}_{1},{E}_{2},\ldots \), are disjoint measurable sets, and \( E = \) \( \mathop{\bigcup }\limits_{{j = 1}}^{\infty }{E}_{j} \), then\n\n\[ m\left( E\right) = \mathop{\sum }\limits_{{j = 1}}^{\infty }m\left( {E}_{j}\right) \] | Proof. First, we assume further that each \( {E}_{j} \) is bounded. Then, for each \( j \), by applying the definition of measurability to \( {E}_{j}^{c} \), we can choose a closed subset \( {F}_{j} \) of \( {E}_{j} \) with \( {m}_{ * }\left( {{E}_{j} - {F}_{j}}\right) \leq \epsilon /{2}^{j} \) . For each fixed \( N \)... | Yes |
Corollary 3.3 Suppose \( {E}_{1},{E}_{2},\ldots \) are measurable subsets of \( {\mathbb{R}}^{d} \). (i) If \( {E}_{k} \nearrow E \), then \( m\left( E\right) = \mathop{\lim }\limits_{{N \rightarrow \infty }}m\left( {E}_{N}\right) \). | Proof. For the first part, let \( {G}_{1} = {E}_{1},{G}_{2} = {E}_{2} - {E}_{1} \), and in general \( {G}_{k} = {E}_{k} - {E}_{k - 1} \) for \( k \geq 2 \). By their construction, the sets \( {G}_{k} \) are measurable, disjoint, and \( E = \mathop{\bigcup }\limits_{{k = 1}}^{\infty }{G}_{k} \). Hence \[ m\left( E\right... | Yes |
Theorem 3.4 Suppose \( E \) is a measurable subset of \( {\mathbb{R}}^{d} \) . Then, for every \( \epsilon > 0 \) :\n\n(i) There exists an open set \( \mathcal{O} \) with \( E \subset \mathcal{O} \) and \( m\left( {\mathcal{O} - E}\right) \leq \epsilon \) . | Proof. Part (i) is just the definition of measurability. | No |
Corollary 3.5 A subset \( E \) of \( {\mathbb{R}}^{d} \) is measurable\n\n(i) if and only if \( E \) differs from a \( {G}_{\delta } \) by a set of measure zero,\n\n(ii) if and only if \( E \) differs from an \( {F}_{\sigma } \) by a set of measure zero. | Proof. Clearly \( E \) is measurable whenever it satisfies either (i) or (ii), since the \( {F}_{\sigma },{G}_{\delta } \), and sets of measure zero are measurable.\n\nConversely, if \( E \) is measurable, then for each integer \( n \geq 1 \) we may select an open set \( {\mathcal{O}}_{n} \) that contains \( E \), and ... | Yes |
Theorem 3.6 The set \( \mathcal{N} \) is not measurable. | The proof is by contradiction, so we assume that \( \mathcal{N} \) is measurable. Let \( {\left\{ {r}_{k}\right\} }_{k = 1}^{\infty } \) be an enumeration of all the rationals in \( \left\lbrack {-1,1}\right\rbrack \), and consider the translates\n\n\[{\mathcal{N}}_{k} = \mathcal{N} + {r}_{k}\]\n\nWe claim that the set... | Yes |
If \( f \) is measurable and finite-valued, and \( \Phi \) is continuous, then \( \Phi \circ f \) is measurable. | In fact, \( \Phi \) is continuous, so \( {\Phi }^{-1}\left( \left( {-\infty, a}\right) \right) \) is an open set \( \mathcal{O} \), and hence \( {\left( \Phi \circ f\right) }^{-1}\left( \left( {-\infty, a}\right) \right) = {f}^{-1}\left( \mathcal{O}\right) \) is measurable. | Yes |
Property 3 Suppose \( {\left\{ {f}_{n}\right\} }_{n = 1}^{\infty } \) is a sequence of measurable functions. Then\n\n\[ \mathop{\sup }\limits_{n}{f}_{n}\left( x\right) ,\;\mathop{\inf }\limits_{n}{f}_{n}\left( x\right) ,\;\mathop{\limsup }\limits_{{n \rightarrow \infty }},{f}_{n}\left( x\right) \;\text{ and }\;\mathop{... | Proving that \( \mathop{\sup }\limits_{n}{f}_{n} \) is measurable requires noting that \( \left\{ {\mathop{\sup }\limits_{n}{f}_{n} > a}\right\} = \) \( \mathop{\bigcup }\limits_{n}\left\{ {{f}_{n} > a}\right\} \) . This also yields the result for \( \mathop{\inf }\limits_{n}{f}_{n}\left( x\right) \), since this quanti... | Yes |
Property 4 If \( {\left\{ {f}_{n}\right\} }_{n = 1}^{\infty } \) is a collection of measurable functions, and\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{f}_{n}\left( x\right) = f\left( x\right) \]\n\nthen \( f \) is measurable. | Since \( f\left( x\right) = \mathop{\limsup }\limits_{{n \rightarrow \infty }}{f}_{n}\left( x\right) = \mathop{\liminf }\limits_{{n \rightarrow \infty }}{f}_{n}\left( x\right) \), this property is a consequence of property 3. | Yes |
Property 5 If \( f \) and \( g \) are measurable, then\n\n(i) The integer powers \( {f}^{k}, k \geq 1 \) are measurable.\n\n(ii) \( f + g \) and \( {fg} \) are measurable if both \( f \) and \( g \) are finite-valued. | For (i) we simply note that if \( k \) is odd, then \( \left\{ {{f}^{k} > a}\right\} = \left\{ {f > {a}^{1/k}}\right\} \), and if \( k \) is even and \( a \geq 0 \), then \( \left\{ {{f}^{k} > a}\right\} = \left\{ {f > {a}^{1/k}}\right\} \cup \left\{ {f < - {a}^{1/k}}\right\} \).\n\nFor (ii), we first see that \( f + g... | Yes |
Theorem 4.1 Suppose \( f \) is a non-negative measurable function on \( {\mathbb{R}}^{d} \). Then there exists an increasing sequence of non-negative simple functions \( {\left\{ {\varphi }_{k}\right\} }_{k = 1}^{\infty } \) that converges pointwise to \( f \), namely,\n\n\[ \n{\varphi }_{k}\left( x\right) \leq {\varph... | Proof. We begin first with a truncation. For \( N \geq 1 \), let \( {Q}_{N} \) denote the cube centered at the origin and of side length \( N \). Then we define\n\n\[ \n{F}_{N}\left( x\right) = \left\{ \begin{matrix} f\left( x\right) & \text{ if }x \in {Q}_{N}\text{ and }f\left( x\right) \leq N, \\ N & \text{ if }x \in... | Yes |
Theorem 4.2 Suppose \( f \) is measurable on \( {\mathbb{R}}^{d} \) . Then there exists a sequence of simple functions \( {\left\{ {\varphi }_{k}\right\} }_{k = 1}^{\infty } \) that satisfies\n\n\[ \left| {{\varphi }_{k}\left( x\right) }\right| \leq \left| {{\varphi }_{k + 1}\left( x\right) }\right| \;\text{ and }\;\ma... | Proof. We use the following decomposition of the function \( f : f\left( x\right) = \) \( {f}^{ + }\left( x\right) - {f}^{ - }\left( x\right) \), where\n\n\[ {f}^{ + }\left( x\right) = \max \left( {f\left( x\right) ,0}\right) \;\text{ and }\;{f}^{ - }\left( x\right) = \max \left( {-f\left( x\right) ,0}\right) . \]\n\nS... | Yes |
Theorem 4.3 Suppose \( f \) is measurable on \( {\mathbb{R}}^{d} \) . Then there exists a sequence of step functions \( {\left\{ {\psi }_{k}\right\} }_{k = 1}^{\infty } \) that converges pointwise to \( f\left( x\right) \) for almost every \( x \) . | Proof. By the previous result, it suffices to show that if \( E \) is a measurable set with finite measure, then \( f = {\chi }_{E} \) can be approximated by step functions. To this end, we recall part (iv) of Theorem 3.4, which states that for every \( \epsilon \) there exist cubes \( {Q}_{1},\ldots ,{Q}_{N} \) such t... | Yes |
Theorem 4.4 (Egorov) Suppose \( {\left\{ {f}_{k}\right\} }_{k = 1}^{\infty } \) is a sequence of measurable functions defined on a measurable set \( E \) with \( m\left( E\right) < \infty \), and assume that \( {f}_{k} \rightarrow f \) a.e on \( E \) . Given \( \epsilon > 0 \), we can find a closed set \( {A}_{\epsilon... | Proof. We may assume without loss of generality that \( {f}_{k}\left( x\right) \rightarrow f\left( x\right) \) for every \( x \in E \) . For each pair of non-negative integers \( n \) and \( k \), let\n\n\[ \n{E}_{k}^{n} = \left\{ {x \in E : \left| {{f}_{j}\left( x\right) - f\left( x\right) }\right| < 1/n,\text{ for al... | Yes |
Theorem 4.5 (Lusin) Suppose \( f \) is measurable and finite valued on \( E \) with \( E \) of finite measure. Then for every \( \epsilon > 0 \) there exists a closed set \( {F}_{\epsilon } \), with\n\n\[ \n{F}_{\epsilon } \subset E,\;\text{ and }\;m\left( {E - {F}_{\epsilon }}\right) \leq \epsilon \n\]\n\nand such tha... | Proof. Let \( {f}_{n} \) be a sequence of step functions so that \( {f}_{n} \rightarrow f \) a.e. Then we may find sets \( {E}_{n} \) so that \( m\left( {E}_{n}\right) < 1/{2}^{n} \) and \( {f}_{n} \) is continuous outside \( {E}_{n} \) . By Egorov’s theorem, we may find a set \( {A}_{\epsilon /3} \) on which \( {f}_{n... | Yes |
Proposition 1.1 The integral of simple functions defined above satisfies the following properties:\n\n(i) Independence of the representation. If \( \varphi = \mathop{\sum }\limits_{{k = 1}}^{N}{a}_{k}{\chi }_{{E}_{k}} \) is any representation of \( \varphi \), then\n\n\[ \int \varphi = \mathop{\sum }\limits_{{k = 1}}^{... | Proof. The only conclusion that is a little tricky is the first, which asserts that the integral of a simple function can be calculated by using any of its decompositions as a linear combination of characteristic functions.\n\nSuppose that \( \varphi = \mathop{\sum }\limits_{{k = 1}}^{N}{a}_{k}{\chi }_{{E}_{k}} \), whe... | Yes |
Lemma 1.2 Let \( f \) be a bounded function supported on a set \( E \) of finite measure. If \( {\left\{ {\varphi }_{n}\right\} }_{n = 1}^{\infty } \) is any sequence of simple functions bounded by \( M \) , supported on \( E \), and with \( {\varphi }_{n}\left( x\right) \rightarrow f\left( x\right) \) for a.e. \( x \)... | Proof. The assertions of the lemma would be nearly obvious if we had that \( {\varphi }_{n} \) converges to \( f \) uniformly on \( E \) . Instead, we recall one of Littlewood's principles, which states that the convergence of a sequence of measurable functions is \ | No |
Proposition 1.3 Suppose \( f \) and \( g \) are bounded functions supported on sets of finite measure. Then the following properties hold.\n\n(i) Linearity. If \( a, b \in \mathbb{R} \), then\n\n\[ \int \left( {{af} + {bg}}\right) = a\int f + b\int g. \] | All these properties follow by using approximations by simple functions, and the properties of the integral of simple functions given in Proposition 1.1. | No |
Theorem 1.4 (Bounded convergence theorem) Suppose that \( \\left\\{ {f}_{n}\\right\\} \) is a sequence of measurable functions that are all bounded by \( M \), are supported on a set \( E \) of finite measure, and \( {f}_{n}\\left( x\\right) \\rightarrow f\\left( x\\right) \) a.e. \( x \) as \( n \\rightarrow \\) \( \\... | Proof. From the assumptions one sees at once that \( f \) is bounded by \( M \) almost everywhere and vanishes outside \( E \), except possibly on a set of measure zero. Clearly, the triangle inequality for the integral implies that it suffices to prove that \( \\int \\left| {{f}_{n} - f}\\right| \\rightarrow 0 \) as \... | Yes |
Proposition 1.6 The integral of non-negative measurable functions enjoys the following properties:\n\n(i) Linearity. If \( f, g \geq 0 \), and \( a, b \) are positive real numbers, then\n\n\[ \n\int \left( {{af} + {bg}}\right) = a\int f + b\int g.\n\] | Proof. Of the first four assertions, only (i) is not an immediate consequence of the definitions, and to prove it we argue as follows. We take \( a = b = 1 \) and note that if \( \varphi \leq f \) and \( \psi \leq g \), where both \( \varphi \) and \( \psi \) are bounded and supported on sets of finite measure, then \(... | Yes |
Lemma 1.7 (Fatou) Suppose \( \left\{ {f}_{n}\right\} \) is a sequence of measurable functions with \( {f}_{n} \geq 0 \) . If \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{f}_{n}\left( x\right) = f\left( x\right) \) for a.e. \( x \), then | Proof. Suppose \( 0 \leq g \leq f \), where \( g \) is bounded and supported on a set \( E \) of finite measure. If we set \( {g}_{n}\left( x\right) = \min \left( {g\left( x\right) ,{f}_{n}\left( x\right) }\right) \), then \( {g}_{n} \) is measurable, supported on \( E \), and \( {g}_{n}\left( x\right) \rightarrow g\le... | Yes |
Corollary 1.8 Suppose \( f \) is a non-negative measurable function, and \( \left\{ {f}_{n}\right\} \) a sequence of non-negative measurable functions with \( {f}_{n}\left( x\right) \leq f\left( x\right) \) and \( {f}_{n}\left( x\right) \rightarrow f\left( x\right) \) for almost every \( x \) . Then\n\n\[ \mathop{\lim ... | Proof. Since \( {f}_{n}\left( x\right) \leq f\left( x\right) \) a.e \( x \), we necessarily have \( \int {f}_{n} \leq \int f \) for all \( n \) ; hence\n\n\[ \mathop{\limsup }\limits_{{n \rightarrow \infty }}\int {f}_{n} \leq \int f \]\n\n## This inequality combined with Fatou's lemma proves the desired limit. | No |
Corollary 1.10 Consider a series \( \mathop{\sum }\limits_{{k = 1}}^{\infty }{a}_{k}\left( x\right) \), where \( {a}_{k}\left( x\right) \geq 0 \) is measurable for every \( k \geq 1 \) . Then\n\n\[ \n\int \mathop{\sum }\limits_{{k = 1}}^{\infty }{a}_{k}\left( x\right) {dx} = \mathop{\sum }\limits_{{k = 1}}^{\infty }\in... | Proof. Let \( {f}_{n}\left( x\right) = \mathop{\sum }\limits_{{k = 1}}^{n}{a}_{k}\left( x\right) \) and \( f\left( x\right) = \mathop{\sum }\limits_{{k = 1}}^{\infty }{a}_{k}\left( x\right) \) . The functions \( {f}_{n} \) are measurable, \( {f}_{n}\left( x\right) \leq {f}_{n + 1}\left( x\right) \), and \( {f}_{n}\left... | Yes |
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