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Proposition 4.5.7 (Hahn Decomposition Theorem). Let \( \nu \) be a signed measure on \( \left( {X,\mathcal{B}}\right) \) . Then there exists a positive set \( A \) and a negative set \( B \) with \( X = A \cup B \) and \( A \cap B = \varnothing \) . | Proof. By definition, \( \nu \) can take on at most one of the values \( + \infty \) , \( - \infty \) . Without loss of generality, assume \( \nu \) never takes on the value \( + \infty \) . Let\n\n\[ \lambda = \sup \{ \nu \left( E\right) \mid E \in \mathcal{B}\text{ and }E\text{ is positive }\} .\](In the case that \(... | Yes |
Let \( C \) be the Cantor set. Again, the Cantor set is a Borel set and so is Hausdorff measurable. To find the Hausdorff dimension of the Cantor set, recall that at the \( k \) th stage of construction, \( {C}_{k} \) consisted of \( {2}^{k} \) intervals of length \( \frac{1}{{3}^{k}} \) . Hence, a candidate for \( {H}... | \[ \mathop{\lim }\limits_{{k \rightarrow \infty }}\frac{{2}^{k}}{{3}^{k\alpha }} = \mathop{\lim }\limits_{{k \rightarrow \infty }}{\left( \frac{2}{{3}^{\alpha }}\right) }^{k}. \] If \( \frac{2}{{3}^{\alpha }} < 1 \), this limit will be 0 . If \( \frac{2}{{3}^{\alpha }} > 1 \), this limit will be \( + \infty \) . As pre... | No |
1.16 Proposition Let \( X \subseteq \mathbb{R} \) and \( f : X \rightarrow E \) be right and left differentiable at \( a \in X \) with \( {\partial }_{ + }f\left( a\right) = {\partial }_{ - }f\left( a\right) \) . Then \( f \) is differentiable at \( a \) and \( \partial f\left( a\right) = {\partial }_{ + }f\left( a\rig... | Proof By hypothesis and Proposition 1.1(iii), there are functions\n\n\[ \n{r}_{ + } : X \cap \lbrack a,\infty ) \rightarrow E\;\text{ and }\;{r}_{ - } : ( - \infty, a\rbrack \cap X \rightarrow E \n\] \n\nwhich are continuous at \( a \) and satisfy \( {r}_{ + }\left( a\right) = {r}_{ - }\left( a\right) = 0 \) and \n\n\[... | Yes |
Let \( - \infty < a < b < \infty \) and \( f \in {C}^{2}\left( {\left\lbrack {a, b}\right\rbrack ,\mathbb{R}}\right) \) be such that \( {f}^{\prime }\left( x\right) \neq 0 \) for all \( x \in \left\lbrack {a, b}\right\rbrack \) . We suppose further that there is some \( \xi \in \left( {a, b}\right) \) such that \( f\le... | Proof (i) By the extreme value theorem (Corollary III.3.8), there are constants \( {M}_{1},{M}_{2}, m > 0 \) such that \( m \leq \left| {{f}^{\prime }\left( x\right) }\right| \leq {M}_{1},\;\left| {{f}^{\prime \prime }\left( x\right) }\right| \leq {M}_{2},\;x \in \left\lbrack {a, b}\right\rbrack \). For the function \(... | Yes |
2.1 Theorem If \( \left( {f}_{n}\right) \) converges uniformly to \( f \) and almost all \( {f}_{n} \) are continuous at \( a \in X \), then \( f \) is also continuous at \( a \) . | Proof Let \( \varepsilon > 0 \) . Because \( {f}_{n} \) converges uniformly to \( f \), there is, by Remark 1.3(e), some \( N \in \mathbb{N} \) such that \( {\begin{Vmatrix}{f}_{n} - f\end{Vmatrix}}_{\infty } < \varepsilon /3 \) for all \( n \geq N \) . Since almost all \( {f}_{n} \) are continuous at \( a \), we can s... | Yes |
6.9 Theorem (Fubini) For \( f \in {\mathcal{L}}_{1}\left( {\mathbb{R}}^{m + n}\right) \), (i) \( f\left( {x, \cdot }\right) \in {\mathcal{L}}_{1}\left( {\mathbb{R}}^{n}\right) \) for \( {\lambda }_{m} \) -almost every \( x \in {\mathbb{R}}^{m} \), \( f\left( {\cdot, y}\right) \in {\mathcal{L}}_{1}\left( {\mathbb{R}}^{m... | Proof (a) For \( f \in {\mathcal{L}}_{1}\left( {{\mathbb{R}}^{m + n},{\mathbb{R}}^{ + }}\right) \), the claim follows from Tonelli’s theorem and Remark 3.3(e). (b) Given the representation \( f = {f}_{1} - {f}_{2} + i\left( {{f}_{3} - {f}_{4}}\right) \), with \( {f}_{j} \in {\mathcal{L}}_{1}\left( {{\mathbb{R}}^{m + n}... | Yes |
Theorem 1. Let \( I \) be a bounded interval. (i) If the bounded functions \( f \) and \( g \) are integrable on \( I \), then so likewise is \( {\alpha f} + {\beta g} \) for any constants \( \alpha \) and \( \beta \) , and | (2.3) \[ m\left( {{\alpha f} + {\beta g}}\right) = {\alpha m}\left( f\right) + {\beta m}\left( g\right) . \] | No |
Theorem 2. If the real functions \( f \) and \( g \) are integrable on \( I \), so are the functions \( \sup \left( {f, g}\right) \) and \( \inf \left( {f, g}\right) \) . | By Theorem 1 and the formula above it is enough to show that if \( f \) is integrable then so is \( {f}^{ + } \) . This follows immediately from the definition,(1) or (1’), and from the inequality \( \left| {{f}^{ + } - {\varphi }^{ + }}\right| \leq \left| {f - \varphi }\right| \) . | No |
Theorem 3. Let \( f \) and \( g \) be two bounded integrable functions on a compact interval \( I \) . Then the function \( f\bar{g} \) is integrable and (Cauchy-Schwarz inequal- \( {it}{y}^{4} \) ) | (2.5) \[ {\left| m\left( f\bar{g}\right) \right| }^{2} \leq m\left( {\left| f\right| }^{2}\right) m\left( {\left| g\right| }^{2}\right) . \] In checking that \( f\bar{g} \) is integrable we may assume \( f \) and \( g \) real, and even positive, since every integrable real function \( f \) is the difference of the inte... | Yes |
Theorem 4. Let \( \left( {f}_{n}\right) \) be a uniformly convergent sequence of integrable functions on a bounded interval \( I \) . Then the function \( f\left( x\right) = \lim {f}_{n}\left( x\right) \) is integrable and\n\n(4.2)\n\n\[ m\left( f\right) = {\int }_{I}f\left( x\right) {dx} = \lim {\int }_{I}{f}_{n}\left... | For \( r > 0 \) given, and for every \( n \), let us choose a step function \( {\varphi }_{n} \) such that \( m\left( \left| {{f}_{n} - {\varphi }_{n}}\right| \right) < r \), and let \( N \) be an integer such that\n\n\[ n > N \Rightarrow {\begin{Vmatrix}f - {f}_{n}\end{Vmatrix}}_{I} < r \]\n\nfrom the definition of un... | Yes |
Consider a power series\n\n\[ f\left( z\right) = \sum {c}_{n}{z}^{n}/n! = \sum {c}_{n}{z}^{\left\lbrack n\right\rbrack } \]\n\nwhich converges on a disc \( \left| z\right| < R \) of nonzero radius, and let us calculate the integral of \( f\left( x\right) \) over an interval \( \left\lbrack {a, b}\right\rbrack \) with \... | We know that the series converges normally on every disc \( \left| z\right| \leq r < R \), so on the interval considered: we can therefore integrate term-by-term. But we also know (Chap. II, \( {\mathrm{n}}^{ \circ }{11} \) ) that\n\n\[ {\int }_{a}^{b}{x}^{n}{dx} = \frac{{b}^{n + 1}}{n + 1} - \frac{{a}^{n + 1}}{n + 1} ... | Yes |
Theorem 5 (Borel-Lebesgue). Let \( K \) be a compact subset of \( \mathbb{R} \) (resp. \( \mathbb{C} \) ) and \( {\left( {U}_{i}\right) }_{i \in I} \) a family of open sets in \( \mathbb{R} \) (resp. \( \mathbb{C} \) ). Suppose that \( K \) is contained in the union of the \( {U}_{i} \) . Then there is a finite subset ... | First we show that if \( K \) is bounded one can, for every \( r > 0 \), find a finite number of points \( {x}_{k} \) of \( K \) such that \( K \) is contained in the union of the open balls \( B\left( {{x}_{k}, r}\right) \) . Since \( K \) is certainly contained in a compact interval or square, it is clear that one ca... | Yes |
Let \( {\left( {K}_{i}\right) }_{i \in I} \) be a family of nonempty compact sets in \( \mathbb{R} \) or \( \mathbb{C} \) . Suppose that the intersection of the \( {K}_{i} \) is empty. Then there is a finite subset \( F \) of \( I \) such that the intersection of the \( {K}_{i}, i \in F \), is empty. | We choose any index \( j \) and replace each \( {K}_{i} \) by \( {K}_{i} \cap {K}_{j} \) . If one of these intersections is empty, the corollary is proved. So assume they are nonempty. This is equivalent to assuming that all the \( {K}_{i} \) are contained in the same compact set \( K \), namely \( {K}_{j} \) . Let \( ... | Yes |
Theorem 6. Let \( f \) be a scalar function defined on an interval \( I \) of \( \mathbb{R} \) . The two following properties are equivalent: (i) \( f \) has left and right limits at every point of \( I \) ; (ii) there is a sequence of step functions on \( I \) which converges to \( f \) uniformly on every compact subs... | The implication (ii) \( \Rightarrow \) (i) was established from Cauchy’s criterion in Chap. III, \( {\mathrm{n}}^{ \circ } 12 (Corollary of Theorem 16). The implication (i) \( \Rightarrow \) (ii) is obtained, when \( I \) is compact, by observing, as at the beginning of the preceding \( {\mathrm{n}}^{ \circ } \), that ... | Yes |
Theorem 7. The integral of a regulated (resp. continuous) positive function \( f \) is zero if and only if the set \( D = \{ f\left( x\right) \neq 0\} \) is countable (resp. empty). | The condition is sufficient. For consider a step function \( \varphi \leq f \) . One can have \( \varphi \left( x\right) > 0 \) only if \( x \in D \) . Since the set of points of a nonsingleton interval is uncountable (Chap. I), the function \( \varphi \) is necessarily negative on all the intervals of nonzero length w... | Yes |
Theorem 8 (Heine \( {}^{12} \) ). Every scalar function defined and continuous on a compact set \( K \subset \mathbb{C} \) is uniformly continuous on \( K \) . | Given \( r > 0 \) let us choose for each \( x \in K \) an open ball \( B\left( x\right) \) with centre \( x \) such that \( f \) is constant to within \( r \) in \( B\left( x\right) \cap K \) . Let \( {B}^{\prime }\left( x\right) \) be the open ball with centre \( x \) and of radius half that of \( B\left( x\right) \) ... | Yes |
Corollary 1. Let \( f \) be a scalar function defined and continuous on \( \mathbb{R} \) (resp. \( \mathbb{C} \) ) and zero for \( \left| x\right| \) large. Then \( f \) is uniformly continuous on \( \mathbb{R} \) (resp. \( \mathbb{C} \) ). | We need only treat the case of \( \mathbb{C} \) . Let \( K \) be a compact set outside which \( f = 0 \), and \( H \) the set of \( x \in \mathbb{C} \) such that \( d\left( {x, K}\right) \leq 1 \) . Since \( d\left( {x, K}\right) \) is a continuous function of \( x \) (Chap. III, \( {\mathrm{n}}^{ \circ }{10} \) ), the... | Yes |
Corollary 2. Let \( f \) be a function defined and continuous on a bounded set \( X \subset \mathbb{C} \) . The following two properties are equivalent: (i) \( f \) is uniformly continuous on \( X \) ; (ii) \( f \) is the restriction to \( X \) of a continuous function on the compact set \( \overline{X} \) . | We have just seen that (i) implies (ii). The converse implication follows from Theorem 8 since \( \bar{X} \) is compact. | Yes |
Corollary 3. Let \( f \) be a scalar function defined and continuous on a compact interval \( I \) . For every \( r > 0 \) there exists an \( {r}^{\prime } > 0 \) such that\n\n\[ \left| {{\int }_{I}f\left( x\right) {dx}-\sum f\left( {\xi }_{k}\right) \left( {{x}_{k + 1} - {x}_{k}}\right) }\right| < r \] \n\nfor any poi... | For example one can decompose \( I \) into \( n \) equal intervals \( {I}_{1},\ldots ,{I}_{n} \) and choose a \( {\xi }_{k} \in {I}_{k} \) at random for each \( k \) . The corresponding Riemann sum is just\n\n\[ m\left( I\right) \frac{f\left( {\xi }_{1}\right) + \ldots + f\left( {\xi }_{n}\right) }{n}. \] \n\nIt tends ... | No |
Theorem 10. Let \( K \) and \( H \) be two compact intervals in \( \mathbb{R} \) and \( f \) a continuous function on \( K \times H \) . Then\n\n\[{\int }_{H}\nu \left( y\right) {dy}{\int }_{K}f\left( {x, y}\right) \mu \left( x\right) {dx} = {\int }_{K}\mu \left( x\right) {dx}{\int }_{H}f\left( {x, y}\right) \nu \left(... | To prove the equality of the two sides of (7), note that, by (3), there exist finite partitions of \( K \) and \( H \) into intervals \( {K}_{p} \) and \( {H}_{q} \) such that \( f \) is constant to within \( r \) on each rectangle \( {K}_{p} \times {H}_{q} \) . Then\n\n\[{\int }_{H}f\left( {x, y}\right) \nu \left( y\r... | No |
Corollary 1. Let \( K \) be a compact interval and \( \left( {f}_{n}\right) ,\left( {g}_{n}\right) \) two everywhere increasing or two everywhere decreasing sequences of real continuous functions on \( K \) . Assume that \( \lim {f}_{n}\left( x\right) = \lim {g}_{n}\left( x\right) \) for every \( x \in K \) . Then\n\n\... | Consider for example the case of increasing sequences, put\n\n\[ \varphi \left( x\right) = \sup {f}_{n}\left( x\right) \leq + \infty \]\n\nand consider the set \( {C}_{\text{inf }}\left( \varphi \right) \) of all real functions \( h \) defined and continuous on \( K \) such that \( h\left( x\right) \leq \varphi \left( ... | Yes |
Corollary 2. Let \( \sum {u}_{n}\left( x\right) \) be a series of continuous functions on a compact interval \( K \) . Assume that the series converges simply to a continuous function \( s\left( x\right) \) and that\n\n(10.3)\n\[ \sum {\int }_{K}\left| {{u}_{n}\left( x\right) }\right| {dx} < + \infty \]\n\nThen\n\n(10.... | To prove (4), one may assume \( s = 0 \) by replacing \( {u}_{1} \) by \( {u}_{1} - s \), which is again continuous. One may also assume the \( {u}_{n} \) real and then use the decomposition \( {u}_{n} = {u}_{n}^{ + } - {u}_{n}^{ - } \) of \( {\mathrm{n}}^{ \circ }2 \) . These positive functions again satisfy the hypot... | Yes |
Theorem 11. Let \( f \) be a regulated function on an interval \( I \) in \( \mathbb{R} \). Then the function \( F \) defined by the relation (3) is continuous and has right and left derivatives equal to \( f\left( {x + }\right) \) and \( f\left( {x - }\right) \) at each point \( x \in I \). | This result, for functions as then understood, is already in Newton in 1665-66 with essentially the same proof, phrased in his language of fluentes and fluxions (Chap. III, \( {\mathrm{n}}^{ \circ }{14} \) ): if \( y \) is the fluent which defines the curve \( \left\lbrack {y = f\left( x\right) \text{in the language of... | No |
For \( x > 0 \) and \( s \in \mathbb{C} \), the function\n\n\[ \n{x}^{s} = \exp \left( {s \cdot \log x}\right)\n\]\n\nhas derivative \( s{x}^{s - 1} \) [Chap. IV, formula (10.10)]. The function \( {x}^{s + 1}/\left( {s + 1}\right) \) is therefore, for \( s \neq - 1 \), a primitive of \( {x}^{s} \) . | Whence the formula\n\n(12.7)\n\n\[ \n{\int }_{a}^{b}{x}^{s}{dx} = \frac{{b}^{s + 1} - {a}^{s + 1}}{s + 1}\;\left( {0 < a, b;s \in \mathbb{C}, s \neq - 1}\right)\n\]\n\nalready obtained for \( s \in \mathbb{N} \) by a direct calculation of the integral (Chap. II, \( {\mathrm{n}}^{ \circ }{11} \) ), but now valid for any... | Yes |
For \( x > 0 \) the function \( \log x \) has derivative \( 1/x \) | \[ {\int }_{a}^{b}{dx}/x = \log b - \log a\;\left( {0 < a, b}\right) \] | No |
The derivative of the function \( \arctan x \) is \( 1/\left( {1 + {x}^{2}}\right) \) ; whence\n\n\[{\int }_{a}^{b}\frac{dx}{1 + {x}^{2}} = \arctan b - \arctan a\] | we must pay attention to the fact that, in this calculation, we take the \ | No |
For \( c \in \mathbb{C}, c \neq 0 \), the derivative of \( {e}^{cx}/c \) is \( {e}^{cx} \) | we again find the formula\n\[ \n{\int }_{a}^{b}{e}^{cx}{dx} = \left( {{e}^{cb} - {e}^{ca}}\right) /c.\n\] | No |
Theorem 13. Let \( I \) and \( J \) be two intervals, \( f \) a continuous function on \( I \times J \) and let \( \varphi ,\psi : I \rightarrow J \) be two differentiable functions. Suppose that \( f \) has a continuous derivative \( {D}_{1}f \) on \( I \times J \) . Then the function\n\n\[ g\left( x\right) = {\int }_... | By subtracting, one need only prove this in the case where \( \varphi \left( x\right) = b \) is constant. Put\n\n\[ F\left( {x, y}\right) = {\int }_{b}^{y}f\left( {x, t}\right) {dt} \]\n\nwhence \( g\left( x\right) = F\left\lbrack {x,\psi \left( x\right) }\right\rbrack \) . Since \( f \) is continuous the FT shows that... | Yes |
Theorem 14. Let \( f\left( {x, y}\right) \) be a function defined and continuous on \( I \times J \) where \( I \) and \( J \) are two intervals in \( \mathbb{R} \) . Assume that \( f \) has continuous second derivatives \( {D}_{1}{D}_{2}f \) and \( {D}_{2}{D}_{1}f \) on \( I \times J \) . Then they are equal. | Since it is enough to verify the statement on a neighbourhood of an arbitrary point of \( I \times J \) one may reduce to the case where \( I = \left\lbrack {a, b}\right\rbrack \) and \( J = \left\lbrack {c, d}\right\rbrack \) are compact. The FT applied to the functions \( y \mapsto {D}_{2}{D}_{1}f\left( {x, y}\right)... | Yes |
A differentiable function \( f \) is convex on an open interval if and only if its derivative is increasing. A twice differentiable function is convex if and only if \( {f}^{\prime \prime }\left( x\right) \geq 0 \) for every \( x \) . | For if \( {f}^{\prime }\left( x\right) \) exists everywhere, and is increasing, so regulated, then \( f \) is a primitive of \( {f}^{\prime } \) and is therefore convex. If \( {f}^{\prime } \) is differentiable it is increasing if and only if \( {f}^{\prime \prime }\left( x\right) \geq 0 \) everywhere, by the mean valu... | Yes |
If \( f \) and \( g \) are regulated functions on a bounded interval \( I \), then | \[ \left| {{\int }_{I}f\left( x\right) g\left( x\right) {dx}}\right| \leq {N}_{p}\left( f\right) {N}_{q}\left( f\right) ,\;{N}_{p}\left( {f + g}\right) \leq {N}_{p}\left( f\right) + {N}_{p}\left( g\right) \]\n\nwhere one puts\n\n\[ {N}_{p}\left( f\right) = {\left( {\int }_{I}{\left| f\left( x\right) \right| }^{p}dx\rig... | No |
Take for \( X \) a finite set and \( {\mu }^{ * }\left( f\right) = \sum \left| {f\left( x\right) }\right| \) . One obtains, in more traditional notation, the original versions of the inequalities: | \[ \left| {\sum {x}_{k}{y}_{k}}\right| \leq {\left( \sum {\left| {x}_{k}\right| }^{p}\right) }^{1/p}{\left( \sum {\left| {y}_{k}\right| }^{q}\right) }^{1/q}, \] \[ {\left( \sum {\left| {x}_{k} + {y}_{k}\right| }^{p}\right) }^{1/p} \leq {\left( \sum {\left| {x}_{k}\right| }^{p}\right) }^{1/p} + {\left( \sum {\left| {y}_... | Yes |
Like the preceding, but with an infinite set \( X \) and, again, \( {\mu }^{ * }\left( f\right) = \) \( \sum \left| {f\left( x\right) }\right| \leq + \infty \) for every function \( f \) with positive values. If the series \( \sum {\left| {x}_{n}\right| }^{p} \) and \( \sum {\left| {y}_{n}\right| }^{q} \) converge then... | All this assumes \( p, q > 1 \) and \( 1/p + 1/q = 1 \) . The case \( p = q = 2 \) is the Cauchy-Schwarz inequality for series, which may be proved much more easily by passing to the limit starting from the case of a finite sum. | No |
Let us calculate the primitive\n\n\[ \int \log \left( x\right) {dx} \] | \[\n\int \log \left( x\right) {dx} = \int \log \left( x\right) {.1} \cdot {dx} = \int \log \left( x\right) \cdot {\left( x\right) }^{\prime } \cdot {dx} = \n\]\n\n\[ \n= \log \left( x\right) x - \int {\log }^{\prime }\left( x\right) {xdx} = x\log x - \int {1dx} \n\]\n\nwhence\n\n(15.3)\n\n\[ \n\int \log \left( x\right)... | Yes |
\[ \int {x}^{5}{e}^{x}{dx} = \] | \[ = \int {x}^{5}{\left( {e}^{x}\right) }^{\prime }{dx} = {x}^{5}{e}^{x} - \int {\left( {x}^{5}\right) }^{\prime }{e}^{x}{dx} = {x}^{5}{e}^{x} - 5\int {x}^{4}{e}^{x}{dx} = \]\n\[ = {x}^{5}{e}^{x} - 5{x}^{4}{e}^{x} + {5.4}\int {x}^{3}{e}^{x}{dx} = \]\n\[ = {x}^{5}{e}^{x} - 5{x}^{4}{e}^{x} + {5.4}{x}^{3}{e}^{x} - {5.4.3}... | Yes |
[{I}_{n} = \int {x}^{n}\cos x \cdot {dx},\;{J}_{n} = \int {x}^{n}\sin x \cdot {dx}.] | [{I}_{n} = \int {x}^{n}{\sin }^{\prime }x \cdot {dx} = {x}^{n}\sin x - n\int {x}^{n - 1}\sin x \cdot {dx}] and continue. It is more economical to observe that [{I}_{n} + i{J}_{n} = \int {x}^{n}{e}^{ix}{dx}] and to calculate as in Example 2 or, for good measure, to calculate \( \int {x}^{n}{e}^{tx}{dx} \) for every \( t... | Yes |
Example 4. Put \( {\log }^{2}x = {\left( \log x\right) }^{2} \) and calculate\n\n\[ \int x.{\log }^{2}x.{dx} \] | \n\[ \int x.{\log }^{2}x.{dx} = \int {\left( \frac{1}{2}{x}^{2}\right) }^{\prime }{\log }^{2}x.{dx} = \frac{1}{2}{x}^{2}{\log }^{2}x - \frac{1}{2}\int {x}^{2}{\left( {\log }^{2}x\right) }^{\prime }{dx} = \]\n\n\[ = \frac{1}{2}{x}^{2}{\log }^{2}x - \frac{1}{2}\int {x}^{2}2\log x\left( {{\log }^{\prime }x}\right) {dx} = ... | Yes |
Theorem 16. Let \( f \) be a function defined and of class \( {C}^{n + 1} \) on an interval \( I \) of \( \mathbb{R} \) . Then, for any \( a, x \in I \) ,\n\n\[ f\left( x\right) = f\left( a\right) + {f}^{\prime }\left( a\right) \left( {x - a}\right) + {f}^{\prime \prime }\left( a\right) {\left( x - a\right) }^{\left\lb... | where\n\n\[ {r}_{n}\left( x\right) = {\int }_{a}^{x}{f}^{\left( n + 1\right) }\left( t\right) {\left( x - t\right) }^{\left\lbrack n\right\rbrack }{dt}. \] | Yes |
Take \( a = 0 \) and \( f\left( x\right) = \sin x \) . For \( n = {2p} \) one finds \( \left| {{r}_{n}\left( x\right) }\right| \leq \) \( {\left| x\right| }^{2p}/\left( {2p}\right) ! since the successive derivatives are everywhere less than 1 in modulus. On passing to the limit one thus recovers the formula | \[ \sin x = \lim \left\lbrack {x - {x}^{3}/3! + \ldots + {\left( -1\right) }^{p - 1}{x}^{{2p} - 1}/\left( {{2p} - 1}\right) !}\right\rbrack . \] | Yes |
Take \( f\left( x\right) = {\left( 1 + x\right) }^{s} \) with \( s \in \mathbb{C} \) and \( - 1 < x \) | Here, by (8),\n\n\[ {r}_{n}\left( x\right) = s\left( {s - 1}\right) \ldots \left( {s - n}\right) \frac{{x}^{n + 1}}{n!}{\int }_{0}^{1}{\left( 1 + ux\right) }^{s - n - 1}{\left( 1 - u\right) }^{n}{dt} \]\n\nor\n\n\[ {r}_{n}\left( x\right) = \frac{s\left( {s - 1}\right) \ldots \left( {s - n}\right) }{n!}{x}^{n + 1}{\int ... | Yes |
Theorem 17. Let \( u \) be a real function defined and of class \( {C}^{1} \) on a compact interval \( I = \left\lbrack {a, b}\right\rbrack \), and \( f \) a function defined and continuous on the interval \( J = u\left( I\right) \) . Then\n\n\[ {\int }_{u\left( a\right) }^{u\left( b\right) }f\left( y\right) {dy} = {\i... | For, let \( F \) be a primitive of \( f \) on \( J \), so that the left hand side is equal to \( F\left\lbrack {u\left( b\right) }\right\rbrack - F\left\lbrack {u\left( a\right) }\right\rbrack \) . The function \( G\left( x\right) = F\left\lbrack {u\left( x\right) }\right\rbrack \) is differentiable on \( I \) and \( {... | Yes |
Calculate the indefinite integral \( \int {\left( {x}^{2} + 1\right) }^{3}{xdx} \) | Putting \( u\left( x\right) = \) \( {x}^{2} + 1 \) we have \( {u}^{\prime }\left( x\right) {dx} = {2xdx} \), so we need to calculate \( \frac{1}{2}\int u{\left( x\right) }^{3}{u}^{\prime }\left( x\right) {dx} \) ; this is situation (2) with \( f\left( y\right) = {y}^{3} \) . Thus\n\n\[ \int {\left( {x}^{2} + 1\right) }... | Yes |
Let \( f \) be a real function of class \( {C}^{1} \) on an interval \( I \), not vanishing on \( I \) . To calculate \( \int {f}^{\prime }\left( x\right) /f\left( x\right) .{dx} \) | one performs the change of variable \( y = f\left( x\right) \), whence \( {dy} = {f}^{\prime }\left( x\right) {dx} \) and\n\n\[\n\int \frac{{f}^{\prime }\left( x\right) }{f\left( x\right) }{dx} = \int {dy}/y\n\]\n\nIt remains to find a primitive of the function \( 1/y \) on the interval \( J = f\left( I\right) \) . Sin... | Yes |
To calculate \( \int {dx}/\sin x \) on an interval where the sine function does not vanish, for example on \( \rbrack 0,\pi \lbrack \) . | If one is inspired, or if one has read all the books, one observes that\n\n\[ 1/\sin x = 1/2\sin \left( {x/2}\right) \cos \left( {x/2}\right) = 1/2\tan \left( {x/2}\right) {\cos }^{2}\left( {x/2}\right) = {f}^{\prime }\left( x\right) /f\left( x\right) \]\n\nwhere \( f\left( x\right) = \tan x/2 \) and \( {f}^{\prime }\l... | Yes |
To calculate\n\n\\[ \int \frac{{x}^{4} + 1}{\sqrt{\left( {x + 1}\right) \left( {x - 5}\right) }}{dx} \\] | we have to work in the interval \( x < - 1 \), or in the interval \( x > 5 \) to obtain a real result. We have \( \left( {x + 1}\right) \left( {x - 5}\right) = {\left( x - 3\right) }^{2} - 4 \), which suggests the change of variable \( x = 3 + {2y} \), whence \( {dx} = {2dy} \) and reduction to\n\n\\[ \int \frac{{\left... | Yes |
Theorem 18. (i) Let \( f \) be a positive regulated function defined on an interval \( X = \\left( {a, b}\\right) \) . Then \( f \) is integrable on \( X \) if and only if the integrals over the compact subsets \( K \subset X \) are bounded above; and then\n\n(22.1)\n\n\[ \n{\\int }_{X}f\\left( x\\right) {dx} = \\matho... | To prove point (i) we observe that, \( f \) being positive, \( s\\left( K\\right) \) is an increasing function of \( K \) :\n\n\[ \nK \subset {K}^{\\prime } \Rightarrow s\\left( K\\right) \leq s\\left( {K}^{\\prime }\\right) \n\]\n\nThe arguments of Chap. II, \( {\\mathrm{n}}^{ \u2060 }9 \) on increasing sequences tran... | No |
Let \( f \) be a bounded regulated function on an interval \( X \), and \( \mu \) an absolutely integrable regulated function on \( X \) . Then the function \( f\left( x\right) \mu \left( x\right) \) is absolutely integrable on \( X \) and \[ \int \left| {f\left( x\right) \mu \left( x\right) {dx}}\right| \leq \parallel... | Obvious. As we shall do on various occasions in the rest of this \( § \), we have used the \( \int \) sign to denote integrals extended over \( X \) . | No |
Corollary 2. Let \( f \) and \( g \) be two regulated square integrable functions on an interval \( X \) ; then the function \( f\left( x\right) \overline{g\left( x\right) } \) is absolutely integrable on \( X \) and\n\n\[{\left| \int f\left( x\right) \overline{g\left( x\right) }dx\right| }^{2} \leq \int {\left| f\left... | One replaces \( f \) and \( g \) by \( \left| f\right| \) and \( \left| g\right| \), writes the Cauchy-Schwarz inequality for every compact interval \( K \subset X \) and notes that the left hand side is, for any \( K \), majorised by the right hand side of the inequality to be established, whence the result on passage... | No |
Consider Euler's ubiquitous Gamma function\n\n\[ \Gamma \left( s\right) = {\int }_{0}^{+\infty }{e}^{-x}{x}^{s - 1}{dx}. \] | Absolute convergence at infinity is automatic, but, at 0, requires \( \operatorname{Re}\left( s\right) > 0 \) . An integration by parts \( {}^{43} \) then shows that\n\n\[ \Gamma \left( {s + 1}\right) = {\int }_{0}^{+\infty }{e}^{-x}{x}^{s}{dx} = - {\left. {e}^{-x}{x}^{s}\right| }_{0}^{+\infty } + s{\int }_{0}^{+\infty... | Yes |
\[ B\left( {x, y}\right) = {\int }_{0}^{1}{t}^{x - 1}{\left( 1 - t\right) }^{y - 1}{dt} \] where \( x \) and \( y \) are a priori complex (and rational for him). Absolute convergence on a neighbourhood of 0 requires \( \operatorname{Re}\left( x\right) > 0 \) and, on a neighbourhood of \( 1,\operatorname{Re}\left( y\rig... | \[ B\left( {x, y}\right) = B\left( {y, x}\right) \] (change of variable \( t \mapsto 1 - t \) ). The change of variable \( t \mapsto {\sin }^{2}t \) shows that \[ B\left( {x, y}\right) = 2{\int }_{0}^{\pi /2}{\sin }^{{2x} - 1}t \cdot {\cos }^{{2y} - 1}t \cdot {dt} \] We shall see later \( \left( {\mathrm{n}}^{ \circ }\... | No |
Theorem 19 (Poor man’s dominated convergence). Let \( \left( {f}_{n}\right) \) be a sequence of regulated functions, absolutely integrable on an interval \( X \subset R \) . Assume that\n\n(i) the \( {f}_{n} \) converge to a limit \( f \) uniformly on every compact \( K \subset X \) ,\n\n(ii) there exists a positive fu... | First of all, it is clear that \( f \), being regulated like the \( {f}_{n} \), is absolutely integrable on \( X \) since \( \left| {f\left( x\right) }\right| \leq p\left( x\right) \) for every \( x \) . Since \( p \) is positive and integrable, for every \( r > 0 \) there exists a compact interval \( K \subset X \) su... | Yes |
Consider the function\n\n\\[ \Gamma \\left( s\\right) = {\\int }_{0}^{+\\infty }{e}^{-x}{x}^{s - 1}{dx},\\;\\operatorname{Re}\\left( s\\right) > 0, \\]\n\nagain, and observe that\n\n\\[ {e}^{-x}{x}^{s - 1} = \\lim {\\left( 1 - x/n\\right) }^{n}{x}^{s - 1}. \\] | We cannot just bluntly apply Theorem 19 since the functions on the right hand side are not integrable between 0 and \\( + \\infty \\) : convergence at 0 presupposes \\( \\operatorname{Re}\\left( s\\right) > 0 \\) and convergence at infinity \\( \\operatorname{Re}\\left( s\\right) < - n \\) . For \\( x < n \\) we always... | Yes |
Theorem 20. Let \( \sum {u}_{n}\left( x\right) \) be a series of absolutely integrable regulated functions on an interval \( X \) . Assume that (i) the series converges uniformly on every compact \( K \subset X \) ; (ii) there exists a positive function \( p\left( x\right) \), integrable on \( X \), such that \( \sum \... | The hypothesis (i) shows that the partial sums \( {s}_{n}\left( x\right) \) converge uniformly on every compact \( K \subset X \) ; since (ii) shows that \( \left| {{s}_{n}\left( x\right) }\right| \leq p\left( x\right) \), one need only apply the preceding theorem to the \( {s}_{n} \). | Yes |
Theorem 21. Let \( X \) be an interval and \( {u}_{n}\left( x\right) \) a series of regulated functions which converges normally on every compact \( K \subset X \) . Assume that\n\n\[ \sum \int \left| {{u}_{n}\left( x\right) }\right| {dx} < + \infty \]\n\nThen the function \( s\left( x\right) = \sum {u}_{n}\left( x\rig... | Let us put,\n\n\[ {m}_{I}\left( f\right) = {\int }_{I}f\left( x\right) {dx} \]\n\nand consider a compact interval \( K \subset X \) . Since the given series converges normally on \( K \) (\ | No |
Theorem 22. Let \( X \) be an interval, \( H \) a compact subset of \( \mathbb{C}, f \) a function defined and continuous on \( X \times H \) and \( \mu \) a function defined and regulated in \( X \) . Assume that there is a positive function \( p \) on \( X \) such that \( \left| {f\left( {x, y}\right) }\right| \leq p... | Hypotheses (i) and (iii) above are clearly satisfied by \( f\left( {x, y}\right) \mu \left( x\right) \) when \( y \) tends to a \( b \in H \) . If \( K \) is a compact subset of \( X \), then the function \( f \) is uniformly continuous on the compact \( K \times H \) ; consequently, the hypothesis (ii) is satisfied al... | No |
Theorem 23. Let \( f \) be a positive regulated function, defined for \( x \geq a > - \infty \) , decreasing, and tending to 0 at infinity. Then the integral\n\n\[ \varphi \left( y\right) = {\int }_{a}^{+\infty }f\left( x\right) \sin \left( {2\pi xy}\right) {dx} \]\n\nconverges for any \( y \neq 0 \), and is a continuo... | To see this, assume \( y > 0 \) and perform the change of variable \( {2xy} = u \), \n\nwhence\n\[ {2y\varphi }\left( y\right) = {\int }_{2ay}^{+\infty }f\left( {u/{2y}}\right) \sin \left( {\pi u}\right) {du}. \]\n\nConvergence is clear, and (3) can now be written\n\n(24.4)\n\[ \left| {{2y\varphi }\left( y\right) - {\i... | Yes |
Theorem 24. Let \( X \) and \( J \) be two intervals in \( \mathbb{R} \), let \( \mu \) be a regulated function on \( X \) and \( f \) a function defined and continuous on \( X \times J \). Assume that\n\n(i) the integral\n\n\[ g\left( y\right) = {\int }_{X}f\left( {x, y}\right) \mu \left( x\right) {dx} \]\n\nconverges... | Example 1. If \( X = Y | No |
If \( X = Y = \mathbb{R} \), if \( \mu \) is an absolutely integrable regulated function on \( \mathbb{R} \) and if \( f\left( {x, y}\right) = {e}^{-{2\pi ixy}} \), then the function \( g\left( y\right) \) is just the Fourier transform \( \widehat{\mu } \) of \( \mu \) . | Here\n\n\[ \n{D}_{2}f\left( {x, y}\right) = - {2\pi ix}{e}^{-{2\pi ixy}} \n\]\n\nand so \( \left| {{D}_{2}f\left( {x, y}\right) }\right| = {2\pi }\left| x\right| = p\left( x\right) \), and this is clearly the smallest positive function which dominates \( x \mapsto {D}_{2}f\left( {x, y}\right) \) for a (or for all) \( y... | Yes |
In particular choose \( \mu \left( x\right) = \exp \left( {-\pi {x}^{2}}\right) \), an integrable function on \( \mathbb{R} \) since it decreases at infinity more rapidly than \( {\left| x\right| }^{-n} \) for any \( n > 0 \) . We have \( - {2\pi ix\mu }\left( x\right) = i{\mu }^{\prime }\left( x\right) \), whence, int... | \[ {\widehat{\mu }}^{\prime }\left( y\right) = i{\int }_{-\infty }^{+\infty }{\mu }^{\prime }\left( x\right) \exp \left( {-{2\pi ixy}}\right) {dx} = - {2\pi y}{\int }_{-\infty }^{+\infty }\mu \left( x\right) \exp \left( {-{2\pi ixy}}\right) {dx} \] since the integrated-out part is zero because of the decrease of \( \mu... | Yes |
If \( \mu \) is a regulated function on the closed interval \( \lbrack 0, + \infty \lbrack \) and is \( O\left( {t}^{N}\right) \) at infinity for some \( N \), its Laplace transform or complex Fourier transform\n\n\[ \n{L}_{\mu }\left( z\right) = {\int }_{0}^{+\infty }{e}^{2\pi itz}\mu \left( t\right) {dt} \n\]\n\nis d... | Here \( f\left( {t, z}\right) = {e}^{2\pi itz} \), whence \( \left| {{f}^{\prime }\left( {t, z}\right) }\right| = {2\pi t}{e}^{-{2\pi ty}} \) . Since every compact subset \( H \) of \( U \) is contained in a half plane \( \operatorname{Im}\left( z\right) \geq \sigma > 0 \), we have, in \( H \) , that \( \left| {{f}^{\p... | Yes |
The function \( \Gamma \left( s\right) = \int {e}^{-x}{x}^{s - 1}{dx} \) is holomorphic in the half plane \( \operatorname{Re}\left( s\right) > 0 \) where it is defined. | It is clear that\n\n(i) the function \( s \mapsto {e}^{-x}{x}^{s - 1} = {e}^{-x}\exp \left\lbrack {\left( {s - 1}\right) \log x}\right\rbrack \) is holomorphic for every \( x > 0 \) since it is the composite of two holomorphic functions;\n\n(ii) its complex derivative \( {}^{48}{e}^{-x}{x}^{s - 1}\log x \) is continuou... | Yes |
\[ \Gamma \left( s\right) = {\int }_{0}^{1}{e}^{-x}{x}^{s - 1}{dx} + {\int }_{1}^{+\infty }{e}^{-x}{x}^{s - 1}{dx}. \] | The second integral converges for any \( s \in \mathbb{C} \) . So, as in the preceding example, is a holomorphic function of \( s \) in all of \( \mathbb{C} \) . In the first integral, term-by-term integration of the exponential series gives, for \( \operatorname{Re}\left( s\right) > 0 \) ,\n\n\[ {\int }_{0}^{1}{e}^{-x... | Yes |
Theorem 25 (Poor man’s Lebesgue-Fubini). Let \( X \) and \( Y \) be two intervals and \( f \) a continuous function on \( X \times Y \) . Suppose that the following conditions are satisfied:\n\n(i) for every compact \( K \subset X \) there exists a positive integrable function \( {q}_{K}\left( y\right) \) on \( Y \) su... | In what follows we shall put\n\n(26.3)\n\n\[{g}_{J}\left( x\right) = {\int }_{J}f\left( {x, y}\right) {dy},\;{h}_{I}\left( y\right) = {\int }_{I}f\left( {x, y}\right) {dx}\]\nfor any intervals \( I \subset X \) and \( J \subset Y \) . We shall also employ the notation \( {m}_{I} \) to denote an integral over \( I \) .\... | Yes |
First note that if the continuous functions \( f\left( x\right) \) and \( g\left( y\right) \) are defined and absolutely integrable on the intervals \( X \) and \( Y \) then the function \( f\left( x\right) g\left( y\right) \) is absolutely integrable on \( X \times Y \), and clearly\n\n\[ \n{\iint }_{X \times Y}f\left... | If, for each \( x \), one effects the change of variable \( y = \left( {{u}^{-1} - 1}\right) x \) in the \( y \) -integration, one finds\n\n(26.15)\n\[ \n\Gamma \left( a\right) \Gamma \left( b\right) = {\int }_{0}^{+\infty }{dx}{\int }_{0}^{1}{e}^{-u/x}{x}^{a + b - 1}{\left( 1 - u\right) }^{b - 1}{u}^{-b - 1}{du} = \n\... | Yes |
Consider on \( \mathbb{R} \) a function \( u\left( x\right) \) which is regulated, positive, with total integral 1, and put \( {u}_{n}\left( x\right) = {nu}\left( {nx}\right) \). The condition (D 1) is satisfied, also (D2) (change of variable \( {nx} = y \) in the integral) and condition (D 3) is satisfied because \[ {... | If \( u \) is of compact support the function \( {u}_{n}\left( x\right) = {nu}\left( {nx}\right) \) is zero for \( \left| x\right| \geq A/n \), i.e. outside an ever-shrinking interval with centre 0 ; the factor \( n \) in its definition shows that, on the other hand, it takes very large values on a neighbourhood of 0 ,... | Yes |
Theorem 26. Let \( \left( {u}_{n}\right) \) be a Dirac sequence. For every function \( f \) defined and continuous on \( \mathbb{R} \) one has\n\n\[ f\left( x\right) = \lim \int f\left( {x - y}\right) {u}_{n}\left( y\right) {dy} \]\n\nuniformly on every compact subset of \( \mathbb{R} \) if \( f \) is bounded or just i... | To establish (8) for \( f \) bounded it is enough to apply the lemma to the function \( y \mapsto f\left( {x - y}\right) \) . The little calculation in Dirac’s lemma shows moreover that\n\n\[ \left| {f\left( x\right) \int {u}_{n}\left( y\right) {dy}-\int f\left( {x - y}\right) {u}_{n}\left( y\right) {dy}}\right| \leq \... | Yes |
Theorem 27. For every function \( f \) defined and continuous on \( \mathbb{R} \) there exists a sequence \( {f}_{n} \) of \( {C}^{\infty } \) functions which converges to \( f \) uniformly on every compact subset of \( \mathbb{R} \) . If \( f \) is of compact support one may assume that the \( {f}_{n} \) are zero outs... | Obvious: one applies Theorem 26 to a Dirac sequence formed by functions of \( \mathcal{D} \), bearing in mind what we want to establish. If \( f \) is zero for \( \left| x\right| > A \) and if one assumes, for example, that the \( {\varphi }_{n} \) vanish for \( \left| x\right| \geq 1/n \), it is clear that the \( {f}_... | No |
Theorem 28 (Weierstrass, 1885). Let \( f \) be a real function defined and continuous on a compact interval \( K \subset \mathbb{R} \). Then there exists a sequence of polynomials which converges to \( f \) uniformly on \( K \). | It is enough to observe that \( f \) can be extended to a continuous function on all \( \mathbb{R} \), zero for \( \left| x\right| \) large: complete the graph of \( f \) by linear functions. | No |
Theorem 29 (Emile Borel,1895). For every sequence \( \left( {a}_{n}\right) \) of complex numbers there exists an indefinitely differentiable function \( f \) of compact support on \( \mathbb{R} \) such that \( {f}^{\left( n\right) }\left( 0\right) = {a}_{n} \) for every \( n \in \mathbb{N} \) . | Our first move, faced by this theorem, is to put\n\n(29.1)\n\n\[ f\left( x\right) = \sum {a}_{n}{x}^{n}/n! \]\n\nin accordance with Maclaurin's formula. Bad idea: the series has every chance of diverging for \( x \neq 0 \) .\n\nAll the same, (1) contains the germ of an idea for a proof. The function \( {a}_{n}{x}^{n}/n... | Yes |
Choose a function \( \mu \left( x\right) \), integrable (in the usual sense) on an interval \( K \subset \mathbb{R} \), and put\n\n\[ \mu \left( f\right) = \int f\left( x\right) \mu \left( x\right) {dx} \]\n\nfor every \( f \in L\left( K\right) \) . | Linearity is obvious and continuity follows from the inequalities\n\n\[ \left| {\mu \left( f\right) }\right| \leq \int \left| {f\left( x\right) }\right| \cdot \left| {\mu \left( x\right) }\right| {dx} \leq \parallel f{\parallel }_{K} \cdot \int \left| {\mu \left( x\right) }\right| {dx}. \]\n\nHere \( \parallel \mu \par... | Yes |
Choose a countable set \( D \) of points of \( K \) and, for every \( \xi \in D \), a number \( c\left( \xi \right) \in \mathbb{C} \) ; assuming \( \sum \left| {c\left( \xi \right) }\right| < + \infty \) one may define | \[ \mu \left( f\right) = \sum c\left( \xi \right) f\left( \xi \right) \] for every continuous function \( f \) on \( K \), the series being taken over \( D \) . No hypothesis on the compact set \( K \) is necessary here. | Yes |
Example 3. Take \( K = A \times B \) where \( A \) and \( B \) are compact intervals in \( \mathbb{R} \) and put\n\n\[ m\left( f\right) = {\iint }_{A \times B}f\left( {x, y}\right) {dxdy} \] | for every continuous function \( f \) on \( K\left( {\mathrm{n}}^{ \circ }\right. \) 9, Theorem 10). | No |
Take for \( X \) the open interval \( \rbrack 0, + \infty \lbrack \) and\n\n\[ \mu \left( f\right) = {\int }_{0}^{+\infty }f\left( x\right) {dx}/x \] | There is no problem with convergence for \( f \in L\left( X\right) \) since \( f\left( x\right) \) is zero on a neighbourhood of 0 and for \( x \) large. If \( f \) is zero outside \( K = \left\lbrack {u, v}\right\rbrack \) then\n\n\[ \left| {\mu \left( f\right) }\right| \leq \left( {\log v - \log u}\right) \parallel f... | Yes |
If one replaces the open interval \( \rbrack 0, + \infty \lbrack \) by the closed interval \( \lbrack 0, + \infty \lbrack \) the formula (6) is no longer meaningful, since, in this case, a function \( f \in L\left( X\right) \) is required to be zero for \( x \) large but not on a neighbourhood of 0, so allowing every c... | But one can replace \( 1/x \) by a function that poses no problem at the origin, and, for example, put\n\n(31.7)\n\n\[ \mu \left( f\right) = {\int }_{0}^{+\infty }f\left( x\right) {x}^{s}{dx}\;\text{ with }\operatorname{Re}\left( s\right) > - 1. \]\n\nIf \( f \) is zero outside \( K = \left\lbrack {0, v}\right\rbrack \... | Yes |
Choose a compact interval \( K \subset X \), a measure \( \mu \) on \( K \) and consider the linear form \( f \mapsto \int f\left( x\right) {d\mu }\left( x\right) \), where one integrates over \( K \), so involving only the values of \( f \) on this fixed compact set. | This example shows that a measure on \( K \) may also be considered as a measure on \( X \) : all the mass is supported by \( K \) . | No |
Example 4. For \( X = \mathbb{R} \) put\n\n(31.8)\n\n\[ \mu \left( f\right) = \sum f\left( n\right) \]\n\nsumming over \( \mathbb{Z} \). If \( f \) is zero outside a compact \( K \), only the \( n \in K \) count, whence \( \left| {\mu \left( f\right) }\right| \leq {M}_{K}\left( \mu \right) \parallel f{\parallel }_{X} \... | \[ \left| {\mu \left( f\right) }\right| \leq {M}_{K}\left( \mu \right) \parallel f{\parallel }_{X}\;\text{ where }\;{M}_{K}\left( \mu \right) = \mathop{\sum }\limits_{{\xi \in K}}\left| {c\left( \xi \right) }\right| \]\n\nsince the \( \xi \notin K \) do not appear in (9). If the total series \( \sum \left| {c\left( \xi... | Yes |
Theorem 31. Let \( X \) and \( Y \) be two intervals, \( \mu \) and \( \nu \) positive measures on \( X \) and \( Y \), and \( \lambda \) the product measure on \( X \times Y \) . Let \( \varphi \) be an lsc (resp. usc) function on \( X \times Y \) which is \( {}^{71} \) the upper (resp. lower) envelope of the \( f \in... | We shall examine the case of an usc function, the other being trivial to deduce from it: multiply the function by -1 . As always, the two crucial points will be that (a) a lower envelope of continuous functions is usc; (b) one may calculate the integral of an usc function from any decreasing philtre of continuous funct... | Yes |
Every absolutely integrable function \( f \) on every compact interval of \( \mathbb{R} \) (for example \( \log \left| x\right| \) despite its singularity at the origin) defines a distribution \( {}^{76} \) which is in fact a measure | \[ {T}_{f}\left( \varphi \right) = \int \varphi \left( x\right) f\left( x\right) {dx} \] | No |
Choose an \( a \in \mathbb{R} \), an integer \( k \in \mathbb{N} \) and put\n\n\[ T\left( \varphi \right) = {\varphi }^{\left( k\right) }\left( a\right) \] | For \( k = 0 \) one obtains the Dirac measure at the point \( a \), denoted by \( {\delta }_{a} \) or \( {\varepsilon }_{a} \) :\n\n\[ {\delta }_{a}\left( \varphi \right) = \varphi \left( a\right) \] | No |
Take for \( T \) the Dirac measure\n\n\[ T\left( \varphi \right) = \varphi \left( 0\right) \]\n\nWe find \( {T}^{\prime }\left( \varphi \right) = - {\varphi }^{\prime }\left( 0\right) \) in accordance with Dirac’s baroque formula to which we alluded at the beginning of \( {\mathrm{n}}^{ \circ }{27} \) . | One might, like Dirac himself, continue:\n\n\[ {T}^{\prime \prime }\left( \varphi \right) = + {\varphi }^{\prime \prime }\left( 0\right) ,\;{T}^{\prime \prime \prime }\left( \varphi \right) = - {\varphi }^{\prime \prime \prime }\left( 0\right) ,\text{ etc. } \] | No |
Take for \( f \) the function equal to 1 for \( x > 0 \) and to 0 for \( x < 0 \). Then \[ {T}_{f}\left( \varphi \right) = {\int }_{0}^{+\infty }\varphi \left( x\right) {dx} \] | \[ {T}_{f}^{\prime }\left( \varphi \right) = - {T}_{f}\left( {\varphi }^{\prime }\right) = - {\int }_{0}^{+\infty }{\varphi }^{\prime }\left( x\right) {dx} = \varphi \left( 0\right) \] since the primitive \( \varphi \) of \( {\varphi }^{\prime } \) is zero for \( x \) large. In other words, the derivative of the distri... | Yes |
Let us multiply term-by-term the relations\n\n\\[ \n{e}^{x} = 1 + x + {x}^{2}/2 + O\\left( {x}^{3}\\right) ,\\;\\sin x = x - {x}^{3}/6 + O\\left( {x}^{5}\\right) \n\\]\n\nvalid for \\( x \\rightarrow 0 \\) ; calculating \\( \\grave{a}{la} \\) Newton one finds | \n\n\\[ \n{e}^{x}\sin x = \\left( {1 + x + {x}^{2}/2}\\right) \\left( {x - {x}^{3}/6}\\right) + \\left( {1 + x + {x}^{2}/2}\\right) O\\left( {x}^{5}\\right) + \n\\]\n\n\\[ \n+ \\left( {x - {x}^{3}/6}\\right) O\\left( {x}^{3}\\right) + O\\left( {x}^{3}\\right) O\\left( {x}^{5}\\right) = \n\\]\n\n\\[ \n= x + {x}^{2} + {x... | Yes |
When \( x \rightarrow 0 \) , what is the expansion of \( {\left( {x}^{4} + {x}^{2}\right) }^{1/3} \)? | \[ {\left( {x}^{4} + {x}^{2}\right) }^{1/3} = {x}^{2/3}{\left( 1 + {x}^{2}\right) }^{1/3} = {x}^{2/3}\left\lbrack {1 + {x}^{2}/3 - {x}^{4}/9 + O\left( {x}^{6}\right) }\right\rbrack \] by the binomial series, whence \[ {\left( {x}^{4} + {x}^{2}\right) }^{1/3} = {x}^{2/3} + {x}^{8/3}/3 - {x}^{{14}/3}/9 + O\left( {x}^{{20... | Yes |
Consider the ratio\n\n\[ \frac{{x}^{2} - x + \log x}{{x}^{2} - {\left( \log x\right) }^{2}} \]\n\n as \( x \) tends to \( + \infty \) . | In the numerator, \( x \) and \( \log x \) are \( o\left( {x}^{2}\right) \), so it is \( \sim {x}^{2} \). In the denominator, \( \log x \) is \( o\left( x\right) \), so \( {\left( \log x\right) }^{2} \) is \( o\left( {x}^{2}\right) \), so that the denominator also is \( \sim {x}^{2} \). The fraction we are considering ... | Yes |
We seek a truncated expansion of order 1 at \( x = 0 \) of the function \( h\left( x\right) = {e}^{x}/{x}^{2}\sin x \). | Here \( h\left( x\right) \sim {x}^{-3} \), so that a relation of the form \( h\left( x\right) = \) \( p\left( x\right) + o\left( x\right) \) can be written as \( {x}^{3}h\left( x\right) = q\left( x\right) + o\left( {x}^{4}\right) \) . We have to find a truncated expansion of order 4 for\n\n\[ {x}^{3}h\left( x\right) = ... | Yes |
Theorem 2. If \( f \) is a continuous function on \( \mathbb{T} \) such that \( \sum \left| {\widehat{f}\left( n\right) }\right| < + \infty \) , then \[ f\left( u\right) = \sum \widehat{f}\left( n\right) {u}^{n}\;\text{ for every }u \in \mathbb{T}. \] | Let us denote the right hand side by \( g\left( u\right) \) . This is the sum of an absolutely convergent Fourier series, whence (Chap. V, \( {\mathrm{n}}^{ \circ }5 \) ) \( \widehat{g}\left( n\right) = \widehat{f}\left( n\right) \) for every \( n \) . Putting \( f = g + h \), we see that all the Fourier coefficients o... | No |
Theorem 3 (Parseval-Bessel \( {}^{14} \) ). Let \( f \) be a regulated periodic function. Then the series \( \sum {\left| \widehat{f}\left( n\right) \right| }^{2} \) is convergent and\n\n\[ \sum {\left| \widehat{f}\left( n\right) \right| }^{2} = \parallel f{\parallel }_{2}^{2} = \int {\left| f\left( u\right) \right| }^... | \[ \mathop{\lim }\limits_{{N \rightarrow \infty }}{\int }_{0}^{1}{\left| f\left( t\right) - \mathop{\sum }\limits_{{\left| n\right| \leq N}}\widehat{f}\left( n\right) {\mathbf{e}}_{n}\left( t\right) \right| }^{2}{dt} = 0. \] | Yes |
Consider the periodic function equal to \( t \) for \( \left| t\right| < \frac{1}{2} \) ; the values at the end-points are immaterial. Integrating by parts, we have, for \( n \neq 0 \) , | \[ \widehat{f}\left( n\right) = {\int }_{-\frac{1}{2}}^{\frac{1}{2}}t\overline{{\mathbf{e}}_{n}\left( t\right) }{dt} = {\left. \frac{t\overline{{\mathbf{e}}_{n}\left( t\right) }}{-{2\pi in}}\right| }_{-\frac{1}{2}}^{\frac{1}{2}} + \frac{1}{2\pi in}{\int }_{-\frac{1}{2}}^{\frac{1}{2}}\overline{{\mathbf{e}}_{n}\left( t\r... | Yes |
Consider the function of period 1 such that\n\n\[ f\left( t\right) = {e}^{2\pi izt}\;\text{ for }\left| t\right| < \frac{1}{2}, \]\n\nwhere \( z \) is a complex number, not an integer, since otherwise the interest of the problem evaporates. | We have\n\n\[ \widehat{f}\left( n\right) = {\int }_{-\frac{1}{2}}^{\frac{1}{2}}{e}^{{2\pi i}\left( {z - n}\right) t}{dt} = {\left. \frac{{e}^{{2\pi i}\left( {z - n}\right) t}}{{2\pi i}\left( {z - n}\right) }\right| }_{-\frac{1}{2}}^{\frac{1}{2}} \]\n\nsince, for every \( \lambda \in \mathbb{C} \), the derivative of \( ... | Yes |
Expansion of \( \cot z \) as a series of rational fractions. | Consider the function of period 1 on \( \mathbb{R} \) given by\n\n\[ f\left( t\right) = \cos {2\pi zt}\;\text{ for }\left| t\right| < \frac{1}{2}, \]\n\nwhere \( z \in \mathbb{C} \) is not a rational integer, for otherwise there would be no problem. Since \( f\left( {-\frac{1}{2}}\right) = f\left( \frac{1}{2}\right) \)... | Yes |
The Bernoulli polynomials are defined by the recurrence relations\n\n\[ \n{B}_{0}\left( x\right) = 1,\;{B}_{k}^{\prime }\left( x\right) = k{B}_{k - 1}\left( x\right) \n\]\n\nand by the condition\n\n\[ \n{B}_{k}\left( 0\right) = {B}_{k}\left( 1\right) \;\text{ for }k \geq 2. \n\] | The inventor was not acquainted with Fourier series, but condition (20) is exactly what one needs to transform the \( {B}_{k} \), for \( k \geq 2 \), into continuous periodic functions \( {B}_{k}^{ * } \), by putting\n\n\[ \n{B}_{k}^{ * }\left( t\right) = {B}_{k}\left( t\right) \;\text{ for }0 \leq t \leq 1 \n\]\n\nas ... | Yes |
For every regulated periodic function \( f \) the arithmetic means of the partial sums of the Fourier series of \( f \) converge to \( \frac{1}{2}\lbrack f\left( {t + }\right) + f\left( {t - }\right) \rbrack \) for any \( t \) . | To establish the first one writes, as in (11.13), (12.7) \[ f \star {F}_{N}\left( t\right) - \frac{1}{2}\left\lbrack {f\left( {t + }\right) + f\left( {t - }\right) }\right\rbrack = \] \[ = \int \left\lbrack {f\left( {t + s}\right) - f\left( {t + }\right) }\right\rbrack {F}_{N}\left( s\right) {ds} + \int \left\lbrack {f... | Yes |
Theorem 10 (Cauchy, 1831). Let \( f \) be a holomorphic function in an open set \( U \) in \( \mathbb{C} \) . Then \( f \) is analytic in \( U \) and, for every \( a \in U \), the Taylor series of \( f \) at a converges and represents \( f \) in the largest open disc with centre a contained in \( U \) . | It is enough, in the preceding arguments, to replace the disc \( \left| z\right| < R \) by the largest disc \( \left| {z - a}\right| < R \) in question or, if one prefers, to consider the function \( f\left( {a + z}\right) \) . Now the only power series that can possibly represent \( f \) on a neighbourhood of \( a \) ... | No |
Theorem 11. Let \( f \) be a holomorphic function in a connected open set \( U \) . Then \( f \) is constant if at a point of \( U \) it has either a local maximum or a non zero local minimum. | The case of a local minimum reduces to the preceding case on considering the function \( 1/f \) : this is defined and holomorphic on a neighbourhood of a local minimum of \( f \) and has a local maximum there; \( 1/f \) (and so \( f \) ) is thus constant on a disc, so \( f \) is constant on \( U \) . | Yes |
Corollary 1. Let \( G \) be a bounded domain in \( \mathbb{C}, K \) its closure, \( F = K - G \) its frontier and \( f \) a function defined and continuous in \( K \) and holomorphic in \( G \) . Then\n\n(15.4)\n\n\[ \parallel f{\parallel }_{G} = \parallel f{\parallel }_{K} = \parallel f{\parallel }_{F} \] | Since \( G \) is bounded, \( K \) is bounded and closed, hence compact. The continuous function \( \left| {f\left( z\right) }\right| \) therefore attains its maximum at a point \( a \in K \) . If \( a \in G \), Theorem 5 shows that \( f \) is constant in \( G \), hence in \( K \), and the corollary is obvious. If \( f ... | Yes |
Corollary 2. Let \( G \) be a bounded domain and \( \left( {f}_{n}\right) \) a sequence of functions defined and continuous on the closure \( K \) of \( G \) and holomorphic in \( G \) . Assume that the \( {f}_{n} \) converge uniformly on the boundary \( F \) of \( G \) to a limit function. Then the \( {f}_{n} \) conve... | Consider the functions \( {f}_{pq} = {f}_{p} - {f}_{q} \) . Cauchy’s criterion for uniform convergence shows that, for every \( r > 0 \), one has \( {\begin{Vmatrix}{f}_{pq}\end{Vmatrix}}_{F} \leq r \) for \( p \) and \( q \) large, and thus (Corollary 1) \( {\begin{Vmatrix}{f}_{pq}\end{Vmatrix}}_{K} \leq r \) . The \(... | No |
Corollary 3 ((H. A.) Schwarz’ lemma). Let \( f \) be a function holomorphic and bounded on a disc \( \left| z\right| < R \) and having a zero of order \( p \) at the origin. Then\n\n\[ \left| {f\left( z\right) }\right| \leq M{\left| z/R\right| }^{p}\;\text{ where }M = \sup \left| {f\left( z\right) }\right| . | The assumption about \( f \) implies that \( f\left( z\right) = {z}^{p}g\left( z\right) \) where \( g \) is, like \( f \) , the sum of a power series in \( \left| z\right| < R \) . The relation \( \left| {{z}^{p}g\left( z\right) }\right| \leq M \) shows that \( \left| {g\left( z\right) }\right| \leq M/{r}^{p} \) for \(... | Yes |
Theorem 13 (Laurent). Let \( f \) be a holomorphic function in an annulus \( C : {R}_{1} < \left| z\right| < {R}_{2} \). Then we have a series expansion\n\n(16.3)\n\n\[ f\left( z\right) = \mathop{\sum }\limits_{{n \in \mathbb{Z}}}{a}_{n}{z}^{n}\;\text{ with }\;{a}_{n} = {r}^{-n}{\int }_{\mathbb{T}}f\left( {ru}\right) {... | An expansion of this type is called a Laurent series; it is the sum of a power series in \( z \) and of a power series in \( 1/z \). The first converges at least for \( \left| z\right| < {R}_{2} \) and the second for \( \left| z\right| > {R}_{1} \) since a power series necessarily converges on a disc. This allows us to... | Yes |
Theorem 14 (Liouville). Let \( f \) be a holomorphic function of period 1 in an open strip \( B : a < \\operatorname{Im}\\left( z\\right) < b \) . We then have a series expansion\n\n\[ f\\left( z\\right) = \\sum {a}_{n}{e}^{2\\pi inz} \]\n\nwhich converges normally on every closed strip \( {B}^{\\prime } \\subset B \) ... | Conversely, if a complex Fourier series, i.e. of the form (5), converges absolutely in a strip \( a < \\operatorname{Im}\\left( z\\right) < b \), the preceding argument shows that the series converges normally on every closed strip (and so on every compact set) contained in the given open strip. Theorem 17 below will s... | No |
Theorem 15 (Liouville). Let \( f \) be an entire function such that\n\n(18.1)\n\n\[ f\left( z\right) = O\left( {z}^{p}\right) \;\text{ when }\;\left| z\right| \rightarrow + \infty ,\]\n\nwhere \( p \) is an integer \( \geq 0 \) . Then \( f \) is a polynomial of degree \( \leq p \) . In particular, a bounded entire func... | By Theorem 10 one has an expansion \( f\left( z\right) = \sum {a}_{n}{z}^{n} \) valid for any \( z \) . The relation (14.10) then shows that, for every \( n \) ,\n\n(18.2)\n\n\[ \left| {a}_{n}\right| {r}^{n} \leq {M}_{f}\left( r\right) \]\n\nwhere \( {M}_{f}\left( r\right) \) is the upper bound of \( \left| {f\left( z\... | Yes |
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