Difference between revisions of "Newman's Tauberian Theorem"

Latest revision as of 07:55, 12 August 2019

Newman's Tauberian Theorem is a tauberian theorem first proven by D.J. Newman in 1980, in his short proof of the prime number theorem.

Statement

Let $f:(0,+\infty)\to\mathbb C$ be a bounded function. Assume that its Laplace transform $F(s) = \int_0^\infty f(t)e^{-st}dt$ (which is well-defined by this formula for $\Re s>0$ ) admits an analytic extension (which we'll denote by the same letter $F$ ) to some open domain $E$ containing the closed half-plane $\{s\in\mathbb C : \Re s\ge 0\}$ . Then, the integral $\int_0^\infty f(t) dt$ converges and its value equals $F(0)$ .

Proof

For every $T>0$ , let $F_T(s) = \int_0^T f(t)e^{-st} dt$ . The function $F_T$ is defined and analytic on the entire complex plane $\mathbb C$ . The conclusion of the theorem is equivalent to the assertion $\lim_{T\to+\infty} F_T(0) = F(0)$ . We choose some large $R>0$ , and some arbitrarily small $\delta > 0$ such that $F$ is defined on the set $\{ z \in \mathbb{C} \mid \Re z \ge - \delta, \lvert z \rvert \le R \} .$ Let $\Gamma$ be the counterclockwise contour on the boundary of this set. Let $\Gamma_+$ be the restriction of this contour to the half-plane $\Re z \ge 0$ . Let $\Gamma_-^1$ be the restriction of the contour to the set $\{ z \in \mathbb{C} \mid \Re z \in (-\delta,0), \lvert z \rvert = R \} ,$ and let $\Gamma_-^2$ be the restriction to the set $\{ z \in \mathbb{C} \mid \Re z = -\delta, \lvert z \rvert \le R\}.$ Let $\Gamma_- = \Gamma_-^1 + \Gamma_-^2$ , as shown in the diagram below.

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By the Cauchy integral formula, we have $F(0)-F_T(0) = \frac{1}{2\pi i} \int_\Gamma K(z) (F(z)-F_T(z)) \frac{dz}{z} ,$ where $K(z) = \left( 1 + \frac{z^2}{R^2}\right) e^{zT} .$ We will estimate this integral separately in the left and right half-planes. In principle, $K$ could be arbitrary, but we have chosen $K$ to make it easier to estimate this integral.

We first estimate the difference $F(z) - F_T(z)$ for $\Gamma_+$ . Let $M$ be an upper bound for $\lvert f(x) \rvert$ . In the the right half-plane $\Re z > 0$ , we note that $\lvert F(z) - F_T(z) \rvert = \biggl\lvert \int_T^{\infty} f(x)e^{-zt} dt \biggr\rvert \le M \int_T^{\infty} e^{-\Re (z) t} dt = \frac{M e^{-\Re(z)T}}{\Re z} .$

Thus, we should kill the denominator $\Re z$ for the integral to converge. On the other hand, we can afford the kernel $K(z)$ growth as $e^{T\Re z}$ in the right half-plane, which will allow us corresponding decay in the left half-plane. Hence our choice $K(z) = \left(1+\frac{z^2}{R^2}\right)e^{Tz}.$ This is convenient because for $\lvert z \rvert = R$ , $K(z) = \frac{2 z \Re z}{R^2} e^{Tz} ,$ so that $K$ kills the unpleasant denominator $\Re z$ on $\Gamma_+$ .

We then have $\biggl\lvert \int_{\Gamma_+} K(z)\bigl[ F(z) - F_T(z) \bigr] \frac{dz}{z} \biggr\rvert \le\int_{\Gamma_+} \frac{2M}{R^2} \lvert dz \rvert = \frac{2\pi M}{R} .$

To estimate the integral over $\Gamma_-$ , we note that $K(z)F_T(z)/z$ is analytic in the left half-plane, so we may change the integration path to the left semicircle $\tilde\Gamma_-$ of radius $R$ . Now, on $\tilde\Gamma_-$ , we have $\lvert F_T(z) \rvert \le M \int_{0}^T \lvert e^{-zt} \rvert dt = M \frac{e^{-T\Re z } - 1}{\lvert\Re z\rvert} < M \frac{e^{-T \Re z}}{\lvert \Re z \rvert} .$ Then as before, $\biggl\lvert \int_{\Gamma_-} F_T(z)K(z) \frac{dz}{z} \biggr\rvert \le \frac{2\pi M}{R}.$

Now, let $N(R)$ be an upper bound for the quantity $\left\lvert F(z) \left(1 + \frac{z^2}{R^2} \right) \frac{1}{z} \right\rvert$ on $\Gamma_-$ . Then for $\delta < R$ , $\biggl\lvert \int_{\Gamma_-^1} F(z)K(z) \frac{dz}{z} \biggr\rvert \le N(R) \biggl\lvert \int_{\Gamma_-^1} e^{zT} dz \biggr\rvert < N(R) \cdot 4 \delta ,$ and $\biggl\lvert \int_{\Gamma_-^2} F(z)K(z) \frac{dz}{z} \biggr\rvert \le N(R) \int_{\Gamma_-^2} e^{-\delta T} \lvert dz \rvert < N(R) \cdot 2R e^{-\delta T} .$ Therefore $\lvert F(0) - F_T(0) \rvert \le \frac{4\pi M}{R} + N(R) \cdot 4\delta + N(R) \cdot 2Re^{-\delta T} .$ But as $T$ becomes arbitrarily large, the last term vanishes, so that $\limsup_{T \to \infty} \lvert F(0) - F_T(0) \rvert \le \frac{4\pi M}{R} + N(R)\cdot 4 \delta .$ We can make $\delta$ arbitrarily small, so that the second term vanishes. Then we pick an arbitrarily large $R$ , so that the first term vanishes, and the theorem follows. $\blacksquare$

@@ Line 1: / Line 1: @@
-{{stub}}
+'''Newman's Tauberian Theorem''' is a [[tauberian theorem]]
+first proven by D.J. Newman in 1980, in his short proof of
+the [[prime number theorem]].
 ==Statement==
-Let <math>f:(0,+\infty)\to\mathbb C</math> be a bounded function. Assume that its [[Laplace transform]] <math>F(s)=\int_0^\infty f(t)e^{-st}\,dt</math> (which is well-defined by this formula for <math>\Re s>0</math>) admits an analytic extension (which we'll denote by the same letter <math>F</math>) to some open domain <math>E</math> containing the closed half-plane <math>\{s\in\mathbb C\,:\,\Re s\ge 0\}</math>. Then the integral <math>\int_0^\infty f(t)\,dt</math> converges and its value equals <math>F(0)</math>.
+Let <math>f:(0,+\infty)\to\mathbb C</math> be a bounded function. Assume that
+its [[Laplace transform]] <math>F(s) = \int_0^\infty f(t)e^{-st}dt</math>
+(which is well-defined by this formula for <math>\Re s>0</math>) admits an
+analytic extension (which we'll denote by the same letter <math>F</math>)
+to some [[open set | open]] domain <math>E</math> containing the closed half-plane
+<math>\{s\in\mathbb C : \Re s\ge 0\}</math>. Then, the integral
+<math>\int_0^\infty f(t) dt</math> converges and its value equals <math>F(0)</math>.
 ==Proof==
-For every <math>T>0</math>, let <math>F_T(s)=\int_0^T f(t)e^{-st}\,dt</math>. The function <math>F_T</math> is defined and analytic on the entire complex plane <math>\mathbb C</math>. The conclusion of the theorem is equivalent to the assertion <math>\lim_{T\to+\infty} F_T(0)=F(0)</math>. Now, choose some big <math>R>0</math> and consider the contour <math>\Gamma=\Gamma_+\cup\Gamma_-</math> as on the picture below.
-[[Image:Newmans_Tauberian_Contour.PNG|Contour picture]]
+For every <math>T>0</math>, let <math>F_T(s) = \int_0^T f(t)e^{-st} dt</math>. The function
+<math>F_T</math> is defined and analytic on the entire complex plane <math>\mathbb C</math>.
+The conclusion of the theorem is equivalent to the assertion
+<math>\lim_{T\to+\infty} F_T(0) = F(0)</math>.  We choose some large <math>R>0</math>,
+and some arbitrarily small <math>\delta > 0</math>
+such that <math>F</math> is defined on the set
+<cmath> \{ z \in \mathbb{C} \mid \Re z \ge - \delta, \lvert z \rvert
+\le R \} . </cmath>
+Let <math>\Gamma</math> be the counterclockwise contour on the boundary
+of this set.  Let <math>\Gamma_+</math> be the restriction of this contour
+to the half-plane <math>\Re z \ge 0</math>.
+Let <math>\Gamma_-^1</math> be the restriction of the contour to the set
+<cmath> \{ z \in \mathbb{C} \mid \Re z \in (-\delta,0), \lvert z \rvert
+= R \} , </cmath>
+and let <math>\Gamma_-^2</math> be the restriction to the set
+<cmath> \{ z \in \mathbb{C} \mid \Re z = -\delta, \lvert z \rvert \le R\}.</cmath>
+Let <math>\Gamma_- = \Gamma_-^1 + \Gamma_-^2</math>, as shown in the diagram
+below.
+<!-- This code replaces the image given by
+[[Image:Newmans_Tauberian_Contour.PNG|Contour picture]] -->
+<asy>
+size(200);
+defaultpen(.7);
+path C=circle((0,0),1);
+path D=((-.3,-1.2)--(-.3,1.2));
+pair d1,d2;
+d1 = intersectionpoints(C,D)[0];
+d2 = intersectionpoints(C,D)[1];
+draw(d2..(1,0)..d1,MidArrow);
+draw(d1--d2);
+draw(d1..(-1,0)..d2,dashed);
+draw((-1.2,0)--(1.5,0),EndArrow);
+draw((0,-1.2)--(0,1.5),EndArrow);
-Here <math>\Gamma_+</math> is a semicircle and <math>\Gamma_-</math> is any smooth curve that lies to the left of the imaginary axis except for its endpoints and such that the domain <math>D</math> bounded by <math>\Gamma</math> is entirely contained in <math>E</math>. By the [[Cauchy Integral Formula|Cauchy integral formula]], we have
+label("\Large{$\Gamma_+$}",(1.1,.5));
+label("\Large{$\Gamma_-$}",(-.45,.4));
+label("\Large{$\tilde\Gamma_-$}",(-1.1,.6));
+</asy>
-<math>F(0)-F_T(0)=\frac 1{2\pi i}\int_\Gamma K(z)(F(z)-F_T(z))\,dz</math>
-where <math>K(z)</math> is any kernel that is analytic in some neighbourhood of <math>D</math> except for the point <math>0</math> where it must have a simple [[pole]] with [[residue]] <math>1</math>.
+By the [[Cauchy Integral Formula|Cauchy integral formula]], we have
+<cmath> F(0)-F_T(0) = \frac{1}{2\pi i} \int_\Gamma K(z) (F(z)-F_T(z))
+\frac{dz}{z} , </cmath>
+where
+<cmath> K(z) = \left( 1 + \frac{z^2}{R^2}\right) e^{zT} . </cmath>
+We will estimate this integral separately in the left and
+right half-planes.  In principle, <math>K</math> could be arbitrary, but
+we have chosen <math>K</math> to make it easier to estimate this
+integral.
-The trick is to choose an appropriate kernel (depending on <math>T</math>) that makes the integral easy to estimate. To make a good choice, let us first estimate the difference
+We first estimate the difference <math>F(z) - F_T(z)</math> for <math>\Gamma_+</math>.
-<math>F(z)-F_T(z)</math> on <math>\Gamma_+</math>. We have
+Let <math>M</math> be an upper bound for <math>\lvert f(x) \rvert</math>.
+In the the right half-plane <math>\Re z > 0</math>, we note that
+<cmath> \lvert F(z) - F_T(z) \rvert = \biggl\lvert \int_T^{\infty}
+f(x)e^{-zt} dt \biggr\rvert \le M \int_T^{\infty} e^{-\Re (z) t} dt
+= \frac{M e^{-\Re(z)T}}{\Re z} . </cmath>
-<math>|F(z)-F_T(z)|= \left|\int_T^{+\infty}f(t)e^{-zt}\,dt\right|
+Thus, we should kill the denominator <math>\Re z</math> for the integral
-\le M\int_T^{+\infty}e^{-\Re z\,t}\,dt=M\frac {e^{-\Re z\,T}}{\Re z}
+to converge. On the other hand, we can afford the kernel <math>K(z)</math>
-</math>
+growth as <math>e^{T\Re z}</math> in the right half-plane, which will allow
+us corresponding decay in the left half-plane.  Hence our choice
+<cmath> K(z) = \left(1+\frac{z^2}{R^2}\right)e^{Tz}.</cmath>
+This is convenient because for <math>\lvert z \rvert = R</math>,
+<cmath> K(z) = \frac{2 z \Re z}{R^2} e^{Tz} , </cmath>
+so that <math>K</math> kills the unpleasant denominator <math>\Re z</math>
+on <math>\Gamma_+</math>.
-where <math>M</math> is a bound for <math>|f|</math> on <math>(0,+\infty)</math>.
+We then have
-Thus, we should kill the denominator <math>\Re z</math> if we want the integral to converge. On the other hand, we can afford the kernel <math>K(z)</math> grow as <math>e^{T\Re z}</math> in the right half-plane (actually, we do not need any growth of <math>K(z)</math> in the right half-plane for its own sake, but we need some decay in the left half-plane to estimate the integral over <math>\Gamma_-</math> and it is impossible to get the latter without the first). This leads us to the choice
+<cmath> \biggl\lvert \int_{\Gamma_+} K(z)\bigl[ F(z) - F_T(z) \bigr]
+\frac{dz}{z} \biggr\rvert
+\le\int_{\Gamma_+} \frac{2M}{R^2} \lvert dz \rvert = \frac{2\pi M}{R} . </cmath>
-<math>K(z)=\left(\frac 1z+\frac z{R^2}\right)e^{Tz}</math>
+To estimate the integral over <math>\Gamma_-</math>, we note that
+<math>K(z)F_T(z)/z</math> is analytic in the left
+half-plane, so we may change the integration path to the left semicircle
+<math>\tilde\Gamma_-</math> of radius <math>R</math>. Now, on <math>\tilde\Gamma_-</math>, we have
+<cmath> \lvert F_T(z) \rvert \le M \int_{0}^T \lvert e^{-zt} \rvert dt
+= M \frac{e^{-T\Re z } - 1}{\lvert\Re z\rvert} < M \frac{e^{-T \Re
+z}}{\lvert \Re z \rvert} . </cmath>
+Then as before,
+<cmath> \biggl\lvert \int_{\Gamma_-} F_T(z)K(z) \frac{dz}{z} \biggr\rvert
+\le \frac{2\pi M}{R}. </cmath>
-Note that <math>K(\pm iR)=0</math>, so the unpleasant denominator <math>\Re z</math> is, indeed, killed by <math>K</math> on <math>\Gamma_+</math>. Also, <math>K</math> decays in the left half-plane as fast as only is allowed by the exponential growth restriction in the right half-plane. This is not the only possible choice that will work, of course, but it is the simplest and the most
-elegant one.
-Once this tricky choice is made, the rest is fairly straightforward.
+Now, let <math>N(R)</math> be an upper bound for the quantity
-The integral over <math>\Gamma_+</math> does not exceed <math>\frac {2\pi M}{R}</math> (just use the standard parametrization <math>z=Re^{i\theta}</math> and notice that <math>\frac 1z+\frac z{R^2}=\frac 2R\cos\theta=\frac 2{R^2}\Re z</math> and that the exponent in the kernel essentially cancels the exponent in the estimate for <math>|F(0)-F_T(0)|</math>). To estimate the integral over <math>\Gamma_-</math>, just write <math>\left|\int_{\Gamma_-}K(z)(F(z)-F_T(z))\,dz\right|\le
+<cmath> \left\lvert F(z) \left(1 + \frac{z^2}{R^2} \right) \frac{1}{z}
-\left|\int_{\Gamma_-}K(z)F(z)\,dz\right|+
+\right\rvert </cmath>
-\left|\int_{\Gamma_-}K(z)F_T(z)\,dz\right|=I_1+I_2 </math>.
+on <math>\Gamma_-</math>.  Then for <math>\delta < R</math>,
+<cmath> \biggl\lvert \int_{\Gamma_-^1} F(z)K(z) \frac{dz}{z} \biggr\rvert
+\le N(R) \biggl\lvert \int_{\Gamma_-^1} e^{zT} dz \biggr\rvert
+< N(R) \cdot 4 \delta , </cmath>
+and
+<cmath> \biggl\lvert \int_{\Gamma_-^2} F(z)K(z) \frac{dz}{z} \biggr\rvert
+\le N(R) \int_{\Gamma_-^2} e^{-\delta T} \lvert dz \rvert
+< N(R) \cdot 2R e^{-\delta T} .</cmath>
+Therefore
+<cmath> \lvert F(0) - F_T(0) \rvert \le \frac{4\pi M}{R} + N(R) \cdot
+\delta + N(R) \cdot 2Re^{-\delta T} . </cmath>
+But as <math>T</math> becomes arbitrarily large, the last term vanishes, so
+that
+<cmath> \limsup_{T \to \infty} \lvert F(0) - F_T(0) \rvert
+\le \frac{4\pi M}{R} + N(R)\cdot 4 \delta . </cmath>
+We can make <math>\delta</math> arbitrarily small, so that the second term
+vanishes.  Then we pick an arbitrarily large <math>R</math>, so that the
+first term vanishes, and the theorem follows.
+<math>\blacksquare</math>
-To estimate <math>I_2</math>, note that <math>K(z)F_T(z)</math>
+==See also==
-is analytic in the left half-plane, so we may change the integration path ot the left semicircle <math>\tilde\Gamma_-</math>. Now, on <math>\tilde\Gamma_-</math>, we have
-<math>|F_T(z)|\le M\int_0^T e^{-\Re z\,t}\,dt= M\frac {e^{-T\,\Re z}-1}{\Re z}\le M\frac {e^{-T\,\Re z}}{\Re z}</math>.
+* [[Tauberian theorem]]
+* [[Prime Number Theorem|Prime number theorem]]
-This yields the estimate <math>I_2\le\frac{2\pi M}{R}</math> in the same way as for the integral over <math>\Gamma_+</math>. At last, note that, for fixed <math>R</math>, the integrand in <math>I_1</math> is uniformly bounded and tends to <math>0</math> at every point of <math>\Gamma_-</math> as <math>T\to +\infty</math>. This allows to conclude (by the Lebesgue [[Dominated Convergence Theorem|dominated convergence theorem]] or in some more elementary way) that <math>I_1\to 0</math> as <math>T\to\infty</math>. Thus, given any <math>\varepsilon>0</math>, we can first fix <math>R>0</math>
+[[Category:Complex analysis]]
-such that <math>\frac{4\pi M}R<\frac\varepsilon 2</math> and then choose <math>T_0>0</math> such that <math>I_1\le\frac\varepsilon 2</math>
-for <math>T>T_0</math>. Then <math>|F(0)-F_T(0)|<\varepsilon</math> for <math>T>T_0</math> and we are done.

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Difference between revisions of "Newman's Tauberian Theorem"

Latest revision as of 07:55, 12 August 2019

Statement

Proof

See also