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Difference between revisions of "Newman's Tauberian Theorem"

 
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{{stub}}
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'''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==
 
==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==
 
==Proof==
  
−
[[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);
 +
 
 +
label("\Large{$\Gamma_+$}",(1.1,.5));
 +
label("\Large{$\Gamma_-$}",(-.45,.4));
 +
label("\Large{$\tilde\Gamma_-$}",(-1.1,.6));
 +
</asy>
 +
 
 +
 
 +
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.
 +
 
 +
We first estimate the difference <math>F(z) - F_T(z)</math> for <math>\Gamma_+</math>.
 +
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>
 +
 
 +
Thus, we should kill the denominator <math>\Re z</math> for the integral
 +
to converge. On the other hand, we can afford the kernel <math>K(z)</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>.
 +
 
 +
We then have
 +
<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>
 +
 
 +
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>
 +
 
 +
 
 +
Now, let <math>N(R)</math> be an upper bound for the quantity
 +
<cmath> \left\lvert F(z) \left(1 + \frac{z^2}{R^2} \right) \frac{1}{z}
 +
\right\rvert </cmath>
 +
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
 +
4\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>
 +
 
 +
==See also==
 +
 
 +
* [[Tauberian theorem]]
 +
* [[Prime Number Theorem|Prime number theorem]]
 +
 
 +
[[Category:Complex analysis]]

Latest revision as of 06:55, 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.

[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); label("\Large{$\Gamma_+$}",(1.1,.5)); label("\Large{$\Gamma_-$}",(-.45,.4)); label("\Large{$\tilde\Gamma_-$}",(-1.1,.6)); [/asy]


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. $\blacksquare$

See also

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