Difference between revisions of "Newman's Tauberian Theorem"

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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==
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
 +
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>
  
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.
 

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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