Difference between revisions of "Newman's Tauberian Theorem"
(→Proof) |
|||
Line 4: | Line 4: | ||
==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]] | [[Image:Newmans_Tauberian_Contour.PNG|Contour picture]] | ||
+ | |||
+ | 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 | ||
+ | |||
+ | <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>. | ||
+ | |||
+ | 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 | ||
+ | <math>F(z)-F_T(z)</math> on <math>\Gamma_+</math>. We have | ||
+ | |||
+ | <math>|F(z)-F_T(z)|= \left|\int_T^{+\infty}f(t)e^{-zt}\,dt\right| | ||
+ | \le M\int_T^{+\infty}e^{-\Re z\,t}\,dt=M\frac {e^{-\Re z\,T}}{\Re z} | ||
+ | </math> | ||
+ | |||
+ | where <math>M</math> is a bound for <math>|f|</math> on <math>(0,+\infty)</math>. | ||
+ | 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 | ||
+ | |||
+ | <math>K(z)=\left(\frac 1z+\frac z{R^2}\right)e^{Tz}</math> | ||
+ | |||
+ | 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. | ||
+ | The integral over <math>\Gamma_+</math> does not exceed | ||
+ | |||
+ | ''To be continued'' |
Revision as of 18:52, 3 July 2006
This article is a stub. Help us out by expanding it.
Statement
Let be a bounded function. Assume that its Laplace transform
(which is well-defined by this formula for
) admits an analytic extension (which we'll denote by the same letter
) to some open domain
containing the closed half-plane
. Then the integral
converges and its value equals
.
Proof
For every , let
. The function
is defined and analytic on the entire complex plane
. The conclusion of the theorem is equivalent to the assertion
. Now, choose some big
and consider the contour
as on the picture below.
Here is a semicircle and
is any smooth curve that lies to the left of the imaginary axis except for its endpoints and such that the domain
bounded by
is entirely contained in
. By the Cauchy integral formula, we have
where is any kernel that is analytic in some neighbourhood of
except for the point
where it must have a simple pole with residue
.
The trick is to choose an appropriate kernel (depending on ) that makes the integral easy to estimate. To make a good choice, let us first estimate the difference
on
. We have
where is a bound for
on
.
Thus, we should kill the denominator
if we want the integral to converge. On the other hand, we can afford the kernel
grow as
in the right half-plane (actually, we do not need any growth of
in the right half-plane for its own sake, but we need some decay in the left half-plane to estimate the integral over
and it is impossible to get the latter without the first). This leads us to the choice
Note that , so the unpleasant denominator
is, indeed, killed by
on
. Also,
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.
The integral over does not exceed
To be continued