Difference between revisions of "Newman's Tauberian Theorem"
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==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 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== | ||
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[[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 | + | 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> | <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>. | + | 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 | 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 | ||
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where <math>M</math> is a bound for <math>|f|</math> on <math>(0,+\infty)</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 | + | 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> | + | <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 | 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 | ||
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Once this tricky choice is made, the rest is fairly straightforward. | Once this tricky choice is made, the rest is fairly straightforward. | ||
| − | 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 | + | 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 |
\left|\int_{\Gamma_-}K(z)F(z)\,dz\right|+ | \left|\int_{\Gamma_-}K(z)F(z)\,dz\right|+ | ||
\left|\int_{\Gamma_-}K(z)F_T(z)\,dz\right|=I_1+I_2 </math>. | \left|\int_{\Gamma_-}K(z)F_T(z)\,dz\right|=I_1+I_2 </math>. | ||
To estimate <math>I_2</math>, note that <math>K(z)F_T(z)</math> | To estimate <math>I_2</math>, note that <math>K(z)F_T(z)</math> | ||
| − | is analytic in the left half-plane, so we may change the integration path | + | is analytic in the left half-plane, so we may change the integration path to 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>. | <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>. | ||
Revision as of 16:55, 5 July 2006
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 
(just use the standard parametrization 
and notice that 
, and that the exponent in the kernel essentially cancels the exponent in the estimate for 
). To estimate the integral over , just write 
.
To estimate 
, note that 
is analytic in the left half-plane, so we may change the integration path to the left semicircle 
. Now, on , we have
.
This yields the estimate 
in the same way as for the integral over . At last, note that, for fixed 
, the integrand in 
is uniformly bounded and tends to at every point of as 
. This allows to conclude (by the Lebesgue dominated convergence theorem or in some more elementary way) that 
as 
. Thus, given any 
, we can first fix
such that 
and then choose 
such that 
for 
. Then 
for and we are done.
