Newman's Tauberian Theorem

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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 $\lim_{T\to+\infty} F_T(0)=F(0)$ . Now, choose some big $R>0$ and consider the contour $\Gamma=\Gamma_+\cup\Gamma_-$ as on the picture below.

Here $\Gamma_+$ is a semicircle and $\Gamma_-$ is any smooth curve that lies to the left of the imaginary axis except for its endpoints and such that the domain $D$ bounded by $\Gamma$ is entirely contained in $E$ . By the Cauchy integral formula, we have

$F(0)-F_T(0)=\frac 1{2\pi i}\int_\Gamma K(z)(F(z)-F_T(z))\,dz$

where $K(z)$ is any kernel that is analytic in some neighbourhood of $D$ except for the point $0$ where it must have a simple pole with residue $1$ .

The trick is to choose an appropriate kernel (depending on $T$ ) that makes the integral easy to estimate. To make a good choice, let us first estimate the difference $F(z)-F_T(z)$ on $\Gamma_+$ . We have

$|F(z)-F_T(z)|= \left|\int_T^{+\infty}f(t)e^{-zt}\,dt\right| \le M\int_T^{+\infty}e^{-\Re z\,t}\,dt=M\frac {e^{-\Re z\,T}}{\Re z}$

where $M$ is a bound for $|f|$ on $(0,+\infty)$ . Thus, we should kill the denominator $\Re z$ if we want the integral to converge. On the other hand, we can afford the kernel $K(z)$ grow as $e^{T\Re z}$ in the right half-plane (actually, we do not need any growth of $K(z)$ in the right half-plane for its own sake, but we need some decay in the left half-plane to estimate the integral over $\Gamma_-$ and it is impossible to get the latter without the first). This leads us to the choice

$K(z)=\left(\frac 1z+\frac z{R^2}\right)e^{Tz}$

Note that $K(\pm iR)=0$ , so the unpleasant denominator $\Re z$ is, indeed, killed by $K$ on $\Gamma_+$ . Also, $K$ 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 $\Gamma_+$ does not exceed $\frac {2\pi M}{R}$ (just use the standard parametrization $z=Re^{i\theta}$ and notice that $\frac 1z+\frac z{R^2}=\frac 2R\cos\theta=\frac 2{R^2}\Re z$ and that the exponent in the kernel essentially cancels the exponent in the estimate for $|F(0)-F_T(0)|$ ). To estimate the integral over $\Gamma_-$ , just write $\left|\int_{\Gamma_-}K(z)(F(z)-F_T(z))\,dz\right|\le \left|\int_{\Gamma_-}K(z)F(z)\,dz\right|+ \left|\int_{\Gamma_-}K(z)F_T(z)\,dz\right|=I_1+I_2$ .

To estimate $I_2$ , note that $K(z)F_T(z)$ is analytic in the left half-plane, so we may change the integration path ot the left semicircle $\tilde\Gamma_-$ . Now, on $\tilde\Gamma_-$ , we have

$|F_T(z)|\le M\int_0^T e^{-\Re z\,t}\,dt= M\frac {e^{-T\,\Re z}-1}{\Re z}\le M\frac {e^{-T\,\Re z}}{\Re z}$ .

This yields the estimate $I_2\le\frac{2\pi M}{R}$ in the same way as for the integral over $\Gamma_+$ . At last, note that, for fixed $R$ , the integrand in $I_1$ is uniformly bounded and tends to $0$ at every point of $\Gamma_-$ as $T\to +\infty$ . This allows to conclude (by the Lebesgue dominated convergence theorem or in some more elementary way) that $I_1\to 0$ as $T\to\infty$ . Thus, given any $\varepsilon>0$ , we can first fix $R>0$ such that $\frac{4\pi M}R<\frac\varepsilon 2$ and then choose $T_0>0$ such that $I_1\le\frac\varepsilon 2$ for $T>T_0$ . Then $|F(0)-F_T(0)|<\varepsilon$ for $T>T_0$ and we are done.

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Newman's Tauberian Theorem

Statement

Proof