Newman's Tauberian Theorem

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.


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 $F(0)$.


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.

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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. $\blacksquare$

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