§12.3.1 The Wiener–Ikehara Theorem

12.3.1 The Wiener–Ikehara Theorem

Previously we have seen the subtle relation between $𝜁$ and $𝜓$ . Given the properties of the Laplace transform of $𝜓 \circ exp$ and certain conditions established above, we apply now prove what is known as a Tauberian theorem to extract sufficient information (such as convergence) of the function itself.

For arbitrary $𝑥 \in ℝ$ and $𝜆 > 0$ , define a triangular kernel cutoff by

𝐾_{𝜆} (𝑥) = {\begin{cases} 1 - \frac{| 𝑥 |}{2 𝜆} & | 𝑥 | < 2 𝜆 \\ 0 & otherwise \end{cases}, supp 𝐾_{𝜆} = [- 2 𝜆, 2 𝜆],

and a normalized Fejér kernel by

𝑘_{𝜆} (𝑥) = {\begin{cases} \frac{2 𝜆}{\sqrt{2 π}} {(\frac{sin (𝜆 𝑥)}{𝜆 𝑥})}^{2} & 𝑥 \neq 0 \\ \frac{2 𝜆}{\sqrt{2 π}} & 𝑥 = 0 \end{cases} .

The two kernels are related by a angular unitary Fourier transform:

Lemma 12.3.1.1.

For every $𝜆 > 0$ , we have that

ℱ︀ {𝐾_{𝜆}} (𝑥) = {ℱ︀}^{- 1} {𝐾_{𝜆}} (𝑥) = 𝑘_{𝜆} (𝑥), ℱ︀ {𝑘_{𝜆}} (𝑥) = {ℱ︀}^{- 1} {𝑘_{𝜆}} (𝑥) = 𝐾_{𝜆} (𝑥),

where we define

ℱ︀ {𝑓} (𝑥) = \frac{1}{\sqrt{2 π}} \int_{- \infty}^{\infty} 𝑓 (𝑡) e^{- i 𝑥 𝑡} d 𝑡, {ℱ︀}^{- 1} {𝑓} (𝑥) = \frac{1}{\sqrt{2 π}} \int_{- \infty}^{\infty} 𝑓 (𝑡) e^{i 𝑥 𝑡} d 𝑡 .

Proof.

The equivalence between the Fourier and inverse Fourier transforms follows from the evenness of both kernels. We hence prove only the forward transform equivalences.

Since $supp 𝐾_{𝜆} = [- 2 𝜆, 2 𝜆]$ , we have that

\begin{aligned} ℱ︀ {𝐾_{𝜆}} (𝑥) & = \frac{1}{\sqrt{2 π}} \int_{- 2 𝜆}^{2 𝜆} (1 - \frac{| 𝑡 |}{2 𝜆}) e^{- i 𝑥 𝑡} d 𝑡 \\ = \frac{1}{\sqrt{2 π}} \int_{0}^{2 𝜆} (1 - \frac{𝑡}{2 𝜆}) (e^{- i 𝑥 𝑡} + e^{i 𝑥 𝑡}) d 𝑡 \\ = \frac{\sqrt{2}}{\sqrt{π}} \int_{0}^{2 𝜆} (1 - \frac{𝑡}{2 𝜆}) cos (𝑥 𝑡) d 𝑡 \\ = \frac{\sqrt{2}}{\sqrt{π}} [{\frac{sin (𝑥 𝑡)}{𝑥} |}_{0}^{2 𝜆} - {\frac{𝑡 sin (𝑥 𝑡)}{2 𝜆 𝑥} |}_{0}^{2 𝜆} + \int_{0}^{2 𝜆} \frac{sin (𝑥 𝑡)}{2 𝜆 𝑥} d 𝑡] \\ = \frac{\sqrt{2}}{\sqrt{π}} [\frac{sin (2 𝜆 𝑥)}{𝑥} - \frac{sin (2 𝜆 𝑥)}{𝑥} - {\frac{cos (𝑥 𝑡)}{2 𝜆 𝑥^{2}} |}_{0}^{2 𝜆}] \\ = \frac{\sqrt{2}}{\sqrt{π}} \frac{[1 - cos (2 𝜆 𝑥)]}{2 𝜆 𝑥^{2}} = \frac{2 𝜆}{\sqrt{2 π}} {(\frac{sin (𝜆 𝑥)}{𝜆 𝑥})}^{2} = 𝑘_{𝜆} (𝑥) . \end{aligned}

On the other hand,

\begin{aligned} ℱ︀ {𝑘_{𝜆}} (𝑥) & = \frac{2 𝜆}{2 π} \int_{- \infty}^{\infty} {(\frac{sin (𝜆 𝑡)}{𝜆 𝑡})}^{2} e^{- i 𝑥 𝑡} d 𝑡 = \frac{2}{π 𝜆} \int_{0}^{\infty} \frac{{sin}^{2} (𝜆 𝑡)}{𝑡^{2}} cos (𝑥 𝑡) d 𝑡 \\ = \frac{1}{π 𝜆} \int_{0}^{\infty} \frac{1 - cos (2 𝜆 𝑡)}{𝑡^{2}} cos (𝑥 𝑡) d 𝑡 . \end{aligned}

Let $𝐼 (𝑥) = \int_{0}^{\infty} \frac{1 - cos (2 𝜆 𝑡)}{𝑡^{2}} cos (𝑥 𝑡) d 𝑡$ . Differentiation under the integral sign yields

\frac{d 𝐼}{d 𝑥} = \int_{0}^{\infty} \frac{𝜕}{𝜕 𝑥} (\frac{1 - cos (2 𝜆 𝑡)}{𝑡^{2}} cos (𝑥 𝑡)) d 𝑡 = \int_{0}^{\infty} (\frac{cos (2 𝜆 𝑡) - 1}{𝑡}) sin (𝑥 𝑡) d 𝑡 .

By the product-to-sum formulas, this can be rewritten as

\begin{aligned} \frac{d 𝐼}{d 𝑥} & = \int_{0}^{\infty} \frac{sin ((𝑥 + 2 𝜆) 𝑡)}{2 𝑡} d 𝑡 + \int_{0}^{\infty} \frac{sin ((𝑥 - 2 𝜆) 𝑡)}{2 𝑡} d 𝑡 - \int_{0}^{\infty} \frac{sin (𝑥 𝑡)}{𝑡} d 𝑡 \\ = I + I I - I I I . \end{aligned}

By substituting $𝑢 = 𝑡 (𝑥 + 2 𝜆)$ , we have that

I = \int_{0}^{\pm \infty} \frac{sin 𝑢}{2 𝑢} d 𝑢, where \pm \infty = {\begin{cases} + \infty & 𝑥 + 2 𝜆 > 0 \\ - \infty & 𝑥 + 2 𝜆 < 0 \end{cases}

and substituting $𝑢 = 𝑡 (𝑥 - 2 𝜆)$ gives

I I = \int_{0}^{\pm \infty} \frac{sin 𝑢}{2 𝑢} d 𝑢, where \pm \infty = {\begin{cases} + \infty & 𝑥 - 2 𝜆 > 0 \\ - \infty & 𝑥 - 2 𝜆 < 0 \end{cases} .

By the Dirichlet integral $\int_{0}^{\infty} \frac{sin 𝑢}{𝑢} d 𝑢 = \frac{π}{2}$ , we have that

I = \frac{π}{4} sgn (𝑥 + 2 𝜆), I I = \frac{π}{4} sgn (𝑥 - 2 𝜆) .

Similarly, $I I I = \frac{π}{2} sgn (𝑥)$ . Hence,

\frac{d 𝐼}{d 𝑥} = \frac{π}{4} [sgn (𝑥 + 2 𝜆) + sgn (𝑥 - 2 𝜆) - 2 sgn (𝑥)] = \frac{π}{2} {\begin{cases} 0 & | 𝑥 | > 2 𝜆 \\ 1 & - 2 𝜆 < 𝑥 < 0 \\ - 1 & 0 < 𝑥 < 2 𝜆 \end{cases},

and

𝐼 (0) + \int_{0}^{𝑥} \frac{d 𝐼}{d 𝑡} d 𝑡 = 𝐼 (0) + {\begin{cases} \int_{0}^{- 2 𝜆} \frac{d 𝐼}{d 𝑡} d 𝑡 + \int_{- 2 𝜆}^{𝑥} \frac{d 𝐼}{d 𝑡} d 𝑡 & 𝑥 < - 2 𝜆 \\ \int_{0}^{𝑥} \frac{d 𝐼}{d 𝑡} d 𝑡 & - 2 𝜆 < 𝑥 < 0 \\ \int_{0}^{𝑥} \frac{d 𝐼}{d 𝑡} d 𝑡 & 0 < 𝑥 < 2 𝜆 \\ \int_{0}^{2 𝜆} \frac{d 𝐼}{d 𝑡} d 𝑡 + \int_{2 𝜆}^{𝑥} \frac{d 𝐼}{d 𝑡} d 𝑡 & 𝑥 > 2 𝜆 \end{cases}

which after simplification, becomes

𝐼 (𝑥) = 𝐼 (0) + π {\begin{cases} - 𝜆 & 𝑥 < - 2 𝜆 \\ \frac{𝑥}{2} & - 2 𝜆 < 𝑥 < 0 \\ - \frac{𝑥}{2} & 0 < 𝑥 < 2 𝜆 \\ - 𝜆 & 𝑥 > 2 𝜆 \end{cases} .

Since

\begin{aligned} 𝐼 (0) & = \int_{0}^{\infty} \frac{1 - cos (2 𝜆 𝑡)}{𝑡^{2}} d 𝑡 = {\frac{cos (2 𝜆 𝑡) - 1}{𝑡} |}_{0}^{\infty} + \int_{0}^{\infty} 2 𝜆 \frac{sin (2 𝜆 𝑡)}{𝑡} d 𝑡 \\ = lim_{𝑡 \to 0} \frac{1 - cos (2 𝜆 𝑡)}{𝑡} + 2 𝜆 \frac{π}{2} = π 𝜆, \end{aligned}

we obtain

𝐼 = {\begin{cases} 0 & 𝑥 < - 2 𝜆 \\ π (𝜆 + \frac{𝑥}{2}) & - 2 𝜆 < 𝑥 < 0 \\ π (𝜆 - \frac{𝑥}{2}) & 0 < 𝑥 < 2 𝜆 \\ 0 & 𝑥 > 2 𝜆 \end{cases} ⟹ ℱ︀ {𝑘_{𝜆}} (𝑥) = {\begin{cases} 0 & 𝑥 < - 2 𝜆 \\ 1 + \frac{𝑥}{2 𝜆} & - 2 𝜆 < 𝑥 < 0 \\ 1 - \frac{𝑥}{2 𝜆} & 0 < 𝑥 < 2 𝜆 \\ 0 & 𝑥 > 2 𝜆 \end{cases},

which confirms the second equivalence. ⁠ $□$

We note that, in a distribution-theoretic treatment, it would suffice to prove one of the above identities, the other then following immediately from Fourier inversion and the Dirac delta. Without the framework of distributions however, the use of the Dirac delta is generally unjustified. For completeness, we have instead given a direct classical proof.

Definition 12.3.1.2.

A function $𝑓 : ℝ \to ℝ$ is said to be slowly decreasing if

\underset{𝛿 \to 0^{+}}{lim inf} \underset{𝑥 \to \infty}{lim inf} [𝑓 (𝑥 + 𝛿) - 𝑓 (𝑥)] \geq 0,

or equivalently, for every $𝜀 > 0$ , $\exists 𝑥_{0}$ , $\exists 𝛿 > 0$ such that

\forall 𝑥, 𝑦 > 𝑥_{0}, 0 < 𝑦 - 𝑥 < 𝛿 ⟹ 𝑓 (𝑦) - 𝑓 (𝑥) > - 𝜀 .

Proposition 12.3.1.3.

Let $𝑓 : ℝ \to ℝ$ be a slowly decreasing function bounded by $𝑀 > 0$ . If the limit of the convolution $𝑓 * 𝑘_{𝜆}$ given by

𝐿 \equiv lim_{𝑥 \to \infty} \frac{1}{\sqrt{2 π}} (𝑓 * 𝑘_{𝜆}) (𝑥) = lim_{𝑥 \to \infty} \frac{1}{\sqrt{2 π}} \int_{- \infty}^{\infty} 𝑓 (𝑡) 𝑘_{𝜆} (𝑥 - 𝑡) d 𝑡

is independent of

𝜆 > 0

, then

{lim}_{𝑥 \to \infty} 𝑓 (𝑥) = 𝐿

Proof.

Let $𝜀 > 0$ be arbitrary. Assume, for the sake of contradiction, that $𝑓$ does not converge to $𝐿$ as $𝑥 \to \infty$ . Then there exists some sequence ${𝑥_{𝑛}}_{𝑛 \in ℕ}$ ( $𝑥_{𝑛} \to \infty$ ) such that $| 𝑓 (𝑥_{𝑛}) - 𝐿 | > 𝜀$ for all $𝑛$ . From here, by the infinite pigeonhole principle we may extract an infinite subsequence (continued to be denoted by ${𝑥_{𝑛}}$ ) such that one of the two cases is assumed:

$𝑓 (𝑥_{𝑛}) - 𝐿 > 𝜀$ for all $𝑛 \in ℕ$ . By the slow decrease of $𝑓$ , there exist $𝛿 > 0$ and $𝑁 \in ℕ$ such that
$𝑓 (𝑦) > 𝑓 (𝑥_{𝑛}) - \frac{𝜀}{2} > 𝐿 + \frac{𝜀}{2}$
for any $𝑛 > 𝑁$ and $0 < 𝑦 - 𝑥_{𝑛} < 2 𝛿$ . Then we have for any such $𝑛$ and $𝑦 = 𝑥_{𝑛} + 𝛿$ , that
$\begin{aligned} \frac{1}{\sqrt{2 π}} (𝑓 * 𝑘_{𝜆}) (𝑥_{𝑛} + 𝛿) & = \frac{1}{\sqrt{2 π}} \int_{- \infty}^{\infty} 𝑓 (𝑡) 𝑘_{𝜆} (𝑥_{𝑛} + 𝛿 - 𝑡) d 𝑡 \\ = \frac{1}{\sqrt{2 π}} (\int_{𝑥_{𝑛}}^{𝑥_{𝑛} + 2 𝛿} + \int_{- \infty}^{𝑥_{𝑛}} + \int_{𝑥_{𝑛} + 2 𝛿}^{\infty}) 𝑓 (𝑡) 𝑘_{𝜆} (𝑥_{𝑛} + 𝛿 - 𝑡) d 𝑡 \\ > \frac{1}{\sqrt{2 π}} \int_{𝑥_{𝑛}}^{𝑥_{𝑛} + 2 𝛿} (𝐿 + \frac{𝜀}{2}) 𝑘_{𝜆} (𝑥_{𝑛} + 𝛿 - 𝑡) d 𝑡 \\ \begin{matrix} \end{matrix} - \frac{𝑀}{\sqrt{2 π}} (\int_{- \infty}^{𝑥_{𝑛}} + \int_{𝑥_{𝑛} + 2 𝛿}^{\infty}) 𝑘_{𝜆} (𝑥_{𝑛} + 𝛿 - 𝑡) d 𝑡 \\ > \frac{2 𝐿 + 𝜀}{2 \sqrt{2 π}} \int_{𝑥_{𝑛}}^{𝑥_{𝑛} + 2 𝛿} 𝑘_{𝜆} (𝑥_{𝑛} + 𝛿 - 𝑡) d 𝑡 \\ \begin{matrix} \end{matrix} - \frac{𝑀}{\sqrt{2 π}} (\int_{𝛿}^{\infty} + \int_{- \infty}^{- 𝛿}) 𝑘_{𝜆} (𝑢) d 𝑢 . \end{aligned}$
Now observe that
$\begin{aligned} \int_{- \infty}^{\infty} {(\frac{sin 𝑡}{𝑡})}^{2} d 𝑡 & = - {\frac{{sin}^{2} 𝑡}{𝑡} |}_{- \infty}^{\infty} + \int_{𝑡 \in ℝ} \frac{d {sin}^{2} 𝑡}{𝑡} \\ = \int_{- \infty}^{\infty} \frac{sin (2 𝑡)}{𝑡} d 𝑡 = π . (12.3.1 . 1) \end{aligned}$
Because
$\begin{aligned} \frac{1}{\sqrt{2 π}} (𝑓 * 𝑘_{𝜆}) (𝑥_{𝑛} + 𝛿) & \geq \frac{2 𝐿 + 𝜀}{2 \sqrt{2 π}} \int_{- 𝛿}^{𝛿} \frac{2 𝜆}{\sqrt{2 π}} {(\frac{sin (𝜆 𝑢)}{𝜆 𝑢})}^{2} d 𝑢 \\ \begin{matrix} \end{matrix} - \frac{2 𝑀}{π} \int_{𝛿 𝜆}^{\infty} {(\frac{sin 𝑡}{𝑡})}^{2} d 𝑡, \end{aligned}$
by letting $𝜆 \to \infty$ (which retains the inequality by the independence assumption), we have from (12.3.1.1) $(12.3.1.8) (12.3.1.1)$ , that
$\begin{aligned} \frac{2 𝐿 + 𝜀}{2 π} \int_{- 𝛿 𝜆}^{𝛿 𝜆} {(\frac{sin 𝑡}{𝑡})}^{2} d 𝑡 - \frac{2 𝑀}{π} \int_{𝛿 𝜆}^{\infty} {(\frac{sin 𝑡}{𝑡})}^{2} d 𝑡 \\ \begin{matrix} \end{matrix} \to \frac{2 𝐿 + 𝜀}{2 π} \int_{- \infty}^{\infty} {(\frac{sin 𝑡}{𝑡})}^{2} d 𝑡 = 𝐿 + \frac{𝜀}{2} . \end{aligned}$
Because ${lim}_{𝑛 \to \infty} \frac{1}{\sqrt{2 π}} (𝑓 * 𝑘_{𝜆}) (𝑥_{𝑛} + 𝛿) = 𝐿 \geq 𝐿 + \frac{𝜀}{2}$ , we reach a contradiction.
$𝑓 (𝑥_{𝑛}) - 𝐿 < - 𝜀$ . Then there exist $𝛿 > 0$ , $𝑁 \in ℕ$ such that
$𝑓 (𝑥_{𝑛}) - 𝑓 (𝑦) > - \frac{𝜀}{2} ⟹ 𝑓 (𝑦) < 𝑓 (𝑥_{𝑛}) + \frac{𝜀}{2} < 𝐿 - \frac{𝜀}{2}$
for any $𝑛 > 𝑁$ and $0 < 𝑥_{𝑛} - 𝑦 < 2 𝛿$ . Splitting the convolution integral into $\int_{- \infty}^{𝑥_{𝑛} - 2 𝛿}$ , $\int_{𝑥_{𝑛} - 2 𝛿}^{𝑥_{𝑛}}$ , and $\int_{𝑥_{𝑛}}^{\infty}$ , we have
$\begin{aligned} \frac{1}{\sqrt{2 π}} (𝑓 * 𝑘_{𝜆}) (𝑥_{𝑛} - 𝛿) & = \frac{1}{\sqrt{2 π}} \int_{- \infty}^{\infty} 𝑓 (𝑡) 𝑘_{𝜆} (𝑥_{𝑛} - 𝛿 - 𝑡) d 𝑡 \\ = \frac{1}{\sqrt{2 π}} (\int_{- \infty}^{𝑥_{𝑛} - 2 𝛿} + \int_{𝑥_{𝑛} - 2 𝛿}^{𝑥_{𝑛}} + \int_{𝑥_{𝑛}}^{\infty}) 𝑓 (𝑡) 𝑘_{𝜆} (𝑥_{𝑛} - 𝛿 - 𝑡) d 𝑡 \\ < \frac{𝑀}{\sqrt{2 π}} (\int_{- \infty}^{𝑥_{𝑛} - 2 𝛿} + \int_{𝑥_{𝑛}}^{\infty}) 𝑘_{𝜆} (𝑥_{𝑛} - 𝛿 - 𝑡) d 𝑡 \\ + \frac{1}{\sqrt{2 π}} \int_{𝑥_{𝑛} - 2 𝛿}^{𝑥_{𝑛}} (𝐿 - \frac{𝜀}{2}) 𝑘_{𝜆} (𝑥_{𝑛} - 𝛿 - 𝑡) d 𝑡 \\ < \frac{𝑀}{\sqrt{2 π}} (\int_{- \infty}^{- 𝛿} + \int_{𝛿}^{\infty}) 𝑘_{𝜆} (𝑢) d 𝑢 + \frac{2 𝐿 - 𝜀}{2 \sqrt{2 π}} \int_{- 𝛿}^{𝛿} 𝑘_{𝜆} (𝑢) d 𝑢 . \end{aligned}$
Letting $𝜆 \to \infty$ , we have similarly that ${lim}_{𝑛 \to \infty} \frac{1}{\sqrt{2 π}} (𝑓 * 𝑘_{𝜆}) (𝑥_{𝑛} - 𝛿) \leq 𝐿 - \frac{𝜀}{2}$ , contradicting ${lim}_{𝑛 \to \infty} (𝑓 * 𝑘_{𝜆}) (𝑥_{𝑛} - 𝛿) = 𝐿$ . $□$

Lemma 12.3.1.4 (Riemann–Lebesgue).

Let $𝑎, 𝑏 \in ℝ$ with $𝑎 < 𝑏$ . Let $𝑓 : [𝑎, 𝑏] \to ℂ$ be continuous. Then

lim_{𝑥 \to \infty} \int_{𝑎}^{𝑏} 𝑓 (𝑡) e^{i 𝑥 𝑡} d 𝑡 = 0 .

Proof.

By the Heine–Cantor Theorem (Theorem 1.2.26 $Theorem 1.2.26 Theorem 1.2.26$ ), $𝑓$ is uniformly continuous. Hence, for any $𝜀 > 0$ , there exists a partition

𝑎 = 𝑡_{0} < 𝑡_{1} < \dots < 𝑡_{𝑛} = 𝑏

such that

| 𝑓 (𝑡) - 𝑓 (𝑡_{𝑘}) | < \frac{𝜀}{2 (𝑏 - 𝑎)}

whenever $𝑡_{𝑘} \leq 𝑡 \leq 𝑡_{𝑘 + 1}$ . Writing

\begin{aligned} \int_{𝑎}^{𝑏} 𝑓 (𝑡) e^{i 𝑥 𝑡} d 𝑡 & = \sum_{𝑘 = 0}^{𝑛 - 1} \int_{𝑡_{𝑘}}^{𝑡_{𝑘 + 1}} 𝑓 (𝑡) e^{i 𝑥 𝑡} d 𝑡 \\ = \sum_{𝑘 = 0}^{𝑛 - 1} \int_{𝑡_{𝑘}}^{𝑡_{𝑘 + 1}} (𝑓 (𝑡) - 𝑓 (𝑡_{𝑘})) e^{i 𝑥 𝑡} d 𝑡 + \sum_{𝑘 = 0}^{𝑛 - 1} 𝑓 (𝑡_{𝑘}) \int_{𝑡_{𝑘}}^{𝑡_{𝑘 + 1}} e^{i 𝑥 𝑡} d 𝑡, \end{aligned}

we obtain

| \sum_{𝑘 = 0}^{𝑛 - 1} \int_{𝑡_{𝑘}}^{𝑡_{𝑘 + 1}} (𝑓 (𝑡) - 𝑓 (𝑡_{𝑘})) e^{i 𝑥 𝑡} d 𝑡 | \leq \sum_{𝑘 = 0}^{𝑛 - 1} \int_{𝑡_{𝑘}}^{𝑡_{𝑘 + 1}} | 𝑓 (𝑡) - 𝑓 (𝑡_{𝑘}) | d 𝑡 < \frac{𝜀}{2} .

Moreover,

\int_{𝑡_{𝑘}}^{𝑡_{𝑘 + 1}} e^{i 𝑥 𝑡} d 𝑡 = \frac{e^{i 𝑥 𝑡_{𝑘 + 1}} - e^{i 𝑥 𝑡_{𝑘}}}{i 𝑥},

and therefore

| \sum_{𝑘 = 0}^{𝑛 - 1} 𝑓 (𝑡_{𝑘}) \int_{𝑡_{𝑘}}^{𝑡_{𝑘 + 1}} e^{i 𝑥 𝑡} d 𝑡 | \leq \frac{2}{𝑥} \sum_{𝑘 = 0}^{𝑛 - 1} | 𝑓 (𝑡_{𝑘}) | .

Since the latter tends to $0$ as $𝑥 \to \infty$ , there exists $𝑥_{0} > 0$ such that

\frac{2}{𝑥} \sum_{𝑘 = 0}^{𝑛 - 1} | 𝑓 (𝑡_{𝑘}) | < \frac{𝜀}{2}

for all $𝑥 > 𝑥_{0}$ . Consequently,

| \int_{𝑎}^{𝑏} 𝑓 (𝑡) e^{i 𝑥 𝑡} d 𝑡 | < 𝜀

for all $𝑥 > 𝑥_{0}$ . Since $𝜀 > 0$ was arbitrary, it follows that

lim_{𝑥 \to \infty} \int_{𝑎}^{𝑏} 𝑓 (𝑡) e^{i 𝑥 𝑡} d 𝑡 = 0 . □

Theorem 12.3.1.5 (Wiener–Ikehara).

Let $𝑓 : ℝ_{\geq 0} \to ℝ$ be a nonnegative, monotonically non-decreasing function such that the Laplace transform

ℒ︀ {𝑓} (𝑠) = \int_{0}^{\infty} 𝑓 (𝑥) e^{- 𝑠 𝑥} d 𝑥 (12.3.1 . 2)

converges for $ℜ𝔢 𝑠 > 1$ . Suppose that there exists some constant $𝑐 > 0$ such that

𝑔 (𝑡) = lim_{𝜎 \to 1^{+}} [ℒ︀ {𝑓} (𝑠) - \frac{𝑐}{𝑠 - 1}], 𝑠 = 𝜎 + i 𝑡 (12.3.1 . 3)

converges locally uniformly with respect to $𝑡$ in $ℝ$ and is continuous. Then

lim_{𝑥 \to \infty} \frac{𝑓 (𝑥)}{e^{𝑥}} = 𝑐 . (12.3.1 . 4)

Proof.

Let

𝑎 (𝑡) = {\begin{cases} 𝑓 (𝑡) e^{- 𝑡} & 𝑡 \geq 0 \\ 0 & 𝑡 < 0 \end{cases}, 𝐴 (𝑡) = {\begin{cases} 𝑐 & 𝑡 \geq 0 \\ 0 & 𝑡 < 0 \end{cases} .

The convergence of (12.3.1.2) $(12.3.1.9) (12.3.1.2)$ implies that for $𝜀, 𝜆 > 0$ , the convolution

𝐼_{𝜆, 𝜀} (𝑥) = \frac{1}{\sqrt{2 π}} \int_{- \infty}^{\infty} 𝑘_{𝜆} (𝑥 - 𝑡) (\frac{𝑎 (𝑡) - 𝐴 (𝑡)}{e^{𝜀 𝑡}}) d 𝑡

converges. Since

\int_{- \infty}^{\infty} (\frac{𝑎 (𝑡) - 𝐴 (𝑡)}{e^{(𝜀 + i 𝑦) 𝑡}}) d 𝑡

converges uniformly for $| 𝑦 | < 2 𝜆$ (as the integrand is uniformly bounded by $| 𝑎 (𝑡) | e^{- 𝜀 𝑡} + | 𝐴 (𝑡) | e^{- 𝜀 𝑡}$ , where both terms have convergent integrals), we have by virtue of Lemma 12.3.1.1 $Lemma 12.3.1.1 Lemma 12.3.1.1$ , that

\begin{aligned} 𝐼_{𝜆, 𝜀} (𝑥) & = \frac{1}{2 π} \int_{- \infty}^{\infty} (\frac{𝑎 (𝑡) - 𝐴 (𝑡)}{e^{𝜀 𝑡}}) \int_{- 2 𝜆}^{2 𝜆} 𝐾_{𝜆} (𝑦) e^{i (𝑥 - 𝑡) 𝑦} d 𝑦 d 𝑡 \\ = \frac{1}{2 π} \int_{- 2 𝜆}^{2 𝜆} 𝐾_{𝜆} (𝑦) e^{i 𝑥 𝑦} \int_{0}^{\infty} \frac{𝑎 (𝑡) - 𝐴 (𝑡)}{e^{(𝜀 + i 𝑦) 𝑡}} d 𝑡 d 𝑦 \\ = \frac{1}{2 π} \int_{- 2 𝜆}^{2 𝜆} 𝐾_{𝜆} (𝑦) e^{i 𝑥 𝑦} [ℒ︀ {𝑓} (1 + 𝜀 + i 𝑦) - \frac{𝑐}{𝜀 + i 𝑦}] d 𝑦 . \end{aligned}

Now $\forall 𝜀^{'} > 0$ , by the local uniform convergence of $𝑔$ in (12.3.1.3) $(12.3.1.10) (12.3.1.3)$ , there exists some $𝛿 > 0$ such that for any $𝜀 < 𝛿$ ,

\forall | 𝑦 | < 2 𝜆, | ℒ︀ {𝑓} (1 + 𝜀 + i 𝑦) - \frac{𝑐}{𝜀 + i 𝑦} - 𝑔 (𝑦) | < \frac{π 𝜀^{'}}{𝜆} .

Hence, we have that

| 𝐼_{𝜆, 𝜀} (𝑥) - \frac{1}{2 π} \int_{- 2 𝜆}^{2 𝜆} 𝐾_{𝜆} (𝑦) e^{i 𝑥 𝑦} 𝑔 (𝑦) d 𝑦 | \leq \frac{1}{2 π} \int_{- 2 𝜆}^{2 𝜆} 𝐾_{𝜆} (𝑦) \frac{π 𝜀^{'}}{𝜆} d 𝑦 = 𝜀^{'} .

Therefore,

lim_{𝜀 \to 0^{+}} 𝐼_{𝜆, 𝜀} (𝑥) = \frac{1}{2 π} \int_{- 2 𝜆}^{2 𝜆} 𝐾_{𝜆} (𝑦) e^{i 𝑥 𝑦} 𝑔 (𝑦) d 𝑦,

(effectively, we have exchanged the limit with the integral). Then by the Riemann–Lebesgue Lemma (Lemma 12.3.1.4 $Lemma 12.3.1.4 Lemma 12.3.1.4$ ),

lim_{𝑥 \to \infty} lim_{𝜀 \to 0^{+}} 𝐼_{𝜆, 𝜀} (𝑥) = 0 . (12.3.1 . 5)

On the other hand, a manual calculation while restricting the integation range yields

\begin{aligned} lim_{𝜀 \to 0^{+}} 𝐼_{𝜆, 𝜀} (𝑥) = lim_{𝜀 \to 0^{+}} \frac{1}{\sqrt{2 π}} ( & \int_{0}^{\infty} 𝑘_{𝜆} (𝑥 - 𝑡) 𝑎 (𝑡) e^{- 𝜀 𝑡} d 𝑡 \\ \begin{matrix}  \end{matrix} - \int_{0}^{\infty} 𝑘_{𝜆} (𝑥 - 𝑡) 𝐴 (𝑡) e^{- 𝜀 𝑡} d 𝑡) . \end{aligned}

Using that $\int_{0}^{\infty} 𝑘_{𝜆} (𝑥 - 𝑡) 𝑐 d 𝑡$ dominates the second integrand and forms a convergent integral, Lebesgue’s Dominated Convergence Theorem then gives

lim_{𝜀 \to 0^{+}} 𝐼_{𝜆, 𝜀} (𝑥) = lim_{𝜀 \to 0^{+}} \frac{1}{\sqrt{2 π}} [\int_{0}^{\infty} 𝑘_{𝜆} (𝑥 - 𝑡) 𝑎 (𝑡) e^{- 𝜀 𝑡} d 𝑡 - (𝐴 * 𝑘_{𝜆}) (𝑥)]

Since for any postive $𝑡_{0}$ ,

\int_{0}^{𝑡_{0}} 𝑘_{𝜆} (𝑥 - 𝑡) 𝑎 (𝑡) e^{- 𝜀 𝑡} d 𝑡 \leq \int_{0}^{\infty} 𝑘_{𝜆} (𝑥 - 𝑡) 𝑎 (𝑡) e^{- 𝜀 𝑡} d 𝑡

where the left-hand side is dominated by the finite integral $\int_{0}^{𝑡_{0}} 𝑘_{𝜆} (𝑥 - 𝑡) 𝑎 (𝑡) d 𝑡$ , letting $𝜀 \to 0^{+}$ (which now may be passed through the left-hand side integral) gives

\int_{0}^{𝑡_{0}} 𝑘_{𝜆} (𝑥 - 𝑡) 𝑎 (𝑡) d 𝑡 \leq lim_{𝜀 \to 0^{+}} \int_{0}^{\infty} 𝑘_{𝜆} (𝑥 - 𝑡) 𝑎 (𝑡) e^{- 𝜀 𝑡} d 𝑡

Since this holds for all $𝑡_{0}$ , letting $𝑡_{0} \to \infty$ gives

\int_{0}^{\infty} 𝑘_{𝜆} (𝑥 - 𝑡) 𝑎 (𝑡) d 𝑡 \leq lim_{𝜀 \to 0^{+}} \int_{0}^{\infty} 𝑘_{𝜆} (𝑥 - 𝑡) 𝑎 (𝑡) e^{- 𝜀 𝑡} d 𝑡

Additionally, the reverse inequality

lim_{𝜀 \to 0^{+}} \int_{0}^{\infty} 𝑘_{𝜆} (𝑥 - 𝑡) 𝑎 (𝑡) e^{- 𝜀 𝑡} d 𝑡 \leq \int_{0}^{\infty} 𝑘_{𝜆} (𝑥 - 𝑡) 𝑎 (𝑡) d 𝑡

is also true as $e^{- 𝜀 𝑡} < 1$ . Thus, we attain an equality. Therefore,

lim_{𝜀 \to 0^{+}} 𝐼_{𝜆, 𝜀} (𝑥) = \frac{1}{\sqrt{2 π}} (𝑎 * 𝑘_{𝜆} - 𝐴 * 𝑘_{𝜆}) (𝑥) \to 0 (by (12.3.1.5) (12.3.1.12) (12.3.1.5))

as $𝑥 \to \infty$ . Since

\begin{aligned} \frac{1}{\sqrt{2 π}} (𝐴 * 𝑘_{𝜆}) (𝑥) & = \frac{𝑐}{\sqrt{2 π}} \int_{- \infty}^{𝑥} 𝑘_{𝜆} (𝑢) d 𝑢 \to \frac{𝑐}{\sqrt{2 π}} \int_{- \infty}^{\infty} 𝑘_{𝜆} (𝑢) d 𝑢 \\ = \frac{𝑐}{\sqrt{2 π}} \int_{- \infty}^{\infty} \sqrt{\frac{2}{π}} {(\frac{sin 𝑢}{𝑢})}^{2} d 𝑢 \\ = 𝑐 (by (12.3.1.1) (12.3.1.8) (12.3.1.1)) \end{aligned}

as $𝑥 \to \infty$ , we have that

lim_{𝑥 \to \infty} \frac{1}{\sqrt{2 π}} (𝑎 * 𝑘_{𝜆}) (𝑥) = 𝑐 . (12.3.1 . 6)

Therefore, $\exists 𝑥_{0} > 0$ such that

\frac{1}{\sqrt{2 π}} (𝑎 * 𝑘_{𝜆}) (𝑥) < 𝑐 + 1 ⟹ \frac{1}{π} \int_{- \infty}^{\infty} {(\frac{sin 𝑡}{𝑡})}^{2} 𝑎 (𝑥 - \frac{𝑡}{𝜆}) d 𝑡 < 𝑐 + 1

for all $𝑥 > 𝑥_{0}$ . Substituting $𝑥 + \frac{2}{\sqrt{𝜆}}$ for $𝑥$ in the integral, we have (after further restricting the integration bounds, which preserves the inequality)

\int_{- \sqrt{𝜆}}^{\sqrt{𝜆}} {(\frac{sin 𝑡}{𝑡})}^{2} 𝑎 (𝑥 + \frac{2 \sqrt{𝜆} - 𝑡}{𝜆}) d 𝑡 < π (𝑐 + 1)

Since $e^{𝑢} 𝑎 (𝑢) = 𝑓 (𝑢)$ (for $𝑢 = 𝑥 + \frac{2 \sqrt{𝜆} - 𝑡}{𝜆} > 0$ ) is non-decreasing in $𝑢$ , it is bounded below by $𝑓 (𝑥 + \frac{\sqrt{𝜆}}{𝜆}) > 𝑓 (𝑥)$ . Thus, we have

\int_{- \sqrt{𝜆}}^{\sqrt{𝜆}} {(\frac{sin 𝑡}{𝑡})}^{2} exp (\frac{𝑡 - 2 \sqrt{𝜆}}{𝜆} - 𝑥) 𝑓 (𝑥) d 𝑡 < π (𝑐 + 1),

implying that

\int_{- \sqrt{𝜆}}^{\sqrt{𝜆}} {(\frac{sin 𝑡}{𝑡})}^{2} e^{\frac{𝑡 - 2 \sqrt{𝜆}}{𝜆}} 𝑎 (𝑥) d 𝑡 < π (𝑐 + 1)

which gives

𝑎 (𝑥) e^{- \frac{3}{\sqrt{𝜆}}} \int_{- \sqrt{𝜆}}^{\sqrt{𝜆}} {(\frac{sin 𝑡}{𝑡})}^{2} d 𝑡 < π (𝑐 + 1),

which is satisfied for all $𝜆 > 0$ and $𝑥 > 𝑥_{0}$ . Letting $𝜆 \to \infty$ , we have that $𝑎 (𝑥) < 𝑐 + 1$ . Compactness shows that $𝑎$ is bounded on $[0, 𝑥_{0}]$ (Theorem 1.2.22 $Theorem 1.2.22 Theorem 1.2.22$ ). Hence, $𝑎$ is bounded (above) on $ℝ$ by some $𝑀 > 0$ .

The final hypothesis required is the slow decrease of $𝑎$ : for any $𝜀 > 0$ , $\exists 𝛿 > 0$ such that $\forall 𝑥, 𝑦 > 0$ with $0 < 𝑦 - 𝑥 < 𝛿$ , we have

\begin{aligned} 𝑎 (𝑦) - 𝑎 (𝑥) & = e^{- 𝑥} (e^{𝑥 - 𝑦} 𝑓 (𝑦) - 𝑓 (𝑥)) > e^{- 𝑥} 𝑓 (𝑥) (e^{- 𝛿} - 1) \\ = 𝑎 (𝑥) (e^{- 𝛿} - 1) > 𝑀 (e^{- 𝛿} - 1) . \end{aligned}

If $𝛿$ is chosen so that $𝑀 (1 - e^{- 𝛿}) < 𝜀$ , namely $0 < 𝛿 < ln (\frac{𝑀}{𝑀 - 𝜀})$ , then $𝑎 (𝑦) - 𝑎 (𝑥) > - 𝜀$ , and $𝑎$ then exhibits slow decrease.

The slow decrease, boundedness, and the condition in (12.3.1.6) $(12.3.1.13) (12.3.1.6)$ are sufficient by Proposition 12.3.1.3 $Proposition 12.3.1.3 Proposition 12.3.1.3$ , to show that ${lim}_{𝑥 \to \infty} 𝑎 (𝑥) = 𝑐$ , or equivalently, ${lim}_{𝑥 \to \infty} \frac{𝑓 (𝑥)}{e^{𝑥}} = 𝑐$ . ⁠ $□$

Theorem 12.3.1.6 (Prime Number Theorem).

The prime counting function

𝜋 (𝑥)

is asymptotically equal to

\frac{𝑥}{log 𝑥}

, i.e.,

{lim}_{𝑥 \to \infty} \frac{𝜋 (𝑥) log 𝑥}{𝑥} = 1

Proof.

By Theorem 12.3.1 $Theorem 12.3.1 Theorem 12.3.1$ , it suffices to show that ${lim}_{𝑥 \to \infty} \frac{𝜓 (𝑥)}{𝑥} = 1$ . Consider the Laplace transform of $𝜓 \circ exp$ in (12.3.6) $(12.3.6) (12.3.6)$ . By Theorem 12.3.2 $Theorem 12.3.2 Theorem 12.3.2$ , the function $𝑔$ defined therein converges uniformly on compact subsets of $ℝ$ and is continuous. Hence, by the Wiener–Ikehara theorem (Theorem 12.3.1.5 $Theorem 12.3.1.5 Theorem 12.3.1.5$ ), we have that

lim_{𝑥 \to \infty} \frac{𝜓 (e^{𝑥})}{e^{𝑥}} = lim_{𝑥 \to \infty} \frac{𝜓 (𝑥)}{𝑥} = 1,

and the Prime Number Theorem follows. ⁠

□