Exercise 9, § 3

Let $\varphi : [0,+\infty [\rightarrow \mathbf{R}$ be a function $\geqslant 0$ increasing, such that the function $t\mapsto e^{-\sigma t}\varphi (t)$ is integrable for Lebesgue measure when $\sigma >1$. For all $s\in \mathbf{C}$ such that $\mathscr{R}(s)>1$, set

(“Laplace transform of $\varphi$”).

Suppose that there exists $A\in \mathbf{C}$ with the following property: when $\sigma$ tends to 1 through values $>1$, the function

converges uniformly towards a function $g$ on every compact subset of $\mathbf{R}$. Then

$_{t\rightarrow}$lim$_{+\infty}\varphi (t)e^{-t}= A$

(“Ikehara’s Tauberian theorem”).

(We set $a(t) =e^{-t}\varphi (t)$ for $t >0,a(t) = 0$ for $t\leqslant 0$, $A(t) = A$ for $t >0$, $A(t) = 0$ for $t\leqslant 0$. Let $\lambda >0$. We set:

2 sin $2\lambda t^2$

$1-$ if $|t|\leqslant 2\lambda$

0 if $|t|>2\lambda$.

Using Exercise 1 of II, p. 262, the Lebesgue–Fubini theorem and the Plancherel theorem, now prove that:

1 $\int+\infty -\varepsilon t$ 1 $\int 2\lambda -ixy$

lim $\surd k_{\lambda}(x-t)(a(t)-A(t))edt=\surd K_{\lambda}(y)eg(y)dy$ $^{\varepsilon\rightarrow 0}2\pi_{-\infty}2\pi_{-2\lambda}$

and deduce that

1 $\int^{+\infty}$

lim $\surd k_{\lambda}(x-t)a(t)dt= A$.

Prove then that $a$ is bounded and slowly decreasing. We can finally apply Exercise 6.)