New characteristic functions from old I am doing an exercise which says: If $f$ is a characteristic function, then show that  $$ F(t):=  \int_0^{\infty} f(ut)e^{-u}du $$ is again a characteristic function. 
Is this answer correct? Let $Z$ be an exponentially distributed random variable with parameter $1$. Let $X$ be a random variable with characteristic function $f$. Then 
\begin{align}
F(t) &= \mathbb{E} \left[ \left( \mathbb{E} e^{itsX} \right) \Bigg{\vert}_{s=Z}\right] \\
&= \mathbb{E} \left[ \mathbb{E}(e^{itZX} \big{\vert} Z ) \right]\\
&=\mathbb{E}(e^{itZX})
\end{align}
where it is the second equality that uses the independence. This shows that $F$ is the characteristic function of the random variable $ZX$. 
Many thanks for your help.
 A: 
Bochner's Theorem:
A function $F:\mathbb R\to \mathbb C$ is the characteristic function for some random variable if, and only if, $F$ is positive-definite, continuous at the origin and $F(0)=1$.

Positive definiteness: The function $f$, being a characteristic function, is positive-definite. Consider, for $\ r,s=1,\ldots,n,\ $ the arbitrary numbers $\ \xi_r, \bar{\xi_s}\in\mathbb C,\ t_r, t_s\in\mathbb R$. Observe that 
$$\begin{array}{crl}\sum\limits_{r,s=1}^{n}\xi_r\bar{\xi_s}F(t_r-t_s) = \sum\limits_{r,s=1}^{n}\xi_r\bar{\xi_s}\int\limits_{0}^{\infty}f(ut_r-ut_s)e^{-u}\text du \\= \int\limits_{0}^{\infty}\sum\limits_{r,s=1}^{n}\xi_r\bar{\xi_s}f(ut_r-ut_s)e^{-u}\text du \geqslant 0\,,
\end{array}$$
which shows that $F$ is positive-definite.
Continuity at the origin: Clearly, $F(0)=1$. Now, for the sequence $\frac{1}{n}$, we see that $\lim\limits_{n\to\infty}|f(u\frac{1}{n})-1|=0$ and $\ |f(u\frac{1}{n})-1|\leqslant 2\ $ for all $n, u>0\ $, by the continuity and boundedness of $f$. Therefore, by Lebesgue dominated convergence,
$$\begin{array}{ccl}\lim\limits_{n\to\infty}|F(\frac{1}{n})-F(0)| &=& \lim\limits_{n\to\infty}|\int_{0}^{\infty}(f(u\frac{1}{n})-1)e^{-u}~\text du| \leqslant \lim\limits_{n\to\infty}\int_{0}^{\infty}|f(u\frac{1}{n})-1|e^{-u}~\text du \\ &=& \int_{0}^{\infty}\lim\limits_{n\to\infty}|f(u\frac{1}{n})-1|e^{-u}~\text du = 0\,,\end{array}$$
so $F(t)$ is continuous at $0$. 
We conclude that $F$ is the characteristic function for some random variable.
