If $\int_0^\infty |f(x)|^2dx<\infty$ then $F(z)=\int_0^\infty f(t)e^{itz}dt$ is well-defined and continuous Suppose $f$ is a complex valued function defined on $(0,\infty)$ such that $\int_0^\infty |f(x)|^2dx<\infty$.  Then for $z\in \Bbb C$ with $\text{Im}(z)>0$, define $F(z)=\int_0^\infty f(t)e^{itz}dt$. I am trying to show that $F$ is continuous, but I can't even see why that the integral is well-defined. How can we show that $F$ is well-defined?
 A: Use the Cauchy-Schwarz inequality (integral form) :
$$\left\lvert \int_0^\infty f(t) \overline{g(t)} \,\mathrm{d}t\right\rvert^2 \leq \left( \int_0^\infty \lvert f(t) \rvert^2 \,\mathrm{d}t\right)\left( \int_0^\infty  \lvert g(t) \rvert^2 \,\mathrm{d}t\right)
$$
A: $\newcommand{\Im}{\operatorname{Im}}\newcommand{\Re}{\operatorname{Re}}$
If $f\in L^2(0,\infty)$, then since $x^2$ is convex on $[k,k+1]$, Jensen's Inequality says
$$
\begin{align}
\|f\|_{L^1(k,k+1)}
&\le\|f\|_{L^2(k,k+1)}\tag{1a}\\
&\le\|f\|_{L^2(0,\infty)}\tag{1b}
\end{align}
$$
Furthermore,
$$
\left\|t^ne^{-t\Im(z)}\right\|_{L^\infty(k,k+1)}\le (k+1)^ne^{-k\Im(z)}\tag2
$$
Thus,
$$
\begin{align}
\left\|f(t)\,t^ne^{-t\Im(z)}\right\|_{L^1(0,\infty)}
&\le\sum_{k=0}^\infty\left\|f(t)\,t^ne^{-t\Im(z)}\right\|_{L^1(k,k+1)}\tag{3a}\\
&\le\sum_{k=0}^\infty\|f\|_{L^1(k,k+1)}\left\|t^ne^{-t\Im(z)}\right\|_{L^\infty(k,k+1)}\tag{3b}\\
&\le\sum_{k=0}^\infty\|f\|_{L^2(0,\infty)}(k+1)^ne^{-k\Im(z)}\tag{3c}\\
&\le\|f\|_{L^2(0,\infty)}e^{2\Im(z)}\int_0^\infty x^ne^{-x\Im(z)}\mathrm{d}x\tag{3d}\\[3pt]
&=\|f\|_{L^2(0,\infty)}\frac{e^{2\Im(z)}n!}{\Im(z)^{n+1}}\tag{3e}
\end{align}
$$
Explanation:
$\text{(3a)}$: Minkowski's Inequality
$\text{(3b)}$: Hölder's Inequality
$\text{(3c)}$: apply $(1)$ and $(2)$
$\text{(3d)}$: $(k+1)^ne^{-k\Im(z)}\le e^{2\Im(z)}\int_{k+1}^{k+2}x^ne^{-x\Im(z)}\mathrm{d}x$
$\text{(3e)}$: evaluate integral
Therefore,
$$
\begin{align}
\left|\,\frac{\mathrm{d}^n}{\mathrm{d}z^n}F(z)\,\right|
&=\left|\int_0^\infty f(t)\,t^ne^{-t\Im(z)}e^{it\Re(z)}\,\mathrm{d}t\right|\tag{4a}\\
&\le\|f\|_{L^2(0,\infty)}\frac{e^{2\Im(z)}n!}{\Im(z)^{n+1}}\tag{4b}
\end{align}
$$
which says that for $\Im(z)\gt0$, $F(z)$ is not only well-defined, but also $C^\infty$.
