How to prove Laplace transform is bounded on $L^2(\mathbb{R}_+)$?

Question

The Laplace transform is defined by \begin{equation} (\mathscr{L}f)(s) \triangleq \int_0^\infty e^{-sx} f(x) dx, \quad s>0, \end{equation} then how can we check that the Laplace transform $\mathscr{L}$ is bounded as an operator from $L^2(\mathbb{R}_+)$ to $L^2(\mathbb{R}_+)$ with norm $\sqrt{\pi}$?

Answer 1

Observe \begin{align} \|\mathscr{L}(f)\|_{L^2(0, \infty)}^2 =&\ \int^\infty_0 \int^\infty_0\int^\infty_0 dsdxdy\ e^{-s(x+y)}f(x)f(y)\\ =&\ \int^\infty_0 \int^\infty_0 dxdy\ \frac{f(x)f(y)}{x+y} = \pi\int^\infty_0 dx\ f(x) H(x) \end{align} where \begin{align} H(f\chi_{[0, \infty)})(x) = \frac{1}{\pi}\int^\infty_{-\infty} dy\ \frac{f(y)\chi_{[0, \infty)}(y)}{x+y} =\frac{1}{\pi}\int^\infty_{-\infty} dy\ \frac{g(y)}{x+y} \end{align} defined for all $x>0$ is the Hilbert transform of $g=f\chi_{[0, \infty)}$. Then it follows \begin{align} \|\mathscr{L}(f)\|_{L^2(0, \infty)}^2 \leq \pi\,\|f\|_{L^2(0, \infty)}\|H(g)\|_{L^2(0, \infty)}. \end{align} Using the $L^2-L^2$ boundedness of the Hilbert transform, we have the estimate \begin{align} \|\mathscr{L}(f)\|_{L^2(0, \infty)}^2 \leq \pi\, \|f\|_{L^2(0, \infty)}^2. \end{align}