There is the claim that Laplace transform is a bounded linear operator on pg 17 of http://lall.stanford.edu/data/engr210a_0102/lecture6_2001_10_17_01.pdf
We assume that the Laplace transform operator $\Lambda: L_2[0,\infty) \to H_2$ where $H_2$ is Hardy space (see link)
We wish to show that $\exists \kappa \geq 0$ s.t. $$\|\Lambda f\|_{H_2} \leq \kappa \cdot \|f\|_{L_2}$$
I am not very familiar with the usual norm that is associated with $H_2$, can someone show how this claim is true?
The Hardy space on the right half-plane $\Pi_{+}$ consists of all holomorphic functions $F$ on $\Pi_{+}$ that are in $L^2(\mathbb{R})$ on every vertical line, with $$ \|F\|_{H^2}=\sup_{r > 0} \left(\int_{-\infty}^{\infty}|F(x+iy)|^2dy < \infty\right)^{1/2} < \infty. $$ If $f\in L^2[0,\infty)$, then $$ F(s) = \int_{0}^{\infty}e^{-ts}f(t)dt \in H^2(\Pi_{+}). $$ To prove this fact, the Plancherel theorem for the Fourier transform is applied to $F(s)$ as follows: \begin{align} \int_{-\infty}^{\infty}|F(x+iy)|^2dy&=\int_{-\infty}^{\infty}\left|\int_{-\infty}^{\infty}e^{-xt-iyt}f(t)dt\right|^2dy \\ &=2\pi\int_{0}^{\infty}e^{-2xt}|f(t)|^2dt \\ &\le 2\pi\int_{0}^{\infty}|f(t)|^2dt < \infty,\;\;\; f\in L^2[0,\infty). \end{align} From this you can show \begin{align} \sup_{y > 0}\int_{-\infty}^{\infty}|F(x+iy)|^2dy&=2\pi\int_{0}^{\infty}|f(t)|^2dt, \\ \|F\|_{H^2}&=2\pi\|f\|_{L^2[0,\infty)}. \end{align} In terms of the Laplace transform $\mathscr{L}$, $$\|\mathscr{L}\{f\}\|_{H^2(\Pi^+)}=2\pi\|f\|_{L^2[0,\infty)}.$$