To find a way of bringing together a minimization problem defined in the time-domain with systems described in the frequency domain via transfer functions, and thus bridge the gap between norms defined for time-domain and frequency-domain quantities, I came across the following relationship in the literature: \begin{equation} \left\|y\left(\;\!\cdot\;\!\right)\right\|_{L_\infty}\leq\left\|G\left(\;\!\cdot\;\!\right)\right\|_{H_2}\left\|u\left(\;\!\cdot\;\!\right)\right\|_{L_2}. \end{equation} The norms $\left\|\;\!\cdot\;\!\right\|_{L_\infty}$ and $\left\|\;\!\cdot\;\!\right\|_{L_2}$ in the above expression refer to the norms on the usual spaces $L_\infty(-\infty,\infty)$ and $L_2(-\infty,\infty)$, respectively. The norm $\left\|\;\!\cdot\;\!\right\|_{H_2}$ is given by \begin{equation}\nonumber \left\|G\left(\;\!\cdot\;\!\right)\right\|_{H_2}=\left(\frac{1}{2\pi}\int^\infty_{-\infty}\left\|G\left(i\omega\right)\right\|^2_F \,d\omega\right)^{1/2}, \end{equation} where $\left\|\;\!\cdot\;\!\right\|_{F}$ is the Frobenius norm. It should be pointed out that, on denoting the Laplace transforms of $y\left(t\right)$ and $u\left(t\right)$ by $\hat y\left(s\right)$ and $\hat u\left(s\right)$, respectively, we have that \begin{equation}\nonumber \hat y\left(s\right)=G\left(s\right)\hat u\left(s\right). \end{equation} My question is: how do we derive the above inequality?
I can show that \begin{equation} \left\|y\left(\;\!\cdot\;\!\right)\right\|_{L_2}\leq\left\|G\left(\;\!\cdot\;\!\right)\right\|_{H_\infty}\left\|u\left(\;\!\cdot\;\!\right)\right\|_{L_2}, \end{equation} since, e.g. in the case that $\hat y\left(s\right)$ and $\hat u\left(s\right)$ are scalars, \begin{equation} \begin{split} \left\|y\left(\;\!\cdot\;\!\right)\right\|_{L_2}&=\left\|\hat y\left(\;\!\cdot\;\!\right)\right\|_{L_2}\\ &=\left(\frac{1}{2\pi}\int^\infty_{-\infty}\left|G\left(i\omega\right)\right|^2\left|\hat u\left(i\omega\right)\right|^2 \,d\omega\right)^{1/2}\\ &\leq\left\|G\left(\;\!\cdot\;\!\right)\right\|_{H_\infty}\left(\frac{1}{2\pi}\int^\infty_{-\infty}\left|\hat u\left(i\omega\right)\right|^2 \,d\omega\right)^{1/2}\\ &=\left\|G\left(\;\!\cdot\;\!\right)\right\|_{H_\infty}\left\|\hat u\left(\;\!\cdot\;\!\right)\right\|_{L_2}\\ &=\left\|G\left(\;\!\cdot\;\!\right)\right\|_{H_\infty}\left\| u\left(\;\!\cdot\;\!\right)\right\|_{L_2}, \end{split} \end{equation} but I am not sure how to do this in the case mentioned at the outset.
Many thanks!
If $\hat u$ is the Laplace transform, then $\hat u(i\omega) = \mathcal F u(\omega)$, and $\mathcal F u$ is the Fourier-transform of $u$. If I understood your post correctly, you are actually working with Fourier transforms most of the time. This is good, since we have a couple of nice properties about norms in this context.
If we start from $\hat y = G\hat u$, but the arguments are $i\omega$ anyway, we can instead look at $\mathcal F(y) = G\mathcal F (u)$.
We could proceed by using the continuity of $\mathcal F^{-1}$ from $L^1$ into the space of continuous functions decaying towards infinity, equipped with the maximum norm (usually you will find this statement only for $\mathcal F$, but since $\mathcal F$ and $\mathcal F^{-1}$ behave exactly the same, this is ok): $$\|y\|_{L^\infty} = \|\mathcal F^{-1}\mathcal F y\|_{L^\infty}\le\frac1{\sqrt{2\pi}}\|\mathcal F y\|_{L^1} = \frac1{\sqrt{2\pi}}\|G\mathcal F (u)\|_{L^1}\le \|G\|_{H^2}\|\mathcal Fu\|_{L^2}= \|G\|_{H^2}\|u\|_{L^2}.$$ The first inequality comes from the aforementioned continuity of $\mathcal F^{-1}:L^1\to (C, \|.\|_{L^\infty})$. The next estimate is due to the Cauchy-Schwarz inequality. The last step uses that the Fourier transform is an isometry on $L^2$, i.e. it does not change the norm (a property you have used in your question already).
Remark 1: In more detail, the first estimate is obtained by the following considerations: $$\sqrt{2\pi}\cdot\|y\|_{L^\infty} = \sup_{x\in\mathbb R}\left|\int_{\mathbb R}e^{i t x} \mathcal F[y(t)]d t\right|\le \sup_{x\in\mathbb R}\int_{\mathbb R}|e^{i t x} |\cdot|\mathcal F[y(t)]|d t = \int_{\mathbb R}|\mathcal F[y] |dx = \|\mathcal Fy\|_{L^1}.$$ The constant $\sqrt{2\pi}$ comes from the choice of the Fourier transform $\mathcal F y = \frac 1{\sqrt{2\pi}}\int_{\mathbb R} e^{-ixt}y(t) dt$. This makes $\mathcal F$ an isometry on $L^2$, but introduces the constant in the first estimate. If you wish to choose another definition of the Fourier transform you should adapt the constants accordingly.
Remark 2: As a mathematician, I would say that what you denote by $\|G\|_{H^2}$ is nothing else than an $L^2$-norm on matrix-values functions. $H^2$ is commonly reserved for a Sobolev space.