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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!

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  • $\begingroup$ Is $\|G\|_{H_\infty}$ defined as $\|\|G(i\omega)\|_{F}\|_{L_\infty}$? $\endgroup$ – DominikS Aug 9 '17 at 16:22
  • $\begingroup$ $\left\|G\left(\;\!\cdot\;\!\right)\right\|_{H_\infty}=\sup\limits_{\omega}\left|G\left(i\omega\right)\right|$. Hope this helps. $\endgroup$ – Trevor3 Aug 9 '17 at 21:27
  • $\begingroup$ I want to make sure I understand correctly: Is the relationship between $u$ and $y$ defined by $\hat y = G \hat u$? Or this there some other defining relation? $\endgroup$ – DominikS Aug 10 '17 at 13:58
  • $\begingroup$ Correct, the relationship between $\hat u\left(s\right)$ and $\hat y\left(s\right)$ is given by $\hat y\left(s\right)=G\left(s\right)\hat u\left(s\right)$. $\endgroup$ – Trevor3 Aug 10 '17 at 14:11
  • $\begingroup$ To me the notation $\|\hat u(\cdot)\|_{L^2}$ is slightly misleading, I'd rather write $\|\hat u(i\cdot)\|_{L^2}$ - but maybe it is more common in the context of control theory? $\endgroup$ – DominikS Aug 10 '17 at 14:12
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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.

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  • $\begingroup$ So you meant $\|G\hat{u}\|_{L^1} \le \|G\|_{L^2}\|\hat{u}\|_{L^2}$ $\endgroup$ – reuns Aug 10 '17 at 14:54
  • $\begingroup$ Thanks very much for your (failry well-explained) answer. I only have one question regarding the first inequality in your estimate above. You are writing that it is a reault of the continuity of the inverse of the Fourier operator. This operator maps into the space of continuous functions, equipped with the maximum norm. Without going into too much detail here, is there a theorem or any other reult in the literature proving this? I am just trying to understand the rationale behind it. $\endgroup$ – Trevor3 Aug 14 '17 at 8:31
  • $\begingroup$ I've added a remark explaining the continuity of $\mathcal F^{-1}$. I have also discovered that I was sloppy with the constants $\sqrt{2\pi}$, an issue I have fixed now as well. $\endgroup$ – DominikS Aug 14 '17 at 14:26
  • $\begingroup$ There is a slight inconsitency that I do not quite understand. As far as I know, the Fourier transform is $\omega\mapsto\check{y}\left(\omega\right)$, which is defined for a function $t\mapsto y\left(t\right)$ which is an element of $L_1\left(\mathbb{R}\right)$. In fact, its definition can be extended to functions which are in $L_2\left(\mathbb{R}\right)$. Yet, we start by assuming a function $t\mapsto y\left(t\right)$ that is in $L_\infty\left(\mathbb{R}\right)$, so we can give meaning to the norm $\left\|y\right\|_\infty$. Is $\left(\mathcal{F}y\right)\left(t\right)$ well-defined then? $\endgroup$ – Trevor3 Aug 18 '17 at 14:41
  • $\begingroup$ @DominikS: I think I have found the answer to my previous question. If we say that $\mathcal{F}^{-1}:L_1\left(\mathbb{R}\right)\rightarrow C_0\left(\mathbb{R}\right)$, then $\mathcal{F}:C_0\left(\mathbb{R}\right)\rightarrow L_1\left(\mathbb{R}\right)$. So, to make sense of $\left\|\mathcal{F}^{-1}\mathcal{F}y\right\|_\infty$, the function $t\mapsto y\left(t\right)$ must be an element of $C_0\left(\mathbb{R}\right)$. But in that case $\left\|y\right\|_\infty$ makes sense to, because $C_0\left(\mathbb{R}\right)\subset L_\infty\left(\mathbb{R}\right)$. Correct? $\endgroup$ – Trevor3 Aug 18 '17 at 21:06

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