convolution operator in $L_{\infty}[0,\infty) \quad$ (from "A Course in Robust Control Theory: A Convex Approach")

Question

I am reading the textbook A Course in Robust Control Theory: A Convex Approach (Ch 3.2, operator) My question is from the following part:

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I am confused about the following:

1. $y(t) = (Fu)(t) \in L_{\infty}[0,\infty)$. And the author prove the upper bound of $|y(t)|=\|f\|_1\|u\|_{\infty}$. My question is does this upper bound derivation for $|y(t)|$ implies that $y(t) = (Fu)(t) \in L_{\infty}[0,\infty)$? $$y(t)\in L_{\infty}[0,\infty) \Rightarrow \int_0^{\infty} |y(t)|_{\infty}^{\infty} \leq \infty,$$ which is the definition from

Since if $F\in L_{\infty}[0,\infty)$, then $F$ is a bounded mapping, right? then we do not have to prove $|y(t)|\leq \|f\|_1\|u\|_{\infty}$

2. How to prove $\|F\|_{L_{\infty}\rightarrow L_{\infty}}\leq \|f\|_1$?

I read and study many times and have no idea to fully understand these.

Answer 1

The norm on $L_p(0,\infty)$ is defined as an integral norm for $p\in (0,\infty)$, as quoted from your book (although the sub-index $p$ confuses me a bit, as it should be ($\|u\|_p=\int_0^\infty|u(t)|^p dt)^{\frac{1}{p}}$).

However, for $p=\infty,$ the norm on $L_\infty(0,\infty)$ is not defined as an integral, but as $\|u\|_\infty = \operatorname{esssup}_{t\in (0,\infty)}|u(t)|$. While I don't want to go into detail, the take home-message is: If a function on $(0,\infty)$ is bounded, then it belongs to $L_\infty(0,\infty)$.

Now to answer 1.: This is concerned with the norm of the function $Fu$. We see that $\max_{t\in(0,\infty)} |y(t)|\leq \max_{t\in(0,\infty)}\|f\|_1 \|u\|_\infty = \|f\|_1 \|u\|_\infty$, i.e. $y$ is bounded and thus belongs to $L_\infty(0,\infty)$.

To answer 2.: This is not concerned with the norm of a function, but with the norm of an operator. For an operator $F:X\rightarrow Y$ from one normed space to another, we say that it's bounded if there's some $M\in \mathbb R$ such that $\|Fu\|_Y \leq M \|u\|_X$. We call the smallest of these $M$ the operator norm of $F$ and your book denotes that by $\|F\|_{X\rightarrow Y}$.

Here we see that $\|Fu\|_\infty = \|y\|_\infty\leq \|f\|_1\|u\|_\infty$ and hence $F$ is bounded and $\|F\|_{X\rightarrow Y}\leq \|f\|_1$ (and we can see equality if we plug in $u(t)=1$).

A closing remark: The vector spaces we work with are usually not vector spaces of functions, but vector spaces of equivalence classes of functions (because of functions which are equal almost everywhere). This doesn't stop us from pretending that $L_p$ is a vector space of functions for all practical purposes.