A question about essential supremum

Question

Definition of the essential supremum is

$$||f||_{\infty}=\text{inf}\{a\geq 0:\mu(\{x:|f(x)|>a\})=0\}.$$

So $||f||_{\infty}$ is not necessarily supremum of $f$.

In http://homepage.ntu.edu.tw/~d04221001/Notes/Problems%20and%20Solutions%20Section/Folland%20Real%20Analysis/Folland%20Real%20Analysis%20Solution%20Chapter%206%20Lp-Spaces.pdf

question 1, It uses that for any $x_n$ belonging to the domain of the functions, $$|f(x_n)+g(x_n)|\leq||f+g||_{\infty}$$ and then takes limit of the both sides.

I cannot understand it. Because $||f+g||_{\infty}$ is not supremum of $f+g$.

Can anyone explain it?

Moreover, in Folland's book page 186, if $f$ is defined on a set $A$ and the measure is counting measure, then it writes $$||f||_{\infty}^p=\text{sup}_{\alpha}|f(\alpha)|^p$$

Answer 1

Let's assume that $\left\Vert f \right\Vert_\infty < \infty$ since otherwise the inequality $$ |f(x)| \leq \left\Vert f\right\Vert_\infty $$ is trivial.

You should interpret $\left\Vert f\right\Vert_\infty$ as the smallest value of $a \geq 0$ such that $|f(x)| \leq a$ for almost every $x \in X$. To make this formal note that by definition of the infimum, we can choose a sequence $(\alpha_n)$ belonging to the set $$ \{a\geq 0:\mu(\{x:|f(x)|>a\})=0\} $$ such that $\alpha_n \to \left\Vert f \right\Vert_\infty$ as $n \to \infty$. By definition, we must have $|f(x)| \leq \alpha_n$ for almost every $x \in X$. That is, for each $n \geq 1$, there exists a set $Z_n$ of measure zero with the property that $$ |f(x)| \leq \alpha_n \quad \forall x \notin Z_n. $$ Now, $Z := \bigcup_{n \geq 1} Z_n$ will also have measure zero and, moreover, for every $n \geq 1$ there holds $$ |f(x)| \leq \alpha_n \quad \forall x \notin Z. $$ For each of these $x$, taking the limit as $n \to \infty$ shows that $$ |f(x)| \leq \left\Vert f\right\Vert_\infty \quad \forall x \notin Z. $$ Since $Z$ has measure zero, this means that $|f(x)| \leq \left\Vert f\right\Vert_\infty$ for almost every $x \in X$. As for when a set is given the counting measure, note that any non-empty set has positive measure. Hence, almost everywhere really becomes "everywhere".

Answer 2

To help clarify what's going on, an example: let $f(x)=\begin{cases}1&x=\frac12\\0&\text{otherwise}\end{cases}$ in a space of functions on $[0,1]$.

What is the $p$-norm of this $f$, for $p\in (1,\infty)$? $0$, of course. That value at $\frac12$ is "invisible" to the integral in $\int_0^1 |f|^p =\int_0^1 f$. When we integrate, it might as well not be there. In fact, if we're working in a $L^p$ space, this function is in the same equivalence class as zero.

So, then, we extend this to $p=\infty$. We'd like there to be a sort of continuity; if $\|f\|_p$ exists for all (sufficiently large) finite $p$, then $\|f\|_{\infty}=\lim_{p\to\infty} \|f\|_p$. For that to be true for our example, we'll have to dismiss that special value $f(\frac12)$ at $p=\infty$ in the same way we do for finite $p$.

Refining that, we reach the idea of the essential supremum. Rather than simply taking the supremum of the values, we ignore measure-zero aberrations and define $\|f\|=\sup (A_f)=\inf (B_f)$, where $$a\in A_f\quad\text{if}\quad \mu\{x\mid |f(x)|\ge a\}\text{ is positive}$$ $$b\in B_f\quad\text{if}\quad \mu\{x\mid |f(x)|\ge b\}\text{ is zero}$$ For a continuous function, the essential supremum is the same as the supremum; if the function exceeds some $a$ at a point, it does so in an interval with positive measure. But of course, continuous functions aren't everything.

Moreover, if ... the measure is counting measure

In counting measure, the only measure-zero set is the empty set. With that, the essential supremum is just the same as the supremum.

It uses that for any $x_n$ belonging to the domain of the functions, $$|f(x_n)+g(x_n)|\le \|f+g\|_{\infty}$$

OK, not "for all $x_n$". That's actually "for almost all $x_n$". And, following the link ... that's not a valid proof. A single sequence of points is insufficient to pin down the essential supremum.