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I am looking at the proof of $\lim_{p\to \infty} ||f||_p = ||f||_{\infty}$ here where we first show that:

$$\liminf_{p\to +\infty}\lVert f\rVert_p\geqslant\lVert f\rVert_\infty.$$

By using:

Fix $\delta>0$ and let $S_\delta:=\{x,|f(x)|\geqslant \lVert f\rVert_\infty-\delta\}$ for $\delta<\lVert f\rVert_\infty$. We have $$\lVert f\rVert_p\geqslant \left(\int_{S_\delta}(\lVert f\rVert_\infty-\delta)^pd\mu\right)^{1/p}=(\lVert f\rVert_\infty-\delta)\mu(S_\delta)^{1/p},$$ since $\mu(S_\delta)$ is finite and positive. This gives $$\liminf_{p\to +\infty}\lVert f\rVert_p\geqslant\lVert f\rVert_\infty.$$

May I know why $S_\delta$ has a strictly positive measure. I am thinking of a discontinuous function $f$ which is a point (which is the sup of $f$) along with a continuous segment such that $S_\delta$ is a single point and its measure is $0$?

  • If $|f|{\infty} =0$, there is nothing to show so assume otherwise. If $S\delta$ had zero measure, this would mean that $|f(x)| \le |f|\infty - \delta$ for a.e. $x$, contradicting the definition of $|f|\infty$ by an essential supremum. In the given link, the underlying measure space is assumed finite; thus $S_\delta$ has finite measure, and what we showed is that the measure of $S_\delta$ is strictly positive. – Calvin Khor Nov 04 '20 at 04:55
  • @CalvinKhor I see, Isn't $||f||\infty = \sup{x\in X}|f(x)|$? So I don't understand why the point where the $\sup$ is attained cannot be an isolated point? –  Nov 04 '20 at 04:59
  • It is an essential supremum, not a supremum. i.e. $|f|\infty = \inf{N} \sup_{x\in X\setminus N} |f(x)|$ where the inf is over null sets $N$. Google essential supremum for alternative definitions etc. In the case of continuous functions, the two coincide – Calvin Khor Nov 04 '20 at 05:00
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    @CalvinKhor Thanks. –  Nov 04 '20 at 05:04

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