If $f:[a,b]\rightarrow \mathbb{R}$ is Riemann integrable does $\lim_{p\rightarrow\infty}(\int_a^b|f(x)|^p)^{\frac{1}{p}}=\sup_{x\in[a,b]}|f(x)|$?

Question

Several posts on this site ask for a proof of the statement

If $f:[a,b]\rightarrow\mathbb{R}$ is continuous, $\lim_{p\rightarrow\infty}(\int_a^b|f(x)|^p)^{\frac{1}{p}}=\sup_{x\in[a,b]}|f(x)|$.

Need $f$ only be Riemann integrable for $\lim_{p\rightarrow\infty}(\int_a^b|f(x)|^p)^{\frac{1}{p}}=\sup_{x\in[a,b]}|f(x)|$ to be true?

If so, where can I find a proof of this that does not involve measure theory?

did you meant to write the $p$ norm of $f$ ? Otherwise, what you wrote is pointwise and is not at all related to the supremum. — dezdichado, Apr 16 '23 at 00:27
Well $\lim_{p\rightarrow\infty}(|f(x)|^p)^{\frac{1}{p}}=\sup_{x\in[a,b]}|f(x)|$ is certainly wrong. The LHS depends on $x$ and the RHS doesn't. I think you want something like: If $f:[a,b]\rightarrow\mathbb{R}$ is continuous, $\lim_{p\rightarrow\infty}(\int_a^b |f(x)|^p,dx)^{\frac{1}{p}}=\sup_{x\in[a,b]}|f(x)|$. Try to prove it yourself: $\varepsilon, \delta$ style. — GEdgar, Apr 16 '23 at 00:28
Yes. This is true even for Lebesgue measurable function (in which case, the supremum norm is slightly modified to essential sup). This is called continuity of norm. I proved it long time ago. You may search my posts and hopefully you can find it. — Danny Pak-Keung Chan, Apr 16 '23 at 00:34
@GEdgar You are correct as to the fact that there was a mistake in the original version of the post. Also, I have proven the case where $f$ is continuous. — FabrizzioMuzz, Apr 16 '23 at 00:35
The statement is not true for Riemann integrable functions. Consider $[a,b]=[0,1]$ and $f(x)=0$ for $0\le x<1$ and $f(1)=1.$ You need essential supremum which coincides with supremum for continuous functions. — Ryszard Szwarc, Apr 16 '23 at 00:36
Does this answer your question? If $f(x)$ is continuous on $[a,b]$ and $M=\max ; |f(x)|$, is $M=\lim \limits_{n\to\infty} \left(\int_a^b|f(x)|^n,\mathrm dx\right)^{1/n}$? — Mittens, Apr 16 '23 at 00:43
Another, more handy definition is $A={\rm essup}f(x)=\sup{ a,:, \mu{x,:, f(x)>a}>0}$. Basing on this definition the proof for Lebesgue integrable function $f$ is easy. — Ryszard Szwarc, Apr 16 '23 at 00:50
You may read
https://math.stackexchange.com/questions/2845992/lim-p-to-r-f-p-f-r/2860957#2860957

and

https://math.stackexchange.com/questions/3878358/how-to-show-that-f-lp-to-f-l-infty-as-p-to-infty-if-f-in/3878434#3878434 — Danny Pak-Keung Chan, Apr 16 '23 at 00:56
I’m voting to close this question because it was answered in comment with a counterexample. — Anne Bauval, Apr 16 '23 at 01:10
@Anne Bauval By replacing the supremum norm with essential supremum norm, the statement is true (even for Lebesgue measurable function). May be it is a good idea to modify the title of the question. — Danny Pak-Keung Chan, Apr 16 '23 at 01:16
A remark about essential sup norm. We say that a real number $M$ is an essential upper bound of $f$ if $f(x)\leq M$ for $x$-a.e.. The essential sup norm of $f$ , denoted by $||f||{\infty}$ , is defined by $||f||{\infty}=\inf{M\mid M\mbox{ is an essential upper bound of }|f|}.$ — Danny Pak-Keung Chan, Apr 16 '23 at 01:20
@DannyPak-KeungChan Yes we know but that was not the point, and the OP said Ryszard's answer was what he was looking for. When he understood his statement was false, he stopped asking for an elementary proof for non continuous Riemann integrable functions. So I think changing his title would betray his intent. — Anne Bauval, Apr 16 '23 at 01:20

If $f:[a,b]\rightarrow \mathbb{R}$ is Riemann integrable does $\lim_{p\rightarrow\infty}(\int_a^b|f(x)|^p)^{\frac{1}{p}}=\sup_{x\in[a,b]}|f(x)|$?

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