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Let $f$ be continuous with maximum $m$ on $[a, b]$ and $f(x) \geq 0$ for all $x \in [a,b]$. Show $$ \lim_{n \to \infty} \sqrt[n]{\int_a^b [f(x)]^n\ dx} = m. $$

This is based on a problem I saw from Courant, and is of course the infinity norm on a continuous function space. Courant asks to prove this via real analysis; measure theory isn't available.

Several posts here use something similar as a definition: $$ \|f\|_\infty := \max_{x \in [0,1]} |f(x)|$$ and some prove it for the discrete case, but none prove it or develop it for continuous functions via integrals.

My work so far below. Criticism or improvements of my work, or alternate solutions, are all requested.

Proof

If $m = 0$, the claim is trivial, so we assume $m > 0$. We will find an upper bound and a lower bound of the limit in question and show they both equal $m$.

Recall that for any $k > 0$, $$ \lim_{n \to \infty} \sqrt[n]{k} = 1. $$ Therefore, we have the upper bound $$ \begin{align*} \lim_{n \to \infty} \sqrt[n]{\int_a^b [f(x)]^n\ dx} &\leq \lim_{n \to \infty} \sqrt[n]{\int_a^b m^n \ dx} \\ &=\lim_{n \to \infty} \sqrt[n]{(b-a)} \cdot \lim_{n \to \infty} \sqrt[n]{m^n}\\ &= m. \end{align*} $$ Since $f$ is continuous and attains its maximum $m$ in $[a, b]$, for any $\varepsilon > 0$, there exists a $c \in (a, b)$ and $\delta > 0$ such that for all $x \in [c, c+\delta]$ $$ f(x) \geq m - \varepsilon $$ so for $0 < \varepsilon < m$, $$ \begin{align*} \int_a^b [f(x)]^n \ dx &\geq \int_c^{c + \delta} [f(x)]^n \ dx \\ &\geq \int_c^{c+\delta} (m - \varepsilon)^n \ dx \\ &= \delta \cdot (m - \varepsilon)^n. \end{align*} $$ By the Binomial Theorem, we have for $0 < \varepsilon < m$ $$ (m - \varepsilon)^n = m^n + \mathcal{O}(m^{n-1}\varepsilon) $$ so that $$ \begin{align*} \lim_{n \to \infty} \sqrt[n]{\delta \cdot (m - \varepsilon)^n} &= \lim_{n \to \infty} \sqrt[n]{\delta} \cdot \lim_{n \to \infty} \sqrt[n]{(m - \varepsilon)^n} \\ &= 1 \cdot \lim_{n \to \infty} \sqrt[n]{ m^n + \mathcal{O}(m^{n-1}\varepsilon)} \\ &= m. \end{align*} $$ giving a lower bound and completing the proof.

Discussion

It's fascinating that the size of the interval where $f$ is near $m$ is irrelevant in the limit. This is the crux of the proof; the rest is just mechanics.

I tried to achieve both rigor and clarity while not getting caught up in arrays of multiple epsilons and deltas. Is the line where I invoke the Binomial Theorem via Big-Oh notation rigorous and sufficient? Is there a more rigorous way to develop that? How could the proof and its exposition be improved?

Update

Based on Asigan's very helpful suggestion, I would revise the proof by deleting the (false!) section stating "By the Binomial Theorem..." and instead writing:

Therefore, for any $\varepsilon > 0$, $$ \begin{align*} \int_a^b [f(x)]^n \ dx &\geq \lim_{n \to \infty} \sqrt[n]{\delta} \cdot \lim_{n \to \infty} \sqrt[n]{(m - \varepsilon)^n} \\ &= m - \varepsilon. \end{align*} $$ This shows $m$ is a lower bound (by the Axiom of Completeness) and completes the proof.

SRobertJames
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    this is the classic result for the limit of $L^p$-norm for $p\to\infty$, see eg https://math.stackexchange.com/questions/242779/limit-of-lp-norm – daw Oct 27 '24 at 18:20
  • @daw The proof you linked to requires measure theory. The result is provable via elementary real analysis, which is what this question is about. I've revised it to make that clearer. – SRobertJames Oct 27 '24 at 19:07

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For a fixed $\epsilon>0$, $$\lim_{n \to \infty} \sqrt[n]{\delta \cdot (m - \epsilon)^n}= \lim_{n \to \infty} \sqrt[n]{\delta} \cdot \lim_{n \to \infty} \sqrt[n]{(m - \epsilon)^n}=\lim_{n \to \infty} \sqrt[n]{\delta}(m-\epsilon)=m-\epsilon.$$ Therefore your proof is incorrect. However you are almost here:

So you have proved that for all $\epsilon>0$, exists $\delta$ subject to, $\int_a^b f^n\ge \delta(m-\epsilon)^n$ for all $n$, i.e. $\sqrt[n]{\int_a^b f^n}\ge \sqrt[n]{\delta}(m-\epsilon)$ for all $n$. As $n\to +\infty$, RHS converges to $m-\epsilon$. Therefore exists $N$, for all $n>N$, $\sqrt[n]{\delta}(m-\epsilon)>m-2\epsilon$. For this choice of $N$, if $n>N$, then $$\sqrt[n]{\int_a^b f^n}\ge \sqrt[n]{\delta}(m-\epsilon)>m-2\epsilon. $$ This and $\sqrt[n]{\int_a^b f^n}\le m$ together gives the desired result.

Asigan
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  • Thanks, this is very helpful. – SRobertJames Oct 27 '24 at 21:18
  • To fully understand this, I rewrote the end of my proof based on your feedback, but in my own words. Can you take a look? – SRobertJames Oct 27 '24 at 21:43
  • I regret to say this is still incorrect. We have for each $n$, $\sqrt[{\color{violet}n}]{\int_a^b f^{\color{violet}n}}\ge\sqrt[{\color{violet}n}]{\delta(m-\epsilon)^{{\color{violet}n}}}$. Both sides depend on $n$ and you can not keep $n$ in LHS fixed while letting $n\to+\infty$ in RHS. However, if you are familiar with the upper/lower limit, then you may know it is reasonable to say $\liminf_{n\to+\infty} \sqrt[n]{\int_a^bf^n}\ge\lim_{n\to+\infty}\sqrt[n]{\delta(m-\epsilon)^n}=m-\epsilon$, which would imply $\liminf_{n\to+\infty}\sqrt[n]{\int_a^bf^n}\ge m$. This may well conclude the proof. – Asigan Oct 27 '24 at 23:12
  • If we come back to the basic property of limits, "taking $n\to+\infty$" is actually using the following fact: If $a_n\ge b_n$ for all $n$, $(a_n)$ and $(b_n)$ converges to $a$ and $b$ respectively, then $a\ge b$. Bearing this in mind would help write more rigorous proof. – Asigan Oct 27 '24 at 23:12
  • Got it - this is extraordinarily helpful. Thank you. – SRobertJames Oct 28 '24 at 01:22