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So I've been trying to find a solution for this all afternoon, but haven't found a good place to start:

Prove that if $f:[a,b]\to\mathbf{R}^+$ is a continuous function with maximum value $M$, then $$ \ \lim_{n\to\infty}\left(\int_a^b f(x)^n\,dx\right)^{1/n} = M $$

Here are some of the paths I've considered, though none have been very successful:

(1) Considering the sequence of functions for all increasing integer $n$ and trying to show that the sequence converges. We've had plenty of work on converging sequences, but with the integral expression, I am not sure how to simplify.

(2) Showing that that sequence is increasing (again, how?) and then showing there to be a supremum at $M$. I'm not sure how the maximum of the function arrives in this problem.

(3) Mean value theorems for integrals

If anyone could give me a solid place to start or perhaps point me to a place where this question has been asked before (I can't seem to find it), I would be very grateful.

Jay Reiter
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  • Completely believable, but I've never seen this fact in my 30 years in calculus. Interesting. – Randall Nov 30 '18 at 03:52
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    Hint: Try squeezing it: For example, $f(x) \leq M$, so $\int_a^bf(x)^n , dx \leq (b-a)M^n$. – Anurag A Nov 30 '18 at 03:53
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    Note that the expression $\left(\int_a^bf(x)^n\right)^\frac{1}{n}$ is the definition of the $L^p$ norm of the function $f$ (since it is positive). For $n\rightarrow\infty$ this converges to the max-norm. – obscurans Nov 30 '18 at 03:55
  • @AnuragA I see how this gives you an upper bound for the limit, but how do you go about squeezing it from below? or show the sequence is increasing? – Jay Reiter Nov 30 '18 at 04:12
  • @obscurans can you point me to a proof of this fact? – Jay Reiter Nov 30 '18 at 04:19
  • https://math.stackexchange.com/questions/242779/limit-of-lp-norm I think this now qualifies as a duplicate? – obscurans Nov 30 '18 at 04:31
  • I agree with obscurans. This question has appeared many times in slightly different formats and is a well known result, – Kavi Rama Murthy Nov 30 '18 at 06:11

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If you have done the following question, then you probably know how to do the one in the post:

Given $m$ positive real numbers $a_1, \dots, a_n$, prove that $$ \lim_{n\to +\infty} (a_1^n + \dots + a_m^n)^{1/n} = \max_{j} a_j. $$

Proof

The trick is to use the squeeze theorem, and note that $$ \max_j a_j \leqslant (a_1^n + \dots + a_m^n)^{1/n} \leqslant m^{1/n} a_j. $$ Now let $n \to +\infty$, and note that $\lim_n m^{1/n} = 1$.

Using the similar trick, we could do the original one. The problem is $f$ is "continuous". Even $f$ could reach $M$ at some point $x_0$, we cannot directly use it. So we only requires that for each $\varepsilon >0$, find those $x$ s.t. $M-\varepsilon \leqslant f(x) \leqslant M$. So we have the following proof.

Proof

Let $x_0 \in [a, b]$ be the point where $f(c) =M$ [this is possible, since $f\in \mathcal C[a,b]$]. For each $\epsilon > 0$, by the definition of continuity, there exists an interval $[c,d]$ that contains $x_0$ s.t. $$ x\in [c,d] \implies M-\varepsilon \leqslant f(x) \leqslant M. $$ Thus $$ (d-c)^{1/n} (M-\varepsilon) \leqslant \left(\int_c^d f^n\right)^{1/n}\leqslant \left(\int_a^b f^n\right)^{1/n} \leqslant M, $$ and then $$M-\varepsilon \leqslant \varliminf_n \left(\int_a^b f^n\right)^{1/n} \leqslant \varlimsup_n \left(\int_a^b f^n\right)^{1/n} \leqslant M. $$ Now let $\varepsilon \to 0^+$, then the upper limit and the lower limit are equal to $M$, hence the limit equation.

Remark

Technically, this is actually not an application of the squeezing theorem [the end of the inequality chains are not the same], but the idea is similar. If no tools of upper/lower limits are allowed, then the whole proof could be completed in $\varepsilon$-$N$ format.

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