# Prove that the sequence $\{v_n\}$ converges to the maximum value of $f$ on $[a,b]$

Question :

Assume that $f$ is a continuous not negative function on any point defined on the interval $[a,b]$.
For each natural number $n$ , Assume that $v_n$ is the $n$'th root of $\int_a^b f^n$.
Prove that the sequence $\{v_n\}$ converges to the maximum value of $f$ on $[a,b]$.

Note : The problem is how to connect these things in a formal way. I know that many important things happen on the roots. ( Specially maxima and minima ) But I don't know how to relate them to convergence of a sequence like $\{v_n\}$.

Please, If you have the time, explain your answer a bit more. I'm new to integration. Thanks in advance.

• $f$ is continuous and does not vanish. Then the sign of $f$ is constant. But if $f$ is negative, then $v_n$ is positive for even $n$ and negative for odd $n$, and $v_n$ converge to $0$ or does not converge. – ajotatxe Dec 27 '16 at 19:23
• Do you mean to include $f$ is nonnegative or to define $v_n=\left(\int_a^b|f|^n\right)^{1/n}$ perchance? – Clayton Dec 27 '16 at 19:24
• If $f$ is known to be nonnegative, then the problem is essentially to show that $\lim_{p \to \infty}\|f\|_p = \|f\|_{\infty}$. See here for example. – Bungo Dec 27 '16 at 19:25
• @Bungo I don't see the problem ... Do you say that this question is wrong ? i can't understand what is it you all are trying to say :) – Arman Malekzadeh Dec 27 '16 at 20:06
• I mean that if $f$ is nonnegative, your question is equivalent to showing that $\lim_{p \to \infty}\|f\|_p = \|f\|_{\infty}$, and this is proved at the link I provided. On the other hand, if $f$ is negative, I don't think the result is true, for the reason given by @ajotatxe – Bungo Dec 27 '16 at 20:09

Suppose $f$ is nonnegative on $[a,b]$ and attains a maximum $M$ at $c \in [a,b]$. For any $\epsilon > 0$, there exists $\delta>0$ such that for all $y \in (c - \delta, c + \delta)$, $f(y) > M - \epsilon$. Therefore $[\int_a^b f^n(x) dx]^{\frac{1}{n}} \ge [\int_{c - \delta}^{c + \delta} f^n(x) dx]^{\frac{1}{n}} > (2\delta)^{\frac{1}{n}} (M - \epsilon)$. As $n \to \infty$, we see that the limit is bounded below by $M - \epsilon$. We also obviously have $(b-a)^{\frac{1}{n}} M \ge [\int_a^b f^n(x) dx]^{\frac{1}{n}}$. As $n \to \infty$, the limit is thus bounded above by $M$. Since $\epsilon$ was arbitrary you can use the squeeze theorem to conclude that the limit is $M$. An essentially identical argument works if $c$ is an endpoint of $[a,b]$.

• And what would happen if $f$ attained its maximum at more than one point? – Mark Viola Sep 7 '18 at 16:38

If $f=0$ the result is clear, so assume that $f \neq 0$.

Let $M=\max f$ and $I_\epsilon = \{ x \in [a,b] | f(x) > M-\epsilon \}$. Note that $m I_\epsilon >0$.

Then $\int f^n \ge (M-\epsilon)^n m I_ \epsilon$ and so $\sqrt[n]{ \int f^n } \ge (M-\epsilon) \sqrt[n]{ I_\epsilon }$ and so $\liminf_n \sqrt[n]{ \int f^n } \ge (M-\epsilon)$ from which we have $\liminf_n \sqrt[n]{ \int f^n } \ge M$.

Similarly, we have $m([a,b]) M^n \ge \int f^n$ and so $\sqrt[n]{m([a,b])} M \ge \sqrt[n]{ \int f^n }$. Taking limits gives $M \ge \limsup \sqrt[n]{ \int f^n }$ from which the answer follows.

• (+1) for the solid answer. I've deleted my previous comments. – Mark Viola Sep 7 '18 at 21:08
• @MarkViola: I'm glad someone is checking! Sometimes I look back at an answer I wrote and wonder what I was thinking :-). Enjoy the weekend! – copper.hat Sep 7 '18 at 21:21
• Thanks Joe. Enjoy yours too. – Mark Viola Sep 7 '18 at 21:22