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.

  • 1
    $\begingroup$ $f$ 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. $\endgroup$ – ajotatxe Dec 27 '16 at 19:23
  • $\begingroup$ Do you mean to include $f$ is nonnegative or to define $v_n=\left(\int_a^b|f|^n\right)^{1/n}$ perchance? $\endgroup$ – Clayton Dec 27 '16 at 19:24
  • 2
    $\begingroup$ 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. $\endgroup$ – Bungo Dec 27 '16 at 19:25
  • $\begingroup$ @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 :) $\endgroup$ – Arman Malekzadeh Dec 27 '16 at 20:06
  • $\begingroup$ 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 $\endgroup$ – 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 $[a,b]$.

  • $\begingroup$ And what would happen if $f$ attained its maximum at more than one point? $\endgroup$ – 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
    $\begingroup$ (+1) for the solid answer. I've deleted my previous comments. $\endgroup$ – Mark Viola Sep 7 '18 at 21:08
  • $\begingroup$ @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! $\endgroup$ – copper.hat Sep 7 '18 at 21:21
  • $\begingroup$ Thanks Joe. Enjoy yours too. $\endgroup$ – Mark Viola Sep 7 '18 at 21:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.