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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.

Arman Malekzadeh
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    $f$ 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
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    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. –  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 –  Dec 27 '16 at 20:09
  • @Bungo thanks :) i didn't say that $f$ is negative :) i just said 'non-zero' :) anyway, excuse me for asking you do this ... can you please write an answer and explain why is my question equivalent to that one ? – Arman Malekzadeh Dec 27 '16 at 20:12
  • You didn't say that $f$ is negative, but your hypothesis does not exclude this possibility. The reason your question is equivalent to that one (when $f$ is nonnegative) is simply because $|f|p = \left( \int_a^b |f(x)|^p\ dx \right)^{1/p}$ for $1 \leq p < \infty$, and $|f|{\infty} = \max_{x \in [a,b]}|f(x)|$ (where the maximum exists and equals $|f|_{\infty}$ because $f$ is continuous on the compact interval $[a,b]$). Note that when $f$ is nonnegative we have $|f(x)| = f(x)$. –  Dec 27 '16 at 20:14
  • @Bungo I checked my book again ... the question was about not negative functions ... excuse me for my mistake ... :) thanks for your help :) – Arman Malekzadeh Dec 27 '16 at 20:20
  • OK, then the result is true and is proved at the link I gave in my first comment. Note, that proof (at least the one with the most upvotes) assumes that you know about Lebesgue integration and measure. You didn't mention your context, so I'm not sure if this background is assumed. If not, you should edit the question to emphasize that you want an elementary proof using Riemann integrals. –  Dec 27 '16 at 20:23

2 Answers2

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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 $[a,b]$.

Vik78
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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.

copper.hat
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