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Let $f(x)$ be a continuous real-valued function on $[a,b]$ and $M=\max\{|f(x)| \; :\; x \in [a,b]\}$. Is it true that: $$ M= \lim_{n\to\infty}\left(\int_a^b|f(x)|^n\,\mathrm dx\right)^{1/n} ? $$

Thanks!

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3  
Did you forget $\lim\limits_{n\to\infty}$ somewhere? –  Ilya Dec 4 '11 at 17:46
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In other words, you are asking if $\| f \|_n \to \| f \|_\infty$ for a continuous $f$. –  Srivatsan Dec 4 '11 at 19:09
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up vote 12 down vote accepted

Put $S_{\delta}:=\{x\in\left[a,b\right], |f(x)|\geq M-\delta\}$ for any $\delta>0$. Then we have for all $n$ $$M\cdot (b-a)^{\frac 1n}\geq\left(\int_a^b|f(x)|^ndx\right)^{\frac 1n}\geq \left(\int_{S_{\delta}}|f(x)|^ndx\right)^{\frac 1n}\geq (M-\delta)\left(\lambda(S_{\delta})\right)^{\frac 1n}.$$ Since $f$ is continuous, the measure of $S_{\delta}$ is positive and taking the $\liminf$ and $\limsup$, we can see that $$M\geq \limsup_n \left(\int_a^b|f(x)|^ndx\right)^{\frac 1n}\geq M-\delta\quad \mbox{and }\quad M\geq \liminf_n\left(\int_a^b|f(x)|^ndx\right)^{\frac 1n}\geq M-\delta$$ for all $\delta>0$, so $\lim_{n\to\infty}\left(\int_a^b|f(x)|^ndx\right)^{\frac 1n}=M$.

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Also, it will be nice to point out where continuity is being used here: to argue that $S_\delta \cap [a,b]$ is of positive measure. –  Srivatsan Dec 4 '11 at 18:16
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You may replace $S_\delta \cap [a,b]$ by $S_\delta$ everywhere. –  Did Dec 4 '11 at 22:21
    
Why do you absolutely need to take the lim inf and lim sup? All the limits exists. –  Patrick Da Silva Jan 5 '12 at 10:42
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@PatrickDaSilva We don't know a priori that the limit $\lim_n\left(\int_a^b|f(x)|^n\right)^{\frac 1n}$ exists. –  Davide Giraudo Jan 5 '12 at 12:36
    
Hm. I thought it was a sandwich argument at first, but I noticed the difference. Thanks for pointing it out –  Patrick Da Silva Jan 5 '12 at 23:13
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