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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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    $\begingroup$ Did you forget $\lim\limits_{n\to\infty}$ somewhere? $\endgroup$ – Ilya Dec 4 '11 at 17:46
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    $\begingroup$ In other words, you are asking if $\| f \|_n \to \| f \|_\infty$ for a continuous $f$. $\endgroup$ – Srivatsan Dec 4 '11 at 19:09
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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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  • $\begingroup$ Why do you absolutely need to take the lim inf and lim sup? All the limits exists. $\endgroup$ – Patrick Da Silva Jan 5 '12 at 10:42
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    $\begingroup$ @PatrickDaSilva We don't know a priori that the limit $\lim_n\left(\int_a^b|f(x)|^n\right)^{\frac 1n}$ exists. $\endgroup$ – Davide Giraudo Jan 5 '12 at 12:36
  • $\begingroup$ Hm. I thought it was a sandwich argument at first, but I noticed the difference. Thanks for pointing it out $\endgroup$ – Patrick Da Silva Jan 5 '12 at 23:13
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It can be seen as a particular case of Laplace's principle. For simplicity assume that $f(x) \ge 0$.

$$\lim_{n\rightarrow\infty} \log \frac{1}{n} \int_a^b \exp(n \log f(x))\mathrm d x = \max_{x\in[a,b]} \log f(x)$$

Taking $\exp$ of both sides gives the result.

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