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Given that $f$ is continuous and non-negative on $[0,1]$, I was asked to find the limit of:

$$\displaystyle\lim_{n \rightarrow \infty}\left(\int_0^1(f(x))^{n} dx\right)^{\frac{1}{n}}$$

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Note that in different notation this corresponds to $\lim_{p \rightarrow \infty} \| f \|_{L^{p}([0,1])}$. –  johnny Aug 29 '12 at 8:14
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It's the maximum value that $f$ takes. Do you see why? –  ronno Aug 29 '12 at 8:14
    
yes ronno thank you - that is what i was thinking (actually i was thinking sup(f) on [0,1] but it is the same as f is continuous there), but i really needed a second opinion. –  JohnnyE. Aug 29 '12 at 8:46

1 Answer 1

up vote 3 down vote accepted

Let $M=\sup_{x\in[0,1]}f(x)$. It's not difficult to see that $$M\geq \displaystyle\lim_{n \rightarrow \infty}\left(\int_0^1(f(x))^{n} dx\right)^{\frac{1}{n}}.$$

On the other hand, there exists $x_0\in[0,1]$ such that $f(x_0)=M$ since $f$ is continuous on $[0,1]$. Again, by the continuity of $f$, for any $\epsilon>0$, there exists $\delta>0$ such that $$\tag{1}f(x)>M-\epsilon\mbox{ whenever }x\in (x_0-\delta,x_0+\delta)\cap[0,1].$$ Hence, we have $$\tag{2}\int_0^1(f(x))^{n} dx\geq \int_{(x_0-\delta,x_0+\delta)\cap[0,1]}(f(x))^{n} dx\geq m((x_0-\delta,x_0+\delta)\cap[0,1])(M-\epsilon)^n,$$ where the first inequality follows from the assumption that $f$ is nonnegative, and the second inequality follows from $(1)$. Also, $m((x_0-\delta,x_0+\delta)\cap[0,1])$ is the "length" of the interval $(x_0-\delta,x_0+\delta)\cap[0,1]$. For example, if $(x_0-\delta,x_0+\delta)\subset[0,1]$, then $m((x_0-\delta,x_0+\delta)\cap[0,1])=2\delta$; if $x_0=0$, then $m((x_0-\delta,x_0+\delta)\cap[0,1])=\delta$. In any case, we always have $$m((x_0-\delta,x_0+\delta)\cap[0,1])\geq\delta>0.$$ (This is important when we take $n$-root and as $n$ goes to infinity.)

Now taking $n$-th root of $(2)$, you will get the answer. I think you can work out the details from here.

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