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Let $f:[0,1]\to\Bbb{R}$ be continuous. How do I show that $\lim_{n\to\infty}\left(\int^1_0{f(x)^n}\right)^{1/n}=\sup_{0\leq x\leq 1}f(x)$?

I tried writing $\left(\int^1_0{f(x)^n}\right)^{1/n}=\left(f(x)^{n+1}/(n+1)\right)^{1/n}$. But I don't know what to do next.

Xena
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1 Answers1

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Assume $f$ is nonnegative.

Let $M:=\sup_{x \in [0,1]} f(x)$. Then notice that for every $n$, we have that $$\bigg(\int_0^1 f(x)^ndx\bigg)^{\frac{1}{n}} \leq \bigg(\int_0^1 M^n dx\bigg)^{\frac{1}{n}} = M$$ Therefore $\limsup_{n \to \infty} \big(\int_0^1 f(x)^ndx\big)^{\frac{1}{n}} \leq M$.

On the other hand, let $\alpha$ be any nonnegative real number strictly less than $M$. Then $f(x_0)=M$ for some $x_0$ in $[0,1]$ by continuity of $f$ hence, again by continuity of $f$, we can find an interval $(c,d) \subset [0,1]$ such that $f(x)> \alpha$ for all $x \in (c,d)$. Then for every $n$, we have that $$ \bigg(\int_0^1 f(x)^ndx\bigg)^{\frac{1}{n}} \geq \bigg(\int_c^d f(x)^ndx\bigg)^{\frac{1}{n}} > \bigg(\int_c^d \alpha^n dx\bigg)^{\frac{1}{n}}= \alpha \; (d-c)^{\frac{1}{n}} $$ Taking limits, we get that $\liminf_{n \to \infty} \big(\int_0^1 f(x)^ndx\big)^{\frac{1}{n}} \geq \lim_{n \to \infty} \alpha \; (d-c)^{\frac{1}{n}} = \alpha$

Since $\alpha$ was an arbitrary real number which was strictly less than $M$, the above calculation implies that $\liminf_{n \to \infty} \big(\int_0^1 f(x)^ndx\big)^{\frac{1}{n}} \geq M$.

We showed that $$M \leq \liminf_{n \to \infty} \bigg(\int_0^1 f(x)^ndx\bigg)^{\frac{1}{n}} \leq \limsup_{n \to \infty} \bigg(\int_0^1 f(x)^ndx\bigg)^{\frac{1}{n}} \leq M$$ which implies that $\lim_{n \to \infty} \big(\int_0^1 f(x)^ndx\big)^{\frac{1}{n}} =M$.

Did
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shalop
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  • In full rigor, in the second step one needs to use liminfs instead of limits, as long as one does not know the limit exists. – Did Mar 30 '15 at 07:35
  • Yeah, you're right. I edited it. – shalop Mar 30 '15 at 15:31
  • Right. If, additionally, you take care to explain that $f$ is assumed to be nonnegative everywhere (an assumption unfortunately forgotten in the question) and that one takes any $\alpha$ nonnegative, this answer will be correct. – Did Mar 30 '15 at 16:02
  • Indeed, my proof would only be true if we replace $f(x)$ with $|f(x)|$ throughout, and replace $M$ with $\sup_{[0,1]} |f|$, but I didn't really worry too much about that subtlety when I was typing this up. – shalop Mar 30 '15 at 16:05
  • No problem (and I added the mention that $\alpha$ should be nonnegative). – Did Mar 30 '15 at 19:02