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Find $$\lim_{n \rightarrow \infty}\left(\int_0^1(f(x))^n\,\mathrm dx\right)^\frac{1}{n}$$if $f:[0,1]\rightarrow(0,\infty)$ is a continuous function.

My attempt:

Say $f(x)$ has a max. value $M$. Then $$\left(\int_0^1(f(x))^ndx\right)^\frac{1}{n}\leq\left(\int_0^1M^ndx\right)^\frac{1}{n} =M$$

I cannot figure out what to do next.

Lord_Farin
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  • Can't do much, there is no universal answer: consider constant $f$. But you might explore what connection there is with the max of $f$. – André Nicolas May 11 '13 at 07:09
  • @ Nicolas Thanks for your efforts. –  May 11 '13 at 07:16
  • If I try to think it as a summation then the maximum value $M$ will be the answer. –  May 11 '13 at 07:18
  • The strategy is to go a little below the max. to $M-\epsilon/2$, and then a little more to $M-\epsilon$. This is for technical reasons, to make the proof go nicely. Show that for large $n$, the part above $M-\epsilon/2$ crushes the part below $M-\epsilon$. – André Nicolas May 11 '13 at 07:28
  • @Umakant Check this question for a more general proof that :$$\lim\limits_{p\rightarrow\infty}||f||p=||f||{\infty}$$http://math.stackexchange.com/questions/242779/limit-of-lp-norm – Dimitris May 11 '13 at 07:43
  • Thank you all for the help. @ Nicolas, @ Dimitris, @ Lord_Farin, @ user17762, @ copper.hat, @ Landscape. –  May 11 '13 at 08:26

2 Answers2

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Your guess that it should be the maximum is a good guess. You have shown that the limit must be $\leq M$. We will now show that the limit must be greater than or equal to $M-\epsilon$ for any $\epsilon$, from which you can conclude that the limit is indeed $M$.

Since $f(x)$ is continuous, given $\epsilon > 0$, there exists a $\delta$ such that $$f(x) > M - \epsilon$$ for all $x \in (x_0 -\delta, x_0 + \delta)$. Hence, we have $$\int_0^1 f(x)^n dx > \int_{x_0 - \delta}^{x_0 + \delta} f(x)^n dx > \int_{x_0 - \delta}^{x_0 + \delta} (M - \epsilon)^n dx = (M-\epsilon)^n 2\delta$$ Hence for any $\epsilon > 0$, $$\left(\int_0^1 f(x)^n dx\right)^{1/n} > (M-\epsilon)(2 \delta)^{1/n}$$ Letting $n \to \infty$, we get what we want, i.e., $$\lim_{n \to \infty}\left(\int_0^1 f(x)^n dx\right)^{1/n} \geq (M-\epsilon)$$

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    "$\int_0^1 f(x)^n dx > \int_{x_0 - \delta}^{x_0 + \delta} f(x)^n dx$" this step have problems when the maximum is at the either ends (for eg: consider a function monotonic in the interval [0,1]). So I would suggest considering three cases (1): Maximum at $x=0$,(2):Maximum at $x=1$,(3):Maximum at $x\in (0,1)$ and proofs of the three cases are very similar to what you have shown . Anyway nice proof!. – hrkrshnn May 11 '13 at 08:58
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Let $M$ be as above, note that $M>0$ by assumption. Choose $\epsilon>0$ and let $E_\epsilon = \{x | f(x) >M-\epsilon \}$. Since $f$ is continuous, we have $m E_\epsilon >0$, and so $M \ge \left(\int_0^1(f(x))^ndx\right)^\frac{1}{n} \ge (M-\epsilon) \ (m E_\epsilon)^{\frac{1}{n}}$. Hence $\liminf_n \left(\int_0^1(f(x))^ndx\right)^\frac{1}{n} \ge M-\epsilon$. Since $\epsilon$ was arbitrary, we have $\liminf_n \left(\int_0^1(f(x))^ndx\right)^\frac{1}{n} \ge M$, from which it follows that $\lim_n \left(\int_0^1(f(x))^ndx\right)^\frac{1}{n} = M = \max_{x \in [0,1]} f(x)$.

copper.hat
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