# Show that $\lim_{n\to \infty}\left[ \int_a^bf(x)^n\,dx \right]^{1/n}=\max_{x\in[a,b]}f(x).$ [duplicate]

Suppose $f:[a,b]\to \mathbb{R}$ is continuous and positive. Show that $$\lim_{n\to \infty}\left[ \int_a^bf(x)^n\,dx \right]^{1/n}=\max_{x\in[a,b]}f(x).$$

My progress: A simpler version of the problem is to suppose $x_1, \dotsc, x_m$ are positive numbers, and show that $$\lim_{n\to\infty}\left( x_1^n + \dotsb + x_m^n \right)^{1/n} = \max_i x_i.$$ This we can do by showing that $\log \left( x_1^n + \dotsb + x_m^n \right)^{1/n} \to \log (\max_i x_i)$. $$\log \left( x_1^n + \dotsb + x_m^n \right)^{1/n} = \frac{1}{n}\log \left( x_1^n + \dotsb + x_m^n \right) = \frac{1}{n}\left( n\log\left( \max_i x_i \right) + \log \left( \left(\frac{x_1}{\max_i x_i} \right)^n + \dotsb + \left(\frac{x_M}{\max_i x_i} \right)^n \right)\right)\\ = \log\left(\max_i x_i\right) + \frac{1}{n}\log\left(\text{bounded}\right)\to \log\left( \max_i x_i\right).$$

Now with the case of a function being integrated, if we tried the same argument, letting $M= \max_{x\in [a,b]}f(x)$, we'd have $$\log \left( \left(\int_a^b f(x)^n\,dx \right) ^{1/n}\right) = \log M + \frac{1}{n}\log \left(\int_a^b (f(x)/M)^n \, dx \right),$$ and now the task is to show that the second term goes to zero.

I see that it goes to zero if $\{x\mid f(x)=M\}$ has positive measure, since then the integral is bounded below by the measure of that set. How can we show the second term goes to zero otherwise?

## marked as duplicate by leo, Davide Giraudo, TZakrevskiy, Daniel Fischer, sdcvvcJan 14 '14 at 21:13

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• But what if $f$ is a constant function? – Brandon Jan 14 '14 at 19:04
• @brandon It seems to work out...what is the problem that you see? Perhaps the result does not hold...I was under the impression that this is why the Lp norms tend to the sup norm – Eric Auld Jan 14 '14 at 19:12
• If $f$ is constant the term in the last inequality doesn't go to zero (I also guess that it should be an inequality since we don't always have the inequality) – Brandon Jan 14 '14 at 19:31
• @brandon I made a typo...should have been f/M. – Eric Auld Jan 14 '14 at 20:56

## 1 Answer

I have a proof that uses a similar approach. Let $c$ be a point in $[a,b]$ such that $f(c)=\sup\limits_{x\in[a,b]} (f(x))=M$. We may assume that $c$ is an inner point of the interval, the other cases are Handel similarly. Note that for every $x$, we have $f(x)^n\leq M^n$. On integrating, we obtain $$\int_a^b f(x)^n \, dx\leq \int_a^b M^n\, dx = M^n(b-a)$$ Hence, by taking the $n$-th root, $$\left(\int_a^b f(x)^n \, dx\right)^{1/n}\leq M(b-a)^{1/n}$$ Using the fact that for every positive real $x$ we have $\lim\limits_{n\to\infty}x^{1/n}=1$, we get $$\lim\limits_{n\to\infty}\left(\int_a^b f(x)^n \, dx\right)^{1/n}\leq M$$

Now, for every $\varepsilon>0$ there exists a $\delta>0$ such that $M-\varepsilon<f(x)$ if $c-\delta\leq x\leq c+\delta$. Rising to the $n$-the power and integrating over the interval $[c-\delta, c+\delta]$ we obtain $$2\delta(M-\varepsilon)^n\leq \int_{c-\delta}^{c+\delta} f(x)^n \, dx\leq \int_a^b f(x)^n\, dx$$ The last inequality follows because $f$ is positive. Again, taking the $n$-the root yields $$(2\delta)^{1/n}(M-\varepsilon)\leq \left( \int_a^b f(x)^n\ , dx\right)^{1/n}$$ Since this is valid for every positive integer $n$, using the same limit that above, it follows that $$M-\varepsilon\leq\lim\limits_{n\to\infty}\left(\int_a^b f(x)^n\right)^{1/n}$$ And since this is valid for every $\varepsilon>0$, we obtain the result.