# 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

• But what if $f$ is a constant function? – Brandon Jan 14 '14 at 19:04
• 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
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.