real analysis - Proving that $lim_{ntoinfty}(int_{a}^{b}f(x)^ndx)^{1/n} = max_{xin [a,b]}f(x)$

I'm trying to prove the following statement:

$$\lim_{n\to\infty}\left(\int_{a}^{b}f(x)^ndx\right)^{1/n} = \max_{x\in [a,b]}f(x)$$

where $[a,b] \subset \mathbb{R}$ and $f$ is non-negative and continuous.

I've tried to prove it in a similar way that we prove that $\displaystyle\lim_{n\to\infty}(a^n+b^n)^{1/n} = b$ if $b>a$.

However, I'm stuck in the end with an iterated limit of the form $\displaystyle\lim_{n\to\infty} \lim_{\epsilon\to 0 } \left(\epsilon^{1/n}\max_{x\in [a,b]}f(x)\right)$.

Is this last expression equal to $\displaystyle\max_{x\in [a,b]}f(x)$? If not, could anyone please give me a hint as to how to go about this proof?

Answer

Here is it, an elementary proof. Let $M=\max \{f(x): x\in [a,b]\}$. Since $f$ is non negative then $M>0$. For $n\geq 1$, define $u_n= \left(\int_{a}^{b} f(x)^n dx \right)^{1/n}$. By monotony of the integral we have
$$u_n\leq \left(\int_{a}^{b} M^n dx \right)^{1/n}= M(b-a)^{1/n}.$$
Let $c\in [a,b]$ such that $f(c)=M$. Then by continuity of $f$, for any $\epsilon \in (0,2M)\; \exists [s,t]\subset [a,b]$ ($[s,t]$ is a neighborhood of $c$) such that for all $x\in [s,t]$ we have $f(x)\geq M-{\epsilon \over 2}$.

Hence for any $n\geq 1$

$$u_n \geq \left(\int_{s}^{t} f(x)^n dx \right)^{1/n}\geq \left(\int_{s}^{t}\left( M-{ \epsilon \over 2 }\right)^n dx \right)^{1/n}= (M-{\epsilon \over 2})(t-s)^{1/n}.$$

Thus we can say that $$\forall \epsilon \in (0,2M) \, \exists [s,t]\subset [a,b] \,\text{ s.t }\, \forall n\geq 1\; (M-{\epsilon \over 2})(t-s)^{1/n}\leq u_n\leq M(b-a)^{1/n}.$$

On the other hand $\lim_{n\to \infty} M(b-a)^{1/n}= M$ and $\lim_{n\to \infty } (M-{\epsilon \over 2})(t-s)^{1/n}=M-{\epsilon \over 2},$ so

$\exists n_1 \geq 1, \forall n\geq n_1, \; M(b-a)^{1/n} and $\exists n_n \geq 1, \forall n\geq n_n, \; (M-{\epsilon \over 2})(t-s)^{1/n}>M-\epsilon$.

Let $n_0=\max\{n_1,n_2\}$. For $n\geq n_0$, $M-\epsilon, and thus we have shown that
$$\forall \epsilon >0 , \;\exists n_0\geq 1,\; \forall n\geq n_0,\; |u_n-M|<\epsilon.$$

Thus $\lim_{n\to \infty} u_n=M.$