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

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?

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