Assume $f:[a,b] \to \mathbb R$ is continuous and for every $x \in \mathbb R :f(x) \ge 0$,if $M:=\sup\left\{f(x): x \in [a,b]\right\}$ Assume $f:[a,b] \to \mathbb R$  is continuous and for every $x \in \mathbb R :f(x) \ge 0$,if $M:=\sup\left\{f(x): x \in [a,b]\right\}$,show that the sequence $$\left\{\left(\int_{a}^{b}f^{n}\left(x\right)dx\right)^{\frac{1}{n}}\right\}_n$$ Is convergent to $M$.

If $M=\sup\left\{f(x): x \in [a,b]\right\}$ then exists $x \in [a,b]$ such that foe every $\epsilon>0:M - \epsilon < f(x) \le M$,so :
$$\left(M-\epsilon\right)^{n}\left(b-a\right)=\int_{a}^{b}\left(M-\epsilon\right)^{n}dx\le\int_{a}^{b}f^{n}\left(x\right)dx\le\int_{a}^{b}M^{n}dx=M^{n}\left(b-a\right)$$
Taking the $n$th root of the both sides yields:
$$\left(M-\epsilon\right)\left(b-a\right)^{\frac{1}{n}}\le\left(\int_{a}^{b}f^{n}\left(x\right)dx\right)^{\frac{1}{n}}\le M\left(b-a\right)^{\frac{1}{n}}$$
Taking the limit of both sides:
$$\left(M-\epsilon\right) \le \lim_{n \to \infty} \left(\int_{a}^{b}f^{n}\left(x\right)dx\right)^{\frac{1}{n}} \le M$$
I have some questions:

*

*How do we know that $M-\epsilon$ is not negative?

*We used that $(M - \epsilon )^n< f^n(x)\le M^n $ and from here $$\int_{a}^{b}\left(M-\epsilon\right)^{n}dx\le\int_{a}^{b}f^{n}\left(x\right)dx\le\int_{a}^{b}M^{n}dx$$
Why that's true??
 A: *

*There are two cases. If $f \equiv 0$ then the claim follows immediately and there's no need to consider the inequality. In all other cases, there exists some point in the interval in which $f$ is positive and so $M > 0$. Hence, for small enough $\varepsilon$, the expression $M - \varepsilon$ will be positive.

*You are using the monotonicity of the integral. If $f(x) \leq g(x)$ for all $x \in [a,b]$ and they are integrable then $\int_{a}^b f(x) \, dx \leq \int_a^b g(x) \, dx$. This follows from the fact that $g(x) - f(x) \geq 0$ for all $x \in [a,b]$ and that the integral of a non-negative function is non-negative.

A: *

*As long as $f$ is not null (which is a degenerate and simple case), one can choose $\epsilon$ positive as little as necessary to $M - \epsilon$ be positive. That's the common use of $\epsilon$.

*Integrals are monotonic. Every good integral theory (Riemann, Lebesgue, Stieltjes, etc ...) should provide monotonicity.

A: You are proving that $||f||_p \rightarrow ||f||_\infty$ as $p\rightarrow \infty$.
See my old solution in this post: How to show that $\|f\|_{L^p} \to \|f\|_{L^\infty}$ as $p \to \infty$ if $f \in L^\infty \cap L^{p_0}$ for some $0 < p_0 < \infty$?
A: First question: if $f(x)=0$ for all $x \in [a,b]$, then you are done.
Hence we assume that $f(x)>0$ for some $x \in [a,b]$. Then we have that $M>0.$
Then consider $ \epsilon >0 $ such that $\epsilon <M.$
Second question: if $f_1$ and $f_2$ are integrable and $f_1 \le f_2$ on $[a,b]$ , then we have
$$\int_a^b f_1(x) dx \le \int_a^b f_2(x) dx.$$
