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 $\epsilon&gt;0:M - \epsilon &lt; f(x) \le M$,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&lt; 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??
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 $M &gt; 0$. 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.
First question: if $f(x)=0$ for all $x \in [a,b]$, then you are done.
Hence we assume that $f(x)&gt;0$ for some $x \in [a,b]$. Then we have that $M&gt;0.$
Then consider $ \epsilon &gt;0 $ such that $\epsilon &lt;M.$
Second question: if $f_1$ and $f_2$ are integrable and $f_1 \le f_2$ on $[a,b]$ , then we have
$$\int_a^b f_1(x) dx \le \int_a^b f_2(x) dx.$$