Let $f:[0,1]\rightarrow\mathbb{R}$ be continuous. Prove that the following limit exists $$\lim_{n\to\infty}\left(\int_{0}^{1}| f(x)|^{n}\,dx\right)^{\!\!1/n}$$
I tried like this: $$\left(\int_{0}^{1}| f(x)|^{n}\,dx\right)^{\!\!1/n}=e^{{\ln}{\left(\int_{0}^{1}| f(x)|^{n}\,dx\right)^{\!1/n}}}=e^{{\frac{1}{n}}\ln\left(\int_{0}^{1}| f(x)|^{n}\,dx\right)}.$$
Now from this how I can proceed?
The limit is $\|f\|_{\infty}=\sup \{|f(x): 0\leq x \leq 1\}$. It is clear that $(\int|f|^{n})^{1/n} \leq \|f\|_{\infty}$. Now there exists $a$ such that $|f(a)| =\|f\|_{\infty}$. By continuity given $\epsilon >0$ there exists $\delta >0$ such that $|f(x)| >\|f\|_{\infty}-\epsilon$ for $|x-a| <\delta$. Hence $(\int|f|^{n})^{1/n} \geq (\int_{(a-\delta,a+\delta)}|f|^{n})^{1/n} >(\|f\|_{\infty}-\epsilon) (2\delta)^{1/n}$. Now let $n \to \infty$. A minor change is required when the supremum is attained at an end point.
The Norm is Increasing
If $n\ge m$, Jensen's Inequality guarantees that $$ \int_0^1\left(|f(x)|^m\right)^{n/m}\,\mathrm{d}x\ge\left(\int_0^1|f(x)|^m\,\mathrm{d}x\right)^{n/m} $$ Therefore, raising both sides to the $1/n$ power, we get $$ \left(\int_0^1|f(x)|^n\,\mathrm{d}x\right)^{1/n}\ge\left(\int_0^1|f(x)|^m\,\mathrm{d}x\right)^{1/m} $$ That is, $$ \left(\int_0^1|f(x)|^n\,\mathrm{d}x\right)^{1/n} $$ is increasing in $n$.
The Norm is Bounded Above
Let $M=\sup\limits_{x\in[0,1]}|f(x)|$, which exists because $f$ is a continuous function on a compact set. Then $$ \begin{align} \left(\int_0^1|f(x)|^n\,\mathrm{d}x\right)^{1/n} &\le\left(\int_0^1M^n\,\mathrm{d}x\right)^{1/n}\\[6pt] &=M \end{align} $$
Existence of the Limit
Thus, $\left(\int_0^1|f(x)|^n\,\mathrm{d}x\right)^{1/n}$ is increasing in $n$ and bounded above by $\sup\limits_{x\in[0,1]}|f(x)|$; therefore, $$ \lim_{n\to\infty}\left(\int_0^1|f(x)|^n\,\mathrm{d}x\right)^{1/n} $$ exists.
