real analysis - If $f(x)$ is continuous on $[a,b]$ and $M=max ; |f(x)|$, is $M=lim limits_{ntoinfty} left(int_a^b|f(x)|^n,mathrm dxright)^{1/n}$?

Let $f(x)$ be a continuous real-valued function on $[a,b]$ and $M=\max\{|f(x)| \; :\; x \in [a,b]\}$. Is it true that:
$$
M= \lim_{n\to\infty}\left(\int_a^b|f(x)|^n\,\mathrm dx\right)^{1/n} ?
$$

Thanks!

Answer

Put $S_{\delta}:=\{x\in\left[a,b\right], |f(x)|\geq M-\delta\}$ for any $\delta>0$. Then we have for all $n$
$$M\cdot (b-a)^{\frac 1n}\geq\left(\int_a^b|f(x)|^ndx\right)^{\frac 1n}\geq \left(\int_{S_{\delta}}|f(x)|^ndx\right)^{\frac 1n}\geq (M-\delta)\left(\lambda(S_{\delta})\right)^{\frac 1n}.$$

Since $f$ is continuous, the measure of $S_{\delta}$ is positive and taking the $\liminf$ and $\limsup$, we can see that
$$M\geq \limsup_n \left(\int_a^b|f(x)|^ndx\right)^{\frac 1n}\geq M-\delta\quad \mbox{and }\quad M\geq \liminf_n\left(\int_a^b|f(x)|^ndx\right)^{\frac 1n}\geq M-\delta$$
for all $\delta>0$, so $\lim_{n\to\infty}\left(\int_a^b|f(x)|^ndx\right)^{\frac 1n}=M$.