Limit of p-norms of functions

Question

$f$ is a continuous function on the closed interval $[a,b]$ such that $f \geq 0$ on $[a,b]$. Let $$ M_n = \left( \int_a^b f (x)^n dx \right)^{1/n}. $$ Show $\lim_n M_n = \sup \{ f(x) \mid x \in [a,b] \}$.

Answer 1

Let $M=\sup\{f(x):x\in [a,b]\}$. We know that

$$ M_{n}\leq \left(\int_{a}^{b}M^{n}\right)^{1/n}=(M^{n}(b-a))^{1/n}=M(b-a)^{1/n}.$$

Let $n\rightarrow\infty$, we have $\lim M_{n}\leq M$.

On the other hand, let $y\in [a,b]$ such that $f(y)=M$. Let $\epsilon>0$ be given. Then there exists $\delta>0$ such that if $|x-y|\leq\delta$, we have $f(x)\geq M-\epsilon$. Then

$$M_{n}=\left(\int_{a}^{b}f^{n}\right)^{1/n}\geq \left(\int_{[y-\delta,y+\delta]}f^{n}\right)^{1/n}\geq \left(\int_{[y-\delta,y+\delta]}(M-\epsilon)^{n}\right)^{1/n}=(M-\epsilon)(2\delta)^{1/n}.$$

As $n\rightarrow \infty$, then $\lim M_{n}\geq M-\epsilon$. Since $\epsilon$ is arbitrary, $\lim M_{n}\geq M$.

Answer 2

Let $s=\sup\limits_{[a,b]}\{f(x):x\in[a,b]\}$, then $$ \begin{align} \left(\int_a^bf(x)^n\,\mathrm{d}x\right)^{1/n} &\le \left(\int_a^bs^n\,\mathrm{d}x\right)^{1/n}\\ &=s(b-a)^{1/n} \end{align} $$ If $c\lt s$, then $$ \begin{align} \left(\int_a^bf(x)^n\,\mathrm{d}x\right)^{1/n} &\ge c\,|\{x:f(x)\ge c\}|^{1/n}\\ \end{align} $$ By the Squeeze theorem $$ c\le\lim_{n\to\infty}\left(\int_a^bf(x)^n\,\mathrm{d}x\right)^{1/n}\le s $$ Since $c\lt s$ was arbitrary, we have $$ \lim_{n\to\infty}\left(\int_a^bf(x)^n\,\mathrm{d}x\right)^{1/n}=\sup_{[a,b]}\{f(x):x\in[a,b]\} $$