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If $f(x)$ is continuous on $[a,b]$ and $M=\max \; |f(x)|$, is $M=\lim \limits_{n\to\infty} \left(\int_a^b|f(x)|^n\,\mathrm dx\right)^{1/n}$?

I'm trying to learn some analysis on my own and I stumbled upon the following proposition.

Suppose $ f:[a,b]\rightarrow \mathbb{R} $ is continuous. Suppose $ f(x) \geq 0 \space$ for all $ x \in [a,b] $. Let $ M=sup\{f(x):x\in[a,b]\} $

Then the sequence $ \{[\int_a^b[f(x)]^n\,dx]^{1/n}\}_{n=1}^\infty $ converges to M.

I'm really not sure where to begin in proving this.

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marked as duplicate by AD., Dylan Moreland, Qiaochu Yuan Feb 22 '12 at 10:27

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Have you seen this ( – azarel Feb 22 '12 at 5:13
Is the same question!!!!!!!!!!!!!!!! – leo Feb 22 '12 at 5:52

2 Answers 2

Since $[a,b]$ is compact, there exists $x_0\in[a,b]$ such that $f(x_0)=M=\max_{[a,b]}f(x)\geq 0$. Since $f$ is continuous at $x_0$, for all $\epsilon>0$, there exists $\delta$ such that $$M-f(x)\leq|f(x)-M|=|f(x)-f(x_0)|<\epsilon\mbox{ whenever }x\in (x_0-\delta,x_0+\delta)\cap[a,b].$$ Hence, $$\left(\int_a^b(f(x))^ndx\right)^{\frac{1}{n}}\geq \left(\int_{(x_0-\delta,x_0+\delta)\cap[a,b]}(f(x))^ndx\right)^{\frac{1}{n}}\geq (M-\epsilon)\cdot\Big( m\big((x_0-\delta,x_0+\delta)\cap[a,b]\big)\Big)^{\frac{1}{n}}$$ which implies that $$\lim_{n\rightarrow\infty}\left(\int_a^b(f(x))^ndx\right)^{\frac{1}{n}}\geq (M-\epsilon)\cdot\lim_{n\rightarrow\infty}\Big( m\big((x_0-\delta,x_0+\delta)\cap[a,b]\big)\Big)^{\frac{1}{n}}=M-\epsilon.$$ On the other hand, $$\lim_{n\rightarrow\infty}\left(\int_a^b(f(x))^ndx\right)^{\frac{1}{n}}\leq M\cdot \lim_{n\rightarrow\infty}\Big( m\big([a,b]\big)\Big)^{\frac{1}{n}}=M.$$ Since $\epsilon>0$ is arbitrary, we have $$\lim_{n\rightarrow\infty}\left(\int_a^b(f(x))^ndx\right)^{\frac{1}{n}}=M.$$

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Thank you, I found the structure of this proof very useful. I'm a little confused about the notation though. What is meant by say m([a,b])? – Shafat Arbaz Alam Feb 22 '12 at 5:40
The Lebesgue measure of $[a,b]$. So $m([a,b])=b-a$. Also $m\big((x_0-\delta,x_0+\delta)\cap[a,b]\big)=2\delta$ if $(x_0-\delta,x_0+\delta)\subset[a,b]$, etc. – Paul Feb 22 '12 at 5:46

The idea is to construct a function $0 \leq g \leq f$ such that $\int_{a}^{b} g^n$ is easy to compute while $\sup g$ is sufficiently close to $\sup f$.

If $M = 0$, there is nothing to prove. So assume $M > 0$. Then $M = f(x)$ is achieved at some point $c \in [a, b]$. Let $M > \epsilon > 0$. Then there is $\delta > 0$ such that $|f(x) - M| \leq \epsilon$ whenever $|x - c| \leq \delta$. Now let $$ g(x) = (M-\epsilon) \chi_{[c-\delta, c+\delta]}(x) = \begin{cases} M-\epsilon & |x - c| \leq \delta \\ 0 & \mathrm{otherwise}.\end{cases}$$ Then $0 \leq g \leq f$. Since $|[c-\delta, c+\delta] \cap [a, b]| \geq \delta$, we have $$\int_{a}^{b} g(x)^n \; dx \geq (M-\epsilon)^n \delta. $$ Let $I_n = \left( \int_{a}^{b} f(x)^n \; dx \right)^{1/n}$. Then we have $$ (M-\epsilon) \delta^{1/n} \leq I_n \leq M(b-a)^{1/n}.$$ Thus we obtain $$ M-\epsilon \leq \liminf_{n\to\infty} I_n \leq \limsup_{n\to\infty} I_n \leq M.$$ Since $\epsilon$ is arbitrary, we obtain $$ \liminf_{n\to\infty} I_n = M = \limsup_{n\to\infty} I_n,$$ completing the proof.

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