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Show that $$\lim(M_n)=\sup\{f(x):x\in[a,b]\}$$ If $f$ is continuous on $[a,b]$, $f(x)\geq0$ for all $x\in[a,b]$, and $M_n=\left(\int_a^b f(x)^n dx\right)^{1/n}$.

I don't have any idea to solve this. What I know is that $\lim(M_n)\leq \sup\{f(x):x\in[a,b]\}$. I don't know how to prove that LHS and RHS are same. Could you give me any clue, please?

EDITED: After I saw comment form Chilango, I want to try to solve this:

I want to prove that $\lim(M_n)\geq\sup\{f(x):x\in[a,b]\}$. Let $M=\sup\{f(x):x\in[a,b]\}$ and $D=\{x:f(x)\geq M-\delta\}$ with $\delta>0$ be arbitrary. Then $$\left(\int_a^b f(x)^n dx\right)^{1/n} \geq \left(\int_D (M-\delta)^n d\mu\right)^{1/n}$$ $$M_n\geq(M-\delta)\mu(D)^{1/n}$$ Then we can say $$\lim(M_n)\geq\lim((M-\delta)\mu(D)^{1/n})=M-\delta$$ Because $\delta$ is arbitrary, It's proved. Is this way correct?

fahadh4ilyas
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2 Answers2

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We can prove this without measure theory. Also we can avoid referring to properties of $\liminf$ or $\limsup$ since it appears you have not covered that yet.

Since $f$ is continuous it attains a maximum on $[a,b]$ at some point $c \in [a,b]$ where $f(c) = M = \sup\{f(x): x \in [a,b]\}.$ Again using continuity, given $\epsilon >0$ there exists an interval $[c_1,c_2] \subset [a,b]$ such that $f(x) > M - \epsilon/2$ for all $x \in [c_1,c_2]$.

Hence, with $f(x) \geqslant 0$ we have

$$(M-\epsilon/2)(c_2-c_1)^{1/n} \leqslant \left(\int_{c_1}^{c_2} [f(x)]^n \, dx\right)^{1/n} \leqslant \left(\int_a^b [f(x)]^n \, dx \right)^{1/n} \leqslant M(b-a)^{1/n}.$$

Since $(c_2 - c_1)^{1/n} \to 1$ and $(b-a)^{1/n} \to 1$ as $ n \to \infty$, for all $n$ sufficiently large we have $(c_2 - c_1)^{1/n} > (M - \epsilon)/(M - \epsilon/2)$ and $(b-a)^{1/n} \leqslant 1 + \epsilon/M,$ and, consequently,

$$M - \epsilon \leqslant \left(\int_a^b [f(x)]^n \, dx \right)^{1/n} \leqslant M + \epsilon.$$

Since $\epsilon$ can be arbitrarily small it follows that

$$\lim_{n \to \infty} \left(\int_a^b [f(x)]^n \, dx \right)^{1/n}= M$$

RRL
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Here is one way to do it: Assume $f\neq 0$ and let $0\le t\le \left \| f \right \|_{\infty }$ . Then $A=\left \{ x:\vert f(x)\vert \ge t \right \}$ has positive measure and $\left \| f \right \|_p\ge \left ( \int_A \vert f\vert ^{p}d\lambda \right )^{1/p}\ge t\lambda (A)^{1/p}$. If $\lambda (A)$ is finite, $\lambda (A)^{1/p}\to 1$ as $p\to \infty .\ $ If $\lambda (A)=\infty $ then RHS is infinite, for all $p$. In either case $\liminf _{p\to \infty }\left \| f \right \|_p\ge t$ so we have by definition of $t$, that

$\tag 1\liminf _{p\to \infty }\left \| f \right \|_p\ge \vert \vert f\vert \vert_{\infty } $.

For the reverse inequality, fix $r<p<\infty$. Then, and here we must assume that $\left \| f \right \|_r<\infty $, observe that $\left \| f \right \|_p\le \left \| f \right \|^{r/p}_r\left \| f \right \|^{1-r/p}_{\infty }$ so that finally we have, since $\left \| f \right \|^{r/p}_r\to 1$ as $p\to \infty $,

$\tag2\limsup_{p\to \infty }\left \| f \right \|_p\le \left \| f \right \|_{\infty }$

Now combine 1) and 2).

Matematleta
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