# Prove that $\lim_{n \to \infty}\bigg[\int_0^1f(t)^n \text{dt}\bigg]^{1/n}=M$

Suppose that $$f$$ is a continuous, non-negative function on the interval $$[0,1]$$. Let $$M$$ be the maximum of $$f$$ on the interval. Prove that $$\lim_{n \to \infty}\bigg[\int_0^1f(t)^n \text{dt}\bigg]^{1/n}=M$$

We wrote out some simple examples to show it worked for functions such as $$x^2$$. We are having trouble finding how to create a general proof. Thanks for any help!

## 2 Answers

Hint: Let $x_0$ be such that $f(x_0) = M$. Fix $\epsilon > 0$. There exists a $\delta > 0$ such that $f(x) > M - \epsilon$ whenever $x \in (x_0-\delta,x_0 + \delta)$. Conclude that $$\lim_{n \to \infty}\bigg[\int_0^1f(t)^n \text{dt}\bigg]^{1/n}\geq M - \epsilon$$ However, $\epsilon$ was arbitrary.

• This technically assumes that $f$ attains its maximum on the interior of the interval. However, I think you can see how this proof might be modified if $f$ attains its maximum at one of the endpoints. – Omnomnomnom Dec 7 '16 at 4:44

A more general statement is that, if $0<\mu(X)<\infty$ and $f\in L^{\infty}(X)$, then $\lim_{p\rightarrow\infty}\|f\|_{L^{p}}=\|f\|_{L^{\infty}}$:

$\|f\|_{L^{\infty}}\\ =\lim_{\epsilon\rightarrow 0^{+}}(\|f\|_{L^{\infty}}-\epsilon)\\ =\lim_{\epsilon\rightarrow 0^{+}}\lim_{p\rightarrow\infty}(\|f\|_{L^{\infty}}-\epsilon)(\mu(S_{\epsilon}))^{1/p},~~~~S_{\epsilon}=\{|f|>\|f\|_{L^{\infty}}-\epsilon\}\\ \leq\liminf_{\epsilon\rightarrow 0^{+}}\liminf_{p\rightarrow\infty}\left(\displaystyle\int_{S_{\epsilon}}|f|^{p}d\mu\right)^{1/p}\\ \leq\liminf_{\epsilon\rightarrow 0^{+}}\liminf_{p\rightarrow\infty}\|f\|_{L^{p}}\\ \leq\limsup_{p\rightarrow\infty}\|f\|_{L^{p}}\\ \leq\limsup_{p\rightarrow\infty}\|f\|_{L^{\infty}}(\mu(X))^{1/p}\\ =\|f\|_{L^{\infty}}$