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Let $f:[0,1]\rightarrow\mathbb{R}$ be continuous. Prove that the following limit exists $$\lim_{n\to\infty}\left(\int_{0}^{1}| f(x)|^{n}\,dx\right)^{\!\!1/n}$$

I tried like this: $$\left(\int_{0}^{1}| f(x)|^{n}\,dx\right)^{\!\!1/n}=e^{{\ln}{\left(\int_{0}^{1}| f(x)|^{n}\,dx\right)^{\!1/n}}}=e^{{\frac{1}{n}}\ln\left(\int_{0}^{1}| f(x)|^{n}\,dx\right)}.$$

Now from this how I can proceed?

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marked as duplicate by Martin R, Lord Shark the Unknown, YuiTo Cheng, Cesareo, Alexander Gruber May 22 at 4:37

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.

  • $\begingroup$ @MartinR: that does answer this question, but this question doesn't ask as much and simpler answers exist. $\endgroup$ – robjohn May 19 at 14:55
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The limit is $\|f\|_{\infty}=\sup \{|f(x): 0\leq x \leq 1\}$. It is clear that $(\int|f|^{n})^{1/n} \leq \|f\|_{\infty}$. Now there exists $a$ such that $|f(a)| =\|f\|_{\infty}$. By continuity given $\epsilon >0$ there exists $\delta >0$ such that $|f(x)| >\|f\|_{\infty}-\epsilon$ for $|x-a| <\delta$. Hence $(\int|f|^{n})^{1/n} \geq (\int_{(a-\delta,a+\delta)}|f|^{n})^{1/n} >(\|f\|_{\infty}-\epsilon) (2\delta)^{1/n}$. Now let $n \to \infty$. A minor change is required when the supremum is attained at an end point.

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The Norm is Increasing

If $n\ge m$, Jensen's Inequality guarantees that $$ \int_0^1\left(|f(x)|^m\right)^{n/m}\,\mathrm{d}x\ge\left(\int_0^1|f(x)|^m\,\mathrm{d}x\right)^{n/m} $$ Therefore, raising both sides to the $1/n$ power, we get $$ \left(\int_0^1|f(x)|^n\,\mathrm{d}x\right)^{1/n}\ge\left(\int_0^1|f(x)|^m\,\mathrm{d}x\right)^{1/m} $$ That is, $$ \left(\int_0^1|f(x)|^n\,\mathrm{d}x\right)^{1/n} $$ is increasing in $n$.


The Norm is Bounded Above

Let $M=\sup\limits_{x\in[0,1]}|f(x)|$, which exists because $f$ is a continuous function on a compact set. Then $$ \begin{align} \left(\int_0^1|f(x)|^n\,\mathrm{d}x\right)^{1/n} &\le\left(\int_0^1M^n\,\mathrm{d}x\right)^{1/n}\\[6pt] &=M \end{align} $$


Existence of the Limit

Thus, $\left(\int_0^1|f(x)|^n\,\mathrm{d}x\right)^{1/n}$ is increasing in $n$ and bounded above by $\sup\limits_{x\in[0,1]}|f(x)|$; therefore, $$ \lim_{n\to\infty}\left(\int_0^1|f(x)|^n\,\mathrm{d}x\right)^{1/n} $$ exists.

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