# Prove that the following limit exists $\lim_{n\to\infty}\left(\int\limits_{0}^{1}\vert f(x)\vert^{n}dx\right)^{\frac{1}{n}}$ [duplicate]

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?

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

• @MartinR: that does answer this question, but this question doesn't ask as much and simpler answers exist. – robjohn May 19 at 14:55

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.

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.