Calculate $\lim_{p \to \infty}\int_{K}\Vert x \Vert_{p}dx = L$

Question

Calculate $$\lim_{p \to \infty}\int_{K}\Vert x \Vert_{p}dx = L$$ where $K$ is compact of $\mathbb{R}^{n}$.

Idea.

First, $$\Vert x \Vert_{p} \to \Vert x \Vert_{\infty}.$$ Moreover, since $K$ is compact, $\Vert x \Vert_{p}$ is uniformly continuous. My first question:

How to ensure that $\Vert x \Vert_{p} \to \Vert x \Vert_{\infty}$ uniformly?

Using it, $\int\Vert x \Vert_{p} \to \int\Vert x \Vert_{\infty}$.

My second question:

Taking $n=2$ and $K = [0,a] \times [0,a]$. Who is $L$?

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Can someone help me?

Answer 1

$\|x\|_{\infty} \leq \|x\|_p\leq n^{1/p}\|x\|_{\infty}$. Use the fact that $n^{1/p} \to 1$ as $ p \to \infty$ to show that the convergence is uniform. [You have to use the boundedness of $K$ here].

Alternatively you can use DCT: $\|x\|_p \leq n^{1/p} \|x\|_{\infty} \leq n \|x\|_{\infty}$ for $p >1$ so DCT can be applied.

The answer to the second part is $\int_0^{a}\int_0^{a} \max \{x,y\} dx dy$ which can be computed by splitting the integral into the parts $x<y$ and $x >y$. I will let you do this computation. The answer is $2a^{3}/3$.