Prove $\|f\|_{L^p(E)} \rightarrow \|f\|_{L^{\infty}(E)}$ as $p \rightarrow \infty$

Question

I'm working on the following question

Suppose m$(E)<\infty$ and $f \in L^\infty(E)$. Prove $\|f\|_{L^p(E)} \rightarrow \|f\|_{L^{\infty}(E)}$ as $p \rightarrow \infty$

I'm trying to show $\|f\|_{L^P(E)} \leq \|f\|_{L^\infty(E)} m(E)^{1/p}$ for getting an upper bound. (checked the way to get a lower bound from the other post)

If I can show $\||f|^p \|_{L^\infty(E)}$ = $\||f| \|_{L^\infty(E)}^p$ then I would get the inequality using Holder's inequality.

Answer 1

Hints : Let $\epsilon<\|f(x)\|_{\infty }$ and define $E_{\epsilon }=\left \{ x:|f(x)|>\|f(x)\|_{\infty }-\epsilon \right \}.$ Then $\|f\|_{p}^{p}\ge \int_{E_{\epsilon }}(\|f\|_{\infty }-\epsilon)^{p}dx=(\|f\|_{\infty }-\epsilon)^{p})\lambda (E_{\epsilon }).$ Now take $pth$ roots and observe that $\lambda (E_{\epsilon })^{1/p}\to 1$ as $p\to \infty.$ This implies that $\|f\|_{p}\ge \|f\|_{\infty} -\epsilon.$

For the other direction, fix $q>1$ and note that $\|f\|_p= \left ( \int_{E} |f|^{p-q}|f|^{q}dx \right )^{1/p}$ whenever $p>q.$