Let $X$ be a measurable subset of $\mathbb{R}$ with $m(X) < \infty$. Let $f \in L^\infty(X)$ with $||f||_\infty > 0$. I'm trying to show that $$\lim_{p \to \infty} \frac{||f||^{p+1}_{p+1}}{||f||_p^p} = ||f||_\infty.$$
So far, I've shown that $$\frac{||f||_{p+1}^{p+1}}{m(X)^{1/{(p+1)}}||f||_{p+1}^p} \leq \frac{||f||^{p+1}_{p+1}}{||f||_p^p}$$ so that when I take the $\liminf$ of both sides the left side tends to $||f||_\infty$. Now for the upper bound, let $\epsilon > 0$ with $\epsilon < ||f||_\infty$. Define $X_\epsilon = \{x \in X : |f(x)| > ||f||_\infty - \epsilon \}$ so that $m(X_\epsilon) > 0$. Then, we have $$\frac{||f||^{p+1}_{p+1}}{||f||_p^p} \leq \frac{m(X) ||f||_\infty^{p+1}}{m(X_\epsilon)(||f||_\infty - \epsilon)^p}.$$ I'm not sure how to proceed from here and would appreciate any help.
