Prove that $\lim_{p\to \infty}\Big(\frac{1}{\mu(A)}\int_{A}|f|^p\Big)^{1/p}=\|f\|_{\infty}$

Question

We assume that $\mu(A)<\infty$. Prove that $$\displaystyle \lim_{p\to \infty}\Big(\frac{1}{\mu(A)}\int_{A}|f|^p\Big)^{1/p}=\|f\|_{\infty}$$

Answer 1

We assume that $f \neq 0$ (otherwise is trivial), so that $|| f||_{\infty} \neq 0 $.

One inequility is clear: using the mean value theorem you have that

$$ \left ( \frac{1}{\mu(A)} \int_A |f|^p \right )^ p = (|f(x_0)|^p)^{1/p} = |f(x_0) | \le || f|| _{\infty}$$

For the other one, let us call $A_{\varepsilon} := \{ x \in A: |f(x)| \ge ||f||_{\infty} (1-\varepsilon) \} $. Note that

$$ \int_A |f|^p = \int_{A_{\varepsilon} } |f|^p + \int_{A \setminus A_{\varepsilon} } |f|^p \ge \mu(A_{\varepsilon}) || f||_{\infty}^{p} (1-\varepsilon)^p$$ Extracting the $p$-th root and dividing by $\mu(A)^{1/p}$ we have $$ \left ( \frac{1}{\mu(A)} \int_A |f|^p \right )^{1/p} \ge \left (\frac{\mu(A_{\varepsilon})}{\mu(A)} \right )^{1/p} ||f||_{\infty} (1-\varepsilon ) $$

By the very definition of $L^{\infty}$ norm, we have that $\mu(A_{\varepsilon}) > 0$: if it was zero, then $f(x) \le ||f||_{\infty} (1-\varepsilon)$ for almost all $x$, which implies that $||f||_{\infty} \le ||f||_{\infty} (1-\varepsilon)$, a contradiction. For any $0 < s \le 1$ we have that $\lim_{p \to \infty} s^{1/p} = 1$, so that taking the liminf of the above expression yields

$$ \liminf_{p \to \infty} \left ( \frac{1}{\mu(A)} \int_A |f|^p \right )^{1/p} \ge || f||_{\infty} ( 1- \varepsilon) $$

for an arbitrary choice of $\varepsilon > 0$, so that

$$ \liminf_{p \to \infty} \left ( \frac{1}{\mu(A)} \int_A |f|^p \right )^{1/p} \ge || f||_{\infty} $$ and we are done.