# Show that $||f||_{\infty} = \lim\limits_{p \to \infty} ||f||_p$

Let $(F,\Omega, \mu)$ be a measure space with $\mu(F) < \infty$. Suppose $f: F \to \mathbb R$ is measurable. Define: $||f||_p = \displaystyle\bigg(\int_F|f|^pd\mu\bigg)^{1/p}$ and $||f||_\infty = \inf \{C \in [0,\infty]: |f| \leq C, \mu-\text{a.e.}\}$. Show that: $||f||_{\infty} = \lim\limits_{p \to \infty} ||f||_p$.

We have: $0 \leq |f| \leq ||f||_\infty \Rightarrow \displaystyle\bigg(\int_F|f|^pd\mu\bigg)^{1/p} \leq \bigg(\int_F||f||_\infty^pd\mu\bigg)^{1/p} \Leftrightarrow ||f||_p \leq \mu(F)^{1/p}||f||_\infty$. When $p \to \infty$, we obtain: $\lim\limits_{p \to \infty} ||f||_p \leq ||f||_\infty$.

How to prove the reverse?

• the problem with the $\|.\|_\infty$ norm is that with $f(x) = 0$ except at $x= 0$ then $\|f\|_p = 0$ and $\max_x |f(x)| = 1$. how do you exclude that case ? do you define $\|f\|_\infty = \max_{A \subset \Omega} \frac{1}{\mu(A)}\int_A |f| d\mu$ ? – reuns Feb 24 '16 at 22:31

By Holder's inequality: $||f||_p \geq \mu(F)^{1/p - 1} ||f||_1$.
Thus, $\sup\limits_{k \geq p} ||f||_k \geq \sup\limits_{k \geq p} \mu(F)^{1/p-1} ||f||_1 = \mu(F)^{1/p-1} ||f||_\infty \mu(F) = \mu(F)^{1/p} ||f||_\infty$.
Let $p \to \infty$: $\limsup ||f||_p \geq ||f||_\infty$.