3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 $\|f\|_\infty = \max_{A \subset \Omega} \frac{1}{\mu(A)}\int_A |f| d\mu$ ? $\endgroup$ – reuns Feb 24 '16 at 22:31
0
$\begingroup$

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 $p \to \infty$: $\limsup ||f||_p \geq ||f||_\infty$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.