4
$\begingroup$

So I am trying to prove that for a set $E$ of finite measure, and for $1 \leq p < \infty$, $||f||_p \leq (m(E))^{1 - 1/p}||f||_{\infty}$. But I think I have proved the wrong thing. Can you help me see where I went wrong?

My proof is something like

$$||f||_p =\left(\int_E |f|^p\right)^{1/p} \leq \left(\int_E ||f||_{\infty}^p\right)^{1/p} = \left(||f||_{\infty}^p \int_E 1\right)^{1/p} = ||f||_{\infty} (m(E))^{1/p},$$

which is not what was asked for in the problem.

Thanks!

1 Answers 1

3

What you get is true but not the wanted inequality. But you can write, assuming that $f\in L^{\infty}$ $|f|^p=|f|^{p-1}|f|\leq ||f||_{\infty}^{p-1}|f|$ then apply Hölder's inequality.

  • 0
    Oh right! Thanks :) For a while I thought the statements contradicted each other.2012-02-23
  • 0
    Wait, but how can my statement be true? If $p \to \infty$, $||f||_p \to 0$, and assuming $||f||_p \to ||f||_\infty$ I think that's a contradiction.2012-02-23
  • 0
    Why $||f||_p\to 0$?2012-02-23
  • 0
    Wait, never mind, it's not :P2012-02-23