3
$\begingroup$

If $\mu$ is a probability measure then

$$\left\| f \right\|_{\infty} = \lim_{p \rightarrow \infty} \left\| f \right\|_{p}$$

where $\left\| f \right\|_{p} = \left( \int f^p d\mu\right) ^{1/p}$ and $\left\| f \right\|_{\infty}$ is the essential supremum of $f$ with respect to $\mu$.

Would someone be so kind as to suggest ways to prove this?

$\endgroup$
0

1 Answer 1

6
$\begingroup$

Here is a proof of the statement. The main ideas are the following:

  • the inequality $\lVert f\rVert_p\leqslant \lVert f\rVert_\infty$ follows from the fact that $|f(x)|\leqslant \lVert f\rVert_\infty$ for $\mu$-almost every $x$;
  • for the converse inequality, we integrate over the set where $f$ is a greater than a number very close to the uniform norm.
$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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