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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?

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1 Answer 1

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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.
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