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This is a well-known problem that has been asked in this website, but I just wanted to get some hints and see if I'm even doing the right thing.

The problem is to show that $\lVert f \rVert_\infty = \lim_{p \to \infty} \lVert f \rVert_p$, with $\mu$ a probability measure. (I believe it works for any finite measure, but the problem says probability measure so we'll go with that.)

My attempt:

First, say $f$ is simple and $f \geq 0$. Then $\lVert f \rVert_p = (\left|a_1 \right|^p \mu(E_1) + \ldots + \left|a_n \right|^p \mu(E_n))^{\frac{1}{p}}$, with $\left|a_1\right| < \ldots < \left|a_n \right|$ ordered from smallest to the largest.

Then $\lim_{p \to \infty} \lVert f \rVert_p = \lim_{p \to \infty} (\left|a_n\right|^p((\frac{a_1}{a_n} \mu(E_1) + \ldots + \mu(E_n)))^{\frac{1}{p}} = \left| a_n \right|$.

Now let $f \geq 0$ and let $f_n$ be simple functions that monotonically increase to $f$.

Then $\lim_{p \to \infty} (\int \left| f \right|^p)^{\frac{1}{p}} = \lim_{p \to \infty}(\int (\lim_{n \to \infty} \left| f_n \right|^p)^{\frac{1}{p}}) = \lim_{n \to \infty}(\lim_{p \to \infty} (\int \left|f_n\right|^p)^{\frac{1}{p}})$.

I'm not sure if I can even switch the limit there, and I'm not too sure what to do from there. But I would think the general idea is $\lim_{p \to \infty}(\int \left|f_n\right|^p)^{\frac{1}{p}} = \max \left|f_n\right|$, and it seems $\lim_{n \to \infty} \max \left|f_n\right| = \lVert f \rVert_\infty$.

I haven't thought about general $f$, not too sure what to do with that.

Thanks!

Bob
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Phil
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1 Answers1

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Some hints:

WLOG, assume $f \geq 0$. We have $\|f\|_{p}\leq\|f\|_{\infty}\mu(X)^{1/p}$.

Also if $M < \|f\|_{\infty}$, then $\mu(f > M) > 0$ and $\|f\|_p \geq M\mu(f > M)^{1/p}$.

Mason
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