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I am attempting a question from Rudin's "Real and Complex Analysis" Chapter 3 question 5. I shall summarise the question as below: Suppose that $f$ is a complex measurable function on $X$, $\mu$ a positive measure on $X$. Also, assume that $\mu(X)=1$.

1) If $0<r<s\leq\infty$, when does $||f||_{r}=||f||_{s}<\infty$ hold? My hunch is that this happens when $f$ is a constant function, but I can't prove it.

2) Assume that $||f||_{r}<\infty$ for some $r>0$, prove that

\begin{eqnarray} \lim_{p\rightarrow 0}||f||_{p}=\exp\left\{\int_{X}\log|f|d\mu\right\} \end{eqnarray} where $\exp(-\infty)$ is defined to be $0$.

For this, I have proven that $||f||_{p}$ is bounded below by the right hand side expression, but I am stuck here.

Any help is appreciated! Thanks!

Davide Giraudo
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enoughsaid05
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1 Answers1

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  1. First assume $s$ finite. Then by equality case in Hölder's inequality, $$|f(x)|^{s-r}=\lVert f\rVert_s^{s-r}.$$ If $s$ is infinite, then $$\lVert f\rVert_{\infty}^r=\int|f(x)|^r,$$ hence $x\mapsto \lVert f\rVert_{\infty}^r-|f(x)|^r$ is a non-negative measurable function of integral $0$. So in both cases $f$ is (almost surely) constant.

  2. It's enough to assume $r=1$, which is done here.

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Davide Giraudo
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