Let $(X, \mathscr{A}, \mu)$ be a finite measure space, and let $f$ be a $\mathscr{A}$- measurable real- or complex- valued function on $X$.

Show that if $f\in \mathscr{L}^\infty$ then $\|f\|_\infty = \lim_{p \rightarrow+\infty} \|f\|_p$.

In my course literature $\|f\|_\infty$ is defined as the infimum of those non-negative numbers $M$ such that $\{x\in X : |f(x)| > M\}$ is locally $\mu$-null. But locally $\mu$- null seems to be equivalent with $\mu$-null subsets when the measure space is $\sigma$-finite?

I know that $f \in \mathscr{L}$ for $p \in [1,+\infty)$ and that $\sup \{|f|_p : 1 \leq p < \infty\}$ is finite. But how do I know that this converge? Thanks

  • $\begingroup$ What is the definition of $\mu$-null in your course? $\endgroup$ Aug 5 '13 at 17:30
  • $\begingroup$ def: A subset $N$ of $X$ is locally $\mu$- null if for each $A\in \mathscr{A}$ satisfying $\mu(A) \leq + \infty$ the set $A\cap N$ is $\mu$ -null $\endgroup$
    – Johan
    Aug 6 '13 at 7:35
  • $\begingroup$ def: A set B is $\mu$-null if there is a subset $A$ of $X$ such that $A\in \mathscr{A}, B \subset A$ and $\mu(A) = 0$ $\endgroup$
    – Johan
    Aug 6 '13 at 7:39

I'll show it for the special case $\mu(X) = 1$. You should be able to derive the general case from that. First of all, we have

$$\lVert f\rVert_p = \left(\int_X \lvert f\rvert^p\, d\mu\right)^{1/p} \leqslant \left(\int_X \lVert f\rVert_\infty^p\, d\mu\right)^{1/p} = \lVert f\rVert_\infty$$

since $\mu(X) = 1$. Hence $\limsup\limits_{p\to\infty} \lVert f\rVert_p \leqslant \lVert f\rVert_\infty$.

Now we need to show that $\liminf\limits_{p\to\infty} \lVert f\rVert_p \geqslant \lVert f\rVert_\infty$. For $\lVert f\rVert_\infty = 0$ that is trivially satisfied, so let's assume $M := \lVert f\rVert_\infty > 0$. For $0 < \varepsilon < M$, let $A_\varepsilon = \{x \in X : \lvert f(x)\rvert \geqslant M - \varepsilon\}$. Then

$$\begin{align} \lVert f\rVert_p &= \left(\int_X \lvert f\rvert^p\,d\mu\right)^{1/p}\\ &\geqslant \left(\int_{A_\varepsilon} \lvert f\rvert^p\,d\mu\right)^{1/p}\\ &\geqslant \left(\int_{A_\varepsilon} (M-\varepsilon)^p\,d\mu\right)^{1/p}\\ &= (M - \varepsilon)\cdot \mu(A_\varepsilon)^{1/p} \to M - \varepsilon, \end{align}$$

hence $\liminf\limits_{p\to\infty} \lVert f\rVert_p \geqslant \lVert f\rVert_\infty - \varepsilon$. Since $\varepsilon$ was arbitrary, the proposition follows.

  • $\begingroup$ great! but what about my question locally $\mu$ -null? How comes that the first inequality is true? $\endgroup$
    – Johan
    Aug 6 '13 at 7:33
  • 1
    $\begingroup$ I knew there was something I'd forgotten :/. Since you now posted the used definitions, I need no longer guess: Yes, for a $\sigma$-finite measure space, $\mu$-null is the same as locally $\mu$-null, since a countable union of sets of measure $0$ still has measure $0$. The first inequality is because $\lvert f(x)\rvert \leqslant \lVert f\rVert_\infty$ almost everywhere. $\endgroup$ Aug 6 '13 at 8:17

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