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I found a proof for this fact that limit of $\|f\|_p$ when $p \to \infty $ is $\|f\|_{\infty}$ in here when $f:X \to R $ and $X \in L^p$ measure space is finite.

But I need a counter example for this when measure space is not finite. Divergent examples will help me more to understand.

Any help will be appreciated.

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SirSaleh
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    Welcome to MSE. Can you share what you've tried and what you're having trouble with? Do you know examples of functions that are $L^{\infty}$ but not $L^p$ for $p < \infty$? –  Jun 23 '16 at 20:05
  • First you need to explain what you mean by "limit of $L^p$ is $L^\infty$"... – David C. Ullrich Jun 23 '16 at 20:55
  • I don't think there is a counterexample. Refer to the proof given in the accepted answer. The proof applies even if $\mu(X) = \infty$, because the set $S_{\delta}$ must still have finite measure (otherwise $f$ would not be in $L^q$). –  Jun 23 '16 at 21:04
  • @Bungo There is a counterexample or not, depending on exactly what it is we're trying to prove! If $||f||p<\infty$ for some $p<\infty$ then $||f||_p\to||f||\infty$; that's true in any measure space. Otoh, if $\mu(X)=\infty$ and $f=1$ then $||f||p\not\to||f||\infty$. – David C. Ullrich Jun 23 '16 at 21:09
  • @DavidC.Ullrich This question is poorly stated, but I interpreted it to mean: if we remove the hypothesis that $\mu(X) < \infty$ in the linked question, is the result still true? The original question assumed $f \in L^{\infty} \cap L^q$ for some $q \in [1,\infty)$. I think the result holds in that case even if $\mu(X) = \infty$. It would be good if the OP would edit this question to clarify. –  Jun 23 '16 at 21:16
  • this question needs rephrasing – zhw. Jun 23 '16 at 21:27
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    @Bungo Whatever the question is, I believe I've answered it. – David C. Ullrich Jun 23 '16 at 23:47
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    @DavidCUllrich What if the question is "Why is there air?" – zhw. Jun 23 '16 at 23:48
  • @zhw. Sorry. Thanks for the correction. Whatever the question is, I believe I've answered it, unless the question is "Why is there air". I can't think of any other counterexamples. – David C. Ullrich Jun 23 '16 at 23:52
  • @DavidC.Ullrich and other friends, Sorry Maybe I was sleepy when I wrote this question. I corrected my question know. I was mean what david says. thanks and sorry again. – SirSaleh Jun 24 '16 at 02:47

1 Answers1

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We assume that $\mu$ is a measure on $X$. Note that is the only standing assumption here; no norms or measures are assumed finite below except where explicitly stated.

The question as stated is unclear. The question at the linked post is clear, but slightly silly because the hypotheses are obviously redundant. Here are the facts, or at least some of the facts:

Theorem 1. If there exists $p<\infty$ with $||f||_p<\infty$ then $||f||_p\to||f||_\infty$.

No need there to assume $\mu(X)<\infty$. I used the word "silly" because if we do assume $\mu(X)<\infty$ then $f\in L^\infty$ implies $f\in L^p$ for all $p$; having all three hypotheses $\mu(X)<\infty$, $f\in L^p$ for some finite $p$ and also $f\in L^\infty$ is just silly when the first and third so obviously imply the second. Proof below.

Another true fact, very easy:

Theorem 2. If $||f||_\infty=\infty$ then $||f||_p\to||f||_\infty$.

What $\mu(X)<\infty$ does get us is this:

Corollary If $\mu(X)<\infty$ then $||f||_p\to||f||_\infty$.

This follows from Theorem 1 or from Theorem 2, depending on whether $f\in L^\infty$.

So if we're going to get $||f||_p\not\to||f||_\infty$ the only possibility is to have $||f||_p=\infty$ for every finite $p$ while $||f||_\infty<\infty$. Which of course requires $\mu(X)=\infty$. And there you are:

Example Suppose $\mu(X)=\infty$. Let $f=1$. Then $||f||_p\not\to||f||_\infty$.

Proofs

Again, note that there are no standing assumptions about anything being finite.

Lemma. $\liminf_{p\to\infty}||f||_p\ge||f||_\infty$.

Proof: Say $0<A<||f||_\infty$. Let $E=\{|f|\ge A\}$. Then $\mu(E)>0$ by definition, while $||f||_p\ge A \mu(E)^{1/p}$. If $\mu(E)<\infty$ then $\mu(E)^{1/p}\to1$, while if $\mu(E)=\infty$ then $\mu(E)=\infty$. In either case we see $\liminf||f||_p\ge A$. QED.

Theorem 2 follows from the lemma.

Proof of Theorem 1: By the lemma we may assume that $||f||_\infty<\infty$, and we need only show that $\limsup||f||_p\le||f||_\infty$. Say $||f||_{p_0}<\infty.$ For $p>p_0$ we have $$||f||_p\le||f||_{p_0}^{p_0/p}||f||_\infty^{(p-p_0)/p}\to||f||_\infty.$$QED.

David C. Ullrich
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  • This is a very nice summary. It would be great if it were included in (or maybe linked from) one of the answers in the original thread, since that one seems to be the "canonical" Q/A on this topic, and this one looks likely to be closed due to the poorly framed question. But I'm not sure what the correct etiquette is for adding supplemental info to an old thread. –  Jun 23 '16 at 23:57
  • Sorry David! My question was written totally wrong. I edit it just now. I just need just one example to see that is the theorem in the link is not true in infinite measure spaces. thanks and sorry again. – SirSaleh Jun 24 '16 at 02:51