Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Possible Duplicate:
Limit of $L^p$ norm

I was asked to show:

Assume $|f|_{r}<\infty$ for some $r<\infty$. Prove that $$ |f|_{p}\rightarrow |f|_{\infty} $$ as $p\rightarrow \infty$.

I am stuck in the situation that $|f|_{p}<\infty$ for all $r<p<\infty$, but $|f|_{\infty}=\infty$ nevertheless. Could this happen? Imagine $f^{-1}(n,n+1)$ has measure $\frac{1}{n^{n}}$, for example. Then $|f|_{p}$ exists for any $p$, but $|f|_{\infty}=\infty$ nevertheless. However, I do not know how to show $f_{p}$ must be monotonely increasing in this case.

Could $f_{p}$ be fluctuating while $|f|_{\infty}=\infty$? I have proved that for $r<p<s$, $|f|_{p}<\max (|f|_{r},|f|_{s})$. But this does not help to show $|f|_{p}$ does not fluctuate.

share|improve this question

marked as duplicate by David Mitra, Martin Argerami, Henry T. Horton, TMM, Davide Giraudo Dec 24 '12 at 20:35

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1  
Could you write down the definition of $\vert f \vert_{\infty}$ in your post? –  user17762 Dec 24 '12 at 18:40
    
The usual definition that $f^{-1}(c,\infty)$ has measure 0, and $c$ is the inf of all such values possible. –  Bombyx mori Dec 24 '12 at 18:41
    
I am sure this is a duplicate question, although cannot find it now. –  sdcvvc Dec 24 '12 at 18:42
    
This is from Rudin, so will not be surprising if it is a duplicate. –  Bombyx mori Dec 24 '12 at 18:42
2  
See this post. –  David Mitra Dec 24 '12 at 18:47

1 Answer 1

I will put a partial answer to my question by combing Davide's proof and my previous work.

Suppose $|f|_{\infty}=a$, then $(r,a)\in E$ and by the continuity of $\phi$ we proved the statement. So we need to prove this under the hypothesis $|f|_{\infty}=\infty$. We have two cases:

(1) $|f|_{p}<\infty,\forall p<\infty$. Then we need to show $|f|_{p}\rightarrow \infty$ as $p\rightarrow \infty$. (2) There exist some $p>r$ such that $|f|_{p}=\infty$. We need to show $\forall q>p$, $|f|_{q}=\infty$ as well.

Now (2) is straightforward since if $|f|_{q}<\infty$, then $p\in [r,q]$ implies $|f|_{p}<\infty$ as well. This contradicts with our hypothesis. So it suffice to prove (1). But Davide already showed via a clever argument that $\lim \inf |f|_{p}\ge |f|_{\infty}$. So this force $\lim \inf |f|_{p}=\infty$ as well.

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.