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.

I am stuck with Rudin's Real and Complex Analysis, Chapter 3 problem 4 question 3. It reads:

Suppose $f$ is a complex measurable function on $X$, $\mu$ is a positive measure on $X$, and $$\phi(p)=\int_{X}|f|^{p}d\mu=|f|^{p}_{p}$$ Let $E=\{p:\phi(p)<\infty\}$. Assume $|f|_{\infty}>0,0<p<\infty$.

Q: By previous question we know $E$ is connected. Is $E$ necessarily open? Closed? Can $E$ consist of a single point? Can $E$ be any connected subset of $(0,\infty)$?

I am really stuck. I want to show $E$ is open and can be any open subset of $X$, but I do not know how to prove it. Suppose $\phi(a)<\infty$, I do not know how to prove $\phi(a\pm \epsilon)<\infty$ as well when $|f|_{\infty}=\infty$ (otherwise I can use Holder's theorem to bridge)

update: Someone asked me why $E$ is connected. here is the proof: Q:If $r<p<s$, $r\in E, s\in E$, then $p\in E$.

Let $X_{1}$ be the part $f$'s absolute value is greater than $1$, and $X_{2}$ be the part $f$'s absolute value is less to $1$. Let $X_{3}$ be the part where $f$'s absolute value equal to 1. Then we have $$ \int_{X}|f|^{p}=\int_{X_{1}}|f|^{p}+\int_{X_{2}}|f|^{p}+\int_{X_{3}}|f|^{p} $$ The first part is less than the part in integrating $|f|^{s}$, the third part is less than that of $|f|^{r}$, and the middle part is not changing at all.

Thoughts:

One way of attacking this is by Holder's inequality. We have $$\int_{X}f^{a+\epsilon}d\mu<\int_{X}f^{a}d\mu *|f^{\epsilon}|_{\infty}$$ And so if I can bound $|f^{\epsilon}|_{\infty}$ then I will be done. However, $|f|_{\infty}=\infty$ is entirely possible (for example $\frac{1}{x^{2}}$ on $\mathbb{R}$), and this proof cannot go through.

Another way is to decompose $X$ into 3 parts as before. Then we only need to prove the statement makes sense for $|f|>1$ and $|f|<1$ respectively. Then further we only need to prove $+\epsilon$ on one direction and $-\epsilon$ on the other direction. In the former case $X_{1}$ must have finite measure, and the proof would work if $|f|_{\infty}<\infty$. But I have no idea how to prove the second case.

share|improve this question
    
How could we prove that $E$ is connected? Did we use $\phi(p)$ in it? –  Babak S. Dec 24 '12 at 9:25
    
This is easy; break $|f|$ into 3 parts and use the inequalities. –  Bombyx mori Dec 24 '12 at 9:27
    
I know that if $1<r<p<s$ then $|f|_s^s<|f|_p^p$. So if $E\neq\emptyset$ then it consist automatically infinite points. I think $E$ cannot be a single point. –  Babak S. Dec 24 '12 at 9:51

1 Answer 1

up vote 1 down vote accepted

Is $(X,\mu)$ a finite measure space? If there are no assumptions on $X$, then I have the following hints:

  1. You can show that $E$ is not necessarily closed; use $X = \mathbb{R}^+$ and think of a function $f$ that behaves like $x^{p_1}$ near $x = 0$ and $x^{p_2}$ for large $x$, for different values of $p_1,p_2$.

  2. I will just go ahead and strongly hint that $E$ also does not necessarily have to be open. Indeed, $E$ can be a single point. To see this, try investigating a function that behaves like $f(x) = x^{-1} (\log{x})^{-2}$ for large $x$, and has the inverse behavior near $0$ (use the substitution $u = \log{x}$). You will find that that such a function only is $p$-integrable for $p = 1$.

P.S. While the above example I hinted at is "famous", I actually did not recall it off the top of my head. I give credit to joriki, who posted the example here.

share|improve this answer
    
There is no specific requirement for $\mu(X)$. Yes, using $\frac{1}{x}$ we can show $E$ is not necessarily closed. The second hint is surprising for me. –  Bombyx mori Dec 24 '12 at 10:38
    
Can I ask if $E$ can be any connected component of $\mathbb{R}$? –  Bombyx mori Dec 24 '12 at 16:13

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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