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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.

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  • $\begingroup$ How could we prove that $E$ is connected? Did we use $\phi(p)$ in it? $\endgroup$ – mrs Dec 24 '12 at 9:25
  • $\begingroup$ This is easy; break $|f|$ into 3 parts and use the inequalities. $\endgroup$ – Bombyx mori Dec 24 '12 at 9:27
  • $\begingroup$ 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. $\endgroup$ – mrs Dec 24 '12 at 9:51
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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.

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  • $\begingroup$ 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. $\endgroup$ – Bombyx mori Dec 24 '12 at 10:38
  • $\begingroup$ Can I ask if $E$ can be any connected component of $\mathbb{R}$? $\endgroup$ – Bombyx mori Dec 24 '12 at 16:13

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