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Wanted to prove the following question since one week but couldn't get even single idea on it . Here is the question : if $m : \mathbb R \to \mathbb R$ a measurable function $1 \le p <\infty $ and the operator $T : L^p (\mathbb R) \to L^p (\mathbb R)$ defined by

$(Tf)(s) := m(s)f(s)$ for all $f \in L^p (\mathbb R) $ with $D \subset L^p(\mathbb R)$ , $D =\{f\in L^p ( \mathbb R) : m.f\in L^p(\mathbb R)\}$

My main aim is to know $T$ is compact if and only if $m=0$ almost everywhere . and also i wonder if it is defined on the whole space $L^p(\mathbb R)$ and if $T$ is bounded .

Thank you for your help .

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I just saw that you also asked this question. The result there can be applied to proving that $T$ is compact $\iff$ $m=0$ a.e. easily. –  23rd Dec 21 '12 at 13:30

1 Answer 1

up vote 7 down vote accepted

Part 1: To show $T$ is compact $\Rightarrow$ $m=0$ a.e, assume that there exists a measurable set $E$ with $\lambda(E)>0$, such that $|m|\ge \epsilon>0$ on $E$. Here $\lambda$ denotes the Lebesgue measure on $\mathbb{R}$. Then it suffices to show that $T$ is not compact.

It is easy to construct a sequence of measurable sets $(E_n)_{n\ge 1}$, such that: (i) they are pairwise disjoint, (2)$\cup_n E_n=E$ and (iii) $\lambda(E_n)=2^{-n}\lambda(E)$.

Now let $f_n=2^{n/p}\chi_{E_n}$. Then $\|f_n\|_p=\lambda(E)^{1/p}$ and $$\|Tf_n-Tf_k\|_p\ge\epsilon \|f_n-f_k\|_p=(2\lambda(E))^{1/p}\epsilon, \quad\forall n\ne k.$$

Therefore, the sequence $Tf_n$ has no Cauchy subsequence, which implies that $T$ is not compact.

Part 2: $T$ is well defined on whole $L^p(\mathbb{R})$ if and only if $\|m\|_\infty<\infty$, and in this situation, $\|T\|=\|m\|_\infty<\infty$.

On the one hand, if $\|m\|_\infty=\infty$, we can find a sequence of measurable sets $(E_n)_{n\ge 1}$, such that: (i) they are pairwise disjoint, (ii) $\lambda_n:=\lambda(E_n)>0$ and (iii) $|m|\ge 2^n$ on $E_n$. Then define $f=\sum_{n=1}^\infty2^{-n}\lambda_n^{-1/p}\chi_{E_n}$. By definition, $$\|f\|_p^p=\sum_{n=1}^\infty\int_{E_n} 2^{-np}\lambda_n^{-1} dx\le \sum_{n=1}^\infty2^{-np}\le 1,$$

and $$\|Tf\|_p^p=\sum_{n=1}^\infty\int_{E_n} 2^{-np}\lambda_n^{-1}|m(x)|^pdx\ge\sum_{n=1}^\infty\int_{E_n} \lambda_n^{-1}dx= \infty.$$ Therefore, $T$ is not well defined on whole $L^p(\mathbb{R})$.

On the one hand, if $\|m\|_\infty<\infty$, then by definition $\|T\|\le\|m\|_\infty<\infty$, i.e. $T$ is well defined on whole $L^p(\mathbb{R})$ and bounded. Additionally, for every $\epsilon>0$, there exists a measurable set $E$, such that $0<\lambda(E)< \infty$ and $|m|\ge \|m\|_\infty-\epsilon$ on $E$. For $f=\chi_E$, it is easy to see that $0<\|f\|_p<\infty$ and $\|Tf\|_p\ge(\|m\|_\infty-\epsilon)\|f\|_p>0$, and it follows that $\|T\|=\|m\|_\infty$.

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can you elaborate more on the construction of the measurable set ? And how can i say that $T$ has to be compact? Why is it not possible for $T$ not to be compact when the measure is $0$. I think it would be nice if you can elaborate more . Thank you :) –  Theorem Dec 20 '12 at 18:51
    
@Theorem:Can you construct a subset $E_1$ of $E$ such that $\lambda(E_1)=\lambda(E)/2$? If so, then replace $E$ with $E\setminus E_1$ and go on. When $m=0$ a.e., $T\equiv 0$, which is obviously compact. –  23rd Dec 20 '12 at 18:56
    
@Theorem: Sorry, there were some small errors in the choice of constants, and I hope all of them are corrected now. Have you filled in all the missing details already? –  23rd Dec 20 '12 at 20:06
    
not yet, i am working on it. finding some difficulty though . –  Theorem Dec 20 '12 at 20:10
    
@Theorem: I believe you can work it out by yourself. –  23rd Dec 20 '12 at 20:14

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