Compact linear operator from $L^p (\mathbb R)$ to $L^p (\mathbb R)$ 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 . 
 A: 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 $\|T\|=\|m\|_\infty$.
