Give necessary and sufficient conditions for a multiplication on $L^p$ to be compact Let $(X, \Omega, \mu)$ be a $\sigma-$ finite measure space and for $\phi \in L^\infty(\mu)$ let $M_\phi:L^p(\mu) \to L^p(\mu)$ defined by $M_\phi f = \phi f $ be the multiplication operator. Give necessary and sufficient conditions on  $(X, \Omega, \mu)$ and $\phi$ for $M_\phi$ to be compact.
I can show if X is infinite,$\mu$ is the lebesgue measure, $\phi=0$ then $M_\phi$ is compact or if X is finite, for every measure and $\phi$, $M_\phi$ is compact. But for other measures I can not show. Please help me. Thanks in advance.
 A: A common way to prove that some operator $T$ is not compact is to exhibit an infinite-dimensional subspace $M$ on which $T$ has a lower bound: that is, there exists $c>0$ such that 
$$\|Tx\|\ge c\|x\|,\quad \forall\ x\in M \tag{1}$$
If (1) holds, then the image of unit ball under $T$ contains a ball of radius $c$ in the infinite-dimensional subspace $TM$, and therefore is not compact. 
To have (1), it suffices to have $|\phi|\ge c$ on the support of $f$. This leads to a necessary condition: for every $c>0$, the subspace $$\{f\in L^p: f=0 \text{ on the set $|\phi|< c$}\}\tag2 $$
is finite-dimensional. 
The finite dimensionality amounts to the requirement that the set $\{|\phi|\ge c\}$ consists of finitely many atoms. 
This necessary condition  is also sufficient: if it holds, the multiplication operator is a norm-limit of finite rank operators, namely multiplication by $\phi$ restricted to $\{|\phi|\ge c\}$. Indeed, let  $\phi_n:= \phi \chi_{\{|\phi|\ge 1/n\}}$. The norm of multiplication by $\phi-\phi_n$ is at most $1/n$. And the multiplication by $\phi_n$ produces functions in the space (2) with $c=1/n$, which is finite-dimensional. 
