# 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 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.

• Thanks. Please introduce to me a reference about it. – niki Sep 2 '14 at 12:29
• I think we should suppose $\{f\in L^p$: f=0 on the set $|\phi|<c\}$ is finite - dimensional, because we want to get a contradiction by $|\phi|\geq c$ on the support of f. Am I right? – niki Sep 3 '14 at 16:19
• @niki Yes, you are right. I corrected the statement. Sorry, I don't have a reference. It's just something that most functional analysis books explain at some point. – user147263 Sep 3 '14 at 16:50
• I think for sufficient part we can not show the multiplication operator is a norm-limit of such finite rank operators. In the other word I can not show $M_{\phi_n}$ is finite rank where $\phi_n:= \phi_{|\{|\phi|<\frac{1}{n}\}}$. Please explain more about this part. – niki Sep 4 '14 at 7:20
• @niki You need $>1/n$ in the definition of $\phi_n$. See the revised answer, last paragraph. – user147263 Sep 4 '14 at 15:15