Your example of a multiplication operator has the property of being weak$^*$-continuous, and in that setting, we can say quite a lot, with an elementary proof.

**Claim:** Let $E$ be a Banach space and $T:E^*\rightarrow E^*$ be a weak$^*$-continuous operator (that is, the adjoint of some $S:E\rightarrow E$). Then the following are equivalent:

- For any
*bounded* net $(a_j)$ in $E^*$ which is weak$^*$-convergent to 0, we have that $\|T(a_j)\| \rightarrow 0$.
- $S$ is a compact operator.
- $T$ is a compact operator.

**Proof:** That $S$ is compact if and only if $T$ is compact is Schauder's theorem. Set $X:=\{S(\omega):\omega\in E, \|\omega\|\leq 1\}$ and notice that (1) is equivalent to:

- For any
*bounded* net $(a_j)$ in $E^*$ which is weak$^*$-convergent to 0, we have that $\langle a_j,\omega\rangle\rightarrow 0$ *uniformly* on $X$.

Then $S$ is compact if and only if $X$ is compact. If $X$ is compact, then a simple $\epsilon$-net argument shows that (4) holds (using that the net $(a_j)$ is assumed bounded). Conversely, suppose that $X$ is not compact, so there is $\epsilon>0$ such that $X$ admits no $2\epsilon$-net. For any finite-dimensional subspace $N\subseteq E$, any closed and bounded subset of $N$ is compact, and so the distance from $N$ to $X$ must be at least $\epsilon$, say. By Hahn-Banach we can find $a_N\in E^*$ and $\omega_N\in X$ with $\langle a_N,\omega \rangle=0$ for all $\omega\in N$, with $|\langle a_N, \omega_N \rangle| \geq \epsilon$ and with $\|a_N\|\leq 1$. The (bounded) net $(a_N)$ hence shows that (4) does not hold.

If $E$ is separable then you can consider (bounded) sequences in place of nets.

Thus, I believe, your original question is equivalent to asking when the (left) multiplication operator given by $h\in M$ is *compact*.

[ I stressed the word "bounded". If you allow *unbounded nets* then the condition becomes that $X$ has finite-dimensional span, i.e. that $S$ (equivalently $T$) is a finite-rank operator. In the sequence case, the Principle of Uniform Boundedness says you have to be bounded anyway. ]