Compactness of multiplication operator on $C[0,1]$

Question

Find a condition on function $a\in C[0,1]$ such that the operator $A:C[0,1]\rightarrow C[0,1]$ $$(Ax)(t) = a(t)x(t)$$ is compact? We are taking uniform norm on $C[0,1]$.

Answer 1

Fix $t_0\in [0,1]$, and consider a sequence of continuous functions $\{x_n\}$ such that $x_n$ is supported in $[t_0-1/n,t_0+1/n]$ and $x_n(t_0)=1$. If $A$ is compact, then we can extract of $\{Ax_n\}$ a convergent subsequence $\{x_{n_k}\}$ to some $f$. As $Ax_{n_k}$ converges pointwise to the function which is $0$ everywhere except at $t_0$, where it takes the value $a(t_0)$, we deduce by continuity that $f\equiv 0$ hence $a(t_0)$.

As $t_0$ was arbitrary...

Answer 2

Hint: Suppose $a$ is not the zero function. Then there is an interval where $a$ is non-zero. Consider the set $X$ of functions in $C[0,1]$ that have support in that interval and have norm less than or equal to $1$. Show that $A(X)$ is not precompact.