Define the map $T(f)=a(x)f(x)$ where $a(x) \in C[-\pi,\pi]$ and $f \in L^2$, show that $T$ is not compact unless $a=0$.
The only idea I had is to consider the orthonormal basis of $L^2$ $\{e^{inx}/\sqrt{2\pi}\}$. We know it converges to $0$ weakly and so since $T$ is compact we should have the image of the sequence converging to $0$ strongly. Unfortunately, this is true by reimann lebesgue lemma. Thus I not sure what to do. I did realize that if we assume it is compact, then it is self-adjoint compact operator on a Hilber space so it has nice properties, but I do not know how to exploit this.
As you pointed out, you have to check that some nice spectral properties about compact operators do not hold for $T$. For example, compact operators have countable spectrum such that any non-zero element is an eigenvalue and an isolated point.
Since $T$ is a multiplication operator, it's easy to see that its spectrum is the closure of the range of $a$. That is, $$\sigma(T)=\overline{a([-\pi,\pi])}.$$ Since $[- \pi, \pi]$ is connected and $a$ continuous, you can argue that $\sigma(T)$ can't be a countable set with isolated non-zero points (unless of course $a=0$).