Multiplication operator with fixed continuous function

Question

Let $u\in C[0,1]$ fixed. The multiplication operator $M_{u}:L^{2}[0,1]\to L^{2}[0,1]$ defined by $u$ is given by $$ M_{u}(f)=u(t)f(t) $$ I want to show that if $M_{u}$ is compact then $u=0$, the zero function. I know I have to proceed by contradiction, nonetheless, I can't see which particular bounded sequence applied to $M_{u}$ contradicts $u\neq 0$. I've seen many related questions but I ask this question because this one is different from the others since the fixed function is on $C[0,1]$ and the arguments are taken on $L^{2}[0,1]$, so I'm not sure if I can proceed the same way.

Answer 1

A very elementary approach exploits the fact that $L^2([0,1])$ has an orthonormal basis $\{e^{int}\}_{n\in \mathbb Z}$.

So, if we write $u(t)=\sum a_ne^{int},$ then, defining \begin{align} g_k(t):=\sum a_ne^{i(n+k)t}\quad \text{gives}\quad g_k(t)-g_{k+m}(t)=\sum a_ne^{i(n+k)t}-\sum a_ne^{i(n+k+m)t}=\sum a_ne^{i(n+k)t}(1-e^{imt})=(1-e^{imt})e^{ikt}\sum a_ne^{int}=(1-e^{imt})e^{ikt}u(t).\end{align}

If $u\neq 0$ then without loss of generality, $u>\delta>0$ on some $[a,b]\subseteq [0,1].$ Then,

\begin{align} |g_k-g_{k+m}|^2\ge \delta^2\int_a^b|(1-e^{imt})e^{ikt}|^2dt=\delta^2\int_a^b|(1-e^{imt})|^2dt=\delta^2\int_a^b(2-2\cos mt)dt=2\delta^2(b-a-\frac{1}{m}(\sin mb-\sin ma))\end{align}

from which we conclude that no subsequence of $(g_k)$ is convergent.