Composition operator compact iff point evaluation

Question

I am interested in how to proof the following claim: Let $T \colon C[0,1] \to C[0,1], f \mapsto f \circ \tau$ be compact, then it must be that $\tau$ is constant. It is clear that if $\tau$ is constant it follows that $T$ is compact. I do not have a clue how to proof the converse. I tried some sequences and tried to prove that there is no convergent subsequence whenever there are $x,y \in [0,1] $ such that $\tau(x) \neq \tau(y)$ but it did not work.

Answer 1

Assume $\tau$ is non-constant. This implies $\tau$ is not locally constant. That means there is a point $a \in [0,1]$ and a sequence $a_n \to a$ such that all $\tau(a_n) \ne \tau(a)$. Assume without loss of generality that $\tau(a)=0$.

Now choose a sequence of continuous functions $f_n$ such that each $f_n(a_n)=1$ and each $f_n(a_m)=0$. Observe this implies every $d(f_n,f_m) \ge 1$. It's an exercise to construct the functions explicitly in the unit ball of the domain.

Now since every $-1 \le f_n(x) \le 1$ we have also that $-1 \le f_n(\tau(x)) \le 1$. Therefore the image of the sequence $f_n \circ \tau$ is contained in the unit ball of the codomain.

But we can see the sequence $f_n \circ \tau$ has no convergent subsequence because that would require a Cauchy subsequence, and every $d(f_n,f_m) \ge 1$ by construction.

Therefore the image of the unit ball fails to be relatively compact.