Equivalent conditions for composition to be compact operator

Question

I did some exercises in Conway's functional analysis book and found the following problem:

Let $\tau:[0,1]\to [0,1]$ be continuous and define $A:C[0,1]\to C[0,1]$ by $Af:= f\circ \tau$. Give necessary and sufficient conditions on $\tau$ for $A$ to be compact (operator).

I am having difficulty to find this condition. What kind of condition should $\tau $ have to have this?

Answer 1

General helpful little fact

If $A$ is a bounded operator and there is an infinite dimensional subspace $M$ on which $A$ has a lower bound $c>0$, i.e., $$\|Ax\|\ge c\|x\|\quad \text{ for all }x\in M \tag{1}$$ then $A$ is not compact. (Proof: pick a bounded uniformly separated sequence in $M$; its image under $M$ has the same properties.)

In your situation

If the range of values of $\tau$ contains a nontrivial interval $[a,b]\subset [0,1]$, then take $M$ to be the set of continuous functions supported on $[a,b]$ and observe that (1) holds with $c=1$.

The other case is $\tau$ being constant. What is the range of $A$ then?