Let $T$ be defined as $Tf(x)=f(\sin(x))$. I've attempted using Arzela Ascoli to check the compactness of this operator. Boundedness is no problem to show and I think I can prove equicontinuity as follows.
Consider $f_n\in B$ where $B$ is some bounded subset of $C([0,1])$
Let $|x-y|<\delta$
$$\implies |\sin(x)-\sin(y)|<\delta$$
By uniform continuity of the continuous functions $f_n$ on a compact set we have
$$|f_n(\sin(x))-f_n(\sin(y))|<\epsilon$$
This delta only depends on $\epsilon$ and our point thus proving the equicontinuity of $Tf_n$. By Arzela-Ascoli this operator appears to be compact, but I believe this is not the case. This appears similar to the identity operator on $C([0,1])$, only with a contraction in the argument of the function. The identity operator is not compact in infinite dimensional Banach spaces. Any thoughts or counter-examples?
Best Answer
Let $f_n(x)=(\frac {h(x)} c)^{n}$ where $c =\sin \, 1$ and $h$ is any continuous function with the following properties:
a) $o\leq h(t) \leq c$ for all $t$
b) There exists a unique point $s$ in $[0,1]$ such that $h(s)=c$ and this point belongs to $[0,c]$
It is easy to construct such a function.
Then $(f_n)$ is norm bounded in $C[0,1]$. If $T$ is compact there exists $n_k$ increasing to $\infty$ such that $(\frac {h(\sin\, x)} c)^{n_k}$ converges uniformly. The limit of this sequence is $0$ when $\sin \, x \neq s$ and $1$ for $\sin \, x=s$. But uniform limit of continuous functions are continuous. [Note that there is a unique $x$ with $\sin \, x=s$.]. Conclusion: $T$ is not compact.