Let $C[0,1]$ be the set of complex valued function on $[0,1]$ and Banach space with sup norm.
Let $g$ be non constant continuous function from $[0,1]$ to $[0,1].$ Define $T$ from $C[0,1]$ to $C[0,1]$ as $T(f(x))=f(g(x)).$
Show that the image of unit ball of $C[0,1]$ under $T$ is not comapct in $C[0,1].$
I want to show this for open unit ball centered at $0.$ Any hint please.
Let $M=\|g\|_{\infty}$. There is a continuous function $h: [0,1] \to [0,1]$ such that $h(M)=1$ and $h(t) \neq 1$ if $t \neq M$. Consider the sequence $f_n(x)=(h(x))^{n}$. This is a bounded sequence in $C[0,1]$. If $T$ is compact there exists $n_k$ increasing to $\infty$ such that $f_{n_k}(g(x))=(h(g(x))^{n_k}$ converges uniformly. But this sequence converges pointwise to $0$ when $h(g(x)) <1$ and to $1$ when $h(g(x))=1$. The limit has to be continuous so we must have $h(g(x)) <1$ for all $x$ or $h(g(x)) =1$ for all $x$. Hence $g(x) <M$ for all $x$ or $g(x) =M$ for all $x$. But $g$ attains the value $M$ at some point. It follows that $g$ is a constant.