show that the image of the unit ball under $T$ is not compact. Let $C[0,1]$ denote the Banach space of continuous real functions from $[0,1]\to \mathbb{R}$. Fix a non constant function $g:[0,1] \to [0,1]$. Define $T: C[0,1] \to C[0,1]$  by $[T(f)](x) = f(g(x))$ for $x \in [0,1]$, ie, $T(f) = f \circ g$. Show image of unit ball under $T$ is not compact.
How does one proceed? A clear solution would be helpful
 A: Observe that if $\{x_1,...,x_n\}$ is a subset of $[0,1]$ with $n$ members, and $\{y_1,...,y_n\}\subset [0,1],$ there exists a continuous $f:[0,1]\to  [0,1]$ such that $f(x_i)=y_i$ for $i\in \{1,...,n\}.$
Since $g$ is not constant, take $a,b$ with $0\leq a<b\leq 1$ and $g(a)\ne g(b).$ Take a strictly increasing or strictly decreasing sequence $(y_n)_{n\in N}$ with $y_1=g(a)$ and $\lim_{n\to \infty}y_n=g(b).$
Since $g$ is continuous,by the IVT we may, for each $n\in N$, choose $x_n\in [a,b]$ with $g(x_n)=y_n.$ 
For each $n\in N$ take a continuous $f_n:[0,1]\to [0,1]$ such that $f_n(y_j)=0$ if $j<n,$ and $f_n(y_n)=1.$ Therefore, for $m<n$ we have: $$(T f_m)(x_m)=f_m(g(x_m))=f_m(y_m)=1.$$ $$ (T f_n)(x_m)=f_n(g(x_m))=f_n(y_m)=0.$$ $$\text {So }\quad m\ne n\implies \|(T f_m)-(T f_n)\|=1.$$ Hence the sequence $(T f_n)_{n\in N}$ has no convergent subsequences,so  the image under $T$ of the closed unit ball is not compact
A: Hint: Partial solution: suppose that there exists $x$ such that $g(x)=1$, consider $f_n(x)=x^n$. $\| f_n\| =1$ and $\lim_n T(f_n)=\lim_ng^n$ is not continuous, since $\lim_ng^n(y)=0$ or $1$. You have $\lim_ng^n(x)=1$, and since $g$ is not constant, there exists $y:g(y)\neq 1$, thus $0\leq g(y)<1$ and $\lim_ng^n(y)=0$. Thus $\lim_ng^n$ is not continuous.
In general, let $M\in [0,1]$ be the maximum of $g$, $M>0$ since $g$ is not constant. Consider $h\colon[0,1]\rightarrow R$ defined by $h(x)={x\over M}$ if $x\leq M$, $h(x)=1-(x-M)$, if $x>M$, $\| h^n\|=1$, $T(h^n)=h^n(g)$. Let $x_M$ such that $g(x_M)=M$. We have $h^n(g(x_M))=1$. Remark that if $g(y)\neq M$, $h(g(y))<1$, thus $\lim_nh^n(g(y))=0$. This implies that $\lim_nT(h^n)(y)=\lim_n h^n(g(y))\in \{0,1\}$. Since $g$ is not constant, we deduce there exists $y$ such that $g(y)\neq M$ and $\lim_nh^n(g(y))=0$. Thus we cannot extract a subsequence of $T(h^n)$ which converges, thus $T$ is not compact.
