Operator that has no fixed point Let $X = \{f \in C[0,1]; \|f\|_{\infty}\leq1, f(0)=0, f(1)=1\}$ be a subset of $C[0,1]$ and define the operator $T:X \rightarrow X$ by $Tf(t)=f(t^{2})$. Show that $T$ is continuous and has no fixed point. 
Could you help me with the later question?

Suppose $f(t)=f(t^2)$ for some $f\in C[0,1]$ such that $f(0)=0$ and $f(1)=1$. Then 
$$f(x^{2^n})=f(x)$$
for all $n\geq 1$. As $f$ is continuous and $x^{2^n}$ converges to $0$ for all $x\in [0,1)$ for $n\to\infty$, we have that $f(0)=f(x)$ for all $x\in [0,1)$ which is a contradiction to continuity and $f(1)=1$.
