# Operator that has no fixed point [duplicate]

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 $$f(1)=1$$.