Assume $ f:[0,1]\to[0,1] $ is continuous, $ f\left(0\right)=0 $ and for any $ x \in (0,1] $ we have $ f(x)<x $.
Define $ f_1=f $ and for any $ n \in \mathbb{N} $ $ f_{n+1}=f\circ f_{n} $.
Prove that $ f $ comverge unifrmly in $[0,1]$
My intuition is that $ f_n $ converge uniformly to $ 0 $. I tried to prove it using that for any $ x $, $ f_n(x) $ is decreasing and also $ f_n $ continuous for any $ n $. So by Dini's Theorem it sufficies to show that $ f_n $ converge pointwise to $ 0 $. But I couldnt see how to prove it.
Thanks in advance.
For $x=0$ everything is clear I think. So you still have to show $\lim_{n\to\infty}f_n(x)=0$ pointwise for all $1\geq x>0$.
Let $1\geq x>0$ be fixded. As you already mentioned the sequence $f_n(x)$ is monotonically decreasing as $$f_{n+1}(x)=f(f_n(x))\leq f_n(x).$$ Since $f$ is bounded from below by $0$ you have a bounded and monotonically decreasing sequence $f_n(x)$ and thus a convergent sequence. Now you still have to show that the limit is $0$:
Assume $\lim_{n\to\infty}f_n(x)=x_0\neq0$. It follows that $x_0>0$. Since $f$ is continuous it holds $$f(x_0)=f(\lim_{n\to\infty}f_n(x))=\lim_{n\to\infty}f_{n+1}(x)=x_0. $$ A contradiction to $f(y)<y$ for all $1\geq y>0$. Thus $\lim_{n\to\infty}f_n(x)=0$ pointwise for all $1\geq x>0$. Now the uniformly convergence follows from Dini's theorem, as you already correctly mentioned.