We have a continuous function $f$ on the interval $[0,1]$ such that $f(0)=1,f(1)=3$. A functions sequence is defined by $f(x^{n})$, and I have a limit function $g\equiv 1$. I need to prove (or disprove) that $f(x^n)$ uniformly converges to $g$ on $[0,\frac{1}{2}]$.
It seems that proving convergence can be done by the fact that $f$ is continuous, so for every sequence $x_n$, and specifically $x^n$, since $\forall x\in[0,\frac{1}{2}]$, $x^n\to0 \Rightarrow f(x^n)\to f(0)=1=g$.
On the other hand, I can construct $f$ such that for example $f(\frac{1}{4})=10^6$ (and even $f([a,b])=10^6$ for some $0<a<b<\frac{1}{2})$ so it remains continuous, and plotting this doesn't seem to convince me that it does uniformly converge to the constant function $g\equiv1$. But then again, the whole sequence is contained within the single $f$. Can I formally argue this? If so, how?
The fact that $f(1/4)$ could be large is immaterial: once $n > 2$, for example,
$$x \in [0, 1/2] \implies x^n \in [0, 1/8]$$
and $f(x^n)$ can no longer "see" that $f(1/2)$ is large. This argument works for any $a > 0$ playing the role of $1/2$.
In fact, this suggests the proof: Given $\epsilon > 0$, there exists $\delta > 0$ such that $f([0, \delta]) \subseteq [1 - \epsilon, 1 + \epsilon]$. Choose $n$ large enough that $2^{-n} < \delta$. Then $f(x^n) \in [1 - \epsilon, 1 + \epsilon]$ as desired.
Notice how the choice of parameters is independent of $x$, so we do indeed have uniform convergence.
Exercise for the interested reader: conclude that we have uniform convergence on any interval $[0, 1 - \epsilon]$, but not on $[0, 1]$.