Does this sequence admit a uniformly convergent subsequence on [0,1)?

Question

Let $(f_n)$ be the functions sequence difined on $[0,1],$ $$f_n(x)=x^n.$$

My question is:

Does this sequence admit a uniformly convergent subsequence on [0,1)?

Can any one help me on this question?

Answer 1

Use this property to show that sequence is not uniformly convergent is$${\rm lim}\ [{\rm sup}\ \{ | f_n(x)-f(x)|\ :\ x\in S\ \}]=1$$, where $S$ is the non zero value interval of the function.

Answer 2

There are two ways:

• Without Arzela-Ascoli: the uniform limit of a subsequence is a pointwise limit. The only possibility is the function which vanishes on $[0,1)$ and takes the value $1$ at $1$. But this function is not continuous. A uniform limit of continuous functions is continuous, hence no subsequence can be uniformly convergent.
• With Arzela-Ascoli: clearly, $\left\lvert f_n(x)\right\rvert\leqslant 1$ for any $n$ and any $x$, so the only thing which can fail is equi-continuity. For any $\delta$ and any $n\gt 1/\delta$, $$\left\lvert f_n(1)-f_n\left(1-\delta\right) \right\rvert=1-\left(1 -\delta\right)^n \geqslant 1-\left(1 -\frac 1n\right)^n $$
  hence $$\sup_{\left\lvert x-y\right\rvert\lt\delta } \left\lvert f_n(x)-f_n\left(y\right) \right\rvert \geqslant 1-\left(1 -\frac 1n\right)^n $$ so that $$\sup_n \sup_{\left\lvert x-y\right\rvert\lt\delta } \left\lvert f_n(x)-f_n\left(y\right) \right\rvert \geqslant 1-e^{-1}\gt 0.$$