Let $f_n:[0,1]\to \mathbb{R}$ be given by $$ f_n(x)=\frac{2x^2}{x^2+(1-2nx)^2},\ n\in \mathbb{N}. $$ Then does the sequence $(f_n)$ have a subsequnce that converges uniformly on $[0,1]$.
I was trying to apply The Arzelà-Ascoli Theorem, but I got the family is not equicontinuous. As if it is equicontinuous then
$$ \forall \ \varepsilon>0, \exists \ \delta(\varepsilon)>0,\ \text{s.t } \forall n\in \mathbb{N}\ \text{and } \forall x,y\in [0,1], |x-y|< \delta \implies |f_n(x)-f_n(y)|<\varepsilon. $$ Take $\varepsilon=1, x=\frac{1}{2n},\ y=\frac{1}{n},$ then $|f_n(x)-f_n(y)|=\frac{2n^2}{n^2+1}\nless 1$.
So, I can not conclude anything. Then how will I say whether the sequence has a uniformly converging subsequence or not?
It does not have a uniformly convergence subsequence. We can see that the limit function $\lim_{n\rightarrow \infty} f_n(x)=0$. However, for any subsequence $f_{n'}(x)$, \begin{equation} \max_{x\in\left[ 0,1 \right]} \left|f_n(x)-0\right|\geq f_{n'}\left( \frac{1}{2n'} \right)=2. \label{} \end{equation} thus, this subsequece is not uniformly convergent.