If a sequence $(f_n)$ do not converge uniformly, then no subsequence of $(f_n)$ converge uniformly

Question

Can someone verify this claim:

If a sequence $(f_n)$ do not converge uniformly, then no subsequence of $(f_n)$ converge uniformly

I saw this in a proof, where $f_n(x) = x^n$ on $x \in [0,1]$, the claim it is well known that the above sequence is not uniformly convergent, therefore no subsequence of $(f_n)$ is uniformly convergent

Answer 1

The claim isn't true in general. Suppose that $f_n\to 0$ pointwise but not uniformly, and define $g_{2n}=f_n$, $g_{2n+1}=0$. Then $g_n$ does not converge to zero uniformly, but a subsequence does.

Answer 2

Consider $f_{2n-1}\equiv 1$ and $f_{2n}=x^{2n}.$ Then $(f_n)$ doesn't converge uniformly on $[0,1]$ but the subsequence $(f_{2n-1})$ do.

Edit to answer the comment

$f_n=x^n$ has no any subsequence converging uniformly on $[0,1].$ Argue by contradiction. If such a subsequence converges uniformly the limit function must be continuous on $[0,1].$ But the pointwise limit is $$f(x)=\begin{cases} 0 & if \: 0\le x<1 \\ 1 & if \: x=1.\end{cases}$$ This is not a continuous function on $[0,1],$ but it should be if there were a uniformly convergent subsequence. So, we are done.

Answer 3

Even if each $f_n$ is continuous and $(f_n)_n$ is a bounded sequence converging pointwise but not uniformly to a continuous $f,$ it may have a uniformly convergent sub-sequence. For example, let $f_{2 n}(x)=0$ for all $x.$ Let $f_{2 n-1}(x)=0$ for $x\not \in [0,2/n],\;$ $f_{2 n-1}(x)=n x$ for $x\in [0,1/n],\;$ and $f_{2 n-1}(x)=2-n x$ for $x\in [1/n,2/n].$ And $f=0.$ Then $(f_{2 n})_n$ is uniformly convergent to $f.$