# Uniform Convergence and Subsequences

Exercise Show that a sequence of functions $$\{f_n\}$$ fails to converge uniformly to a function $$f$$ on a set $$E$$ if and only if there exists some positive $$\varepsilon$$ such that a sequence $$\{x_k\}$$ of points in $$E$$ and a subsequence $$\{f_{n_k}\}$$ can be found such that

$$|f_{n_k}(x_k)-f(x_k)| \geq \varepsilon$$

I am struggling to prove the $$(\Leftarrow)$$ direction.

For the $$(\Leftarrow)$$ direciton, we suppose that there is $$\varepsilon > 0$$ such that a sequence $$\{x_k\} \subset E$$ and $$\{f_{n_k}\} \subset \{f_n\}$$ such that

$$|f_{n_k}(x_k)-f(x_k)| \geq \varepsilon$$

It is at this point that I get stuck. In the previous direction $$\big[(\Rightarrow)\big]$$ if you assume that the $$f_n$$'s are not uniformly convergent on the entire sequence of functions and the entire set $$E$$, then it follows immediately that any sequence $$\{x_k\}$$ and subsequence $$\{f_{n_k}\}$$ also satisfy the conclusion (not uniformly convergent).

But working from the bottom and generalizing outward is tougher. How does one start with a non-uniformly convergent subsequence $$\{f_{n_k}\}$$ of a sequence of functions $$\{f_n\}$$ and generalize this to the sequence of functions $$\{f_n\}$$?

• For uniform convergence limit of supremum of $|f_n(x)-f(x)|$ should be zero. Commented Sep 6, 2021 at 3:18
• @zkutch could you please clarify your comment? Commented Sep 6, 2021 at 3:24
• It's classical equivalent condition for uniform convergence and is used in José Carlos Santos answer below. Commented Sep 6, 2021 at 3:35

Suppose that, for some $$\varepsilon>0$$, there is a sequence $$(x_n)_{n\in\Bbb N}$$ of points of $$E$$ and a subsequence $$(f_{n_k})_{k\in\Bbb N}$$ of $$(f_n)_{n\in\Bbb N}$$ such that $$(\forall k\in\Bbb N):\left|f_{n_k}(x_k)-f(x_k)\right|\geqslant\varepsilon$$. Then$$(\forall k\in\Bbb N):\sup|f_{n_k}-f|\geqslant\varepsilon.$$Therefore, the sequence $$(f_n)_{n\in\Bbb N}$$ does not converge uniformly to $$f$$ because, if it did, $$\lim_{n\to\infty}\sup|f_n-f|=0$$ and so $$\sup|f_n-f|<\varepsilon$$ if $$n\gg1$$.