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 $\{f_n\}$?

  • $\begingroup$ For uniform convergence limit of supremum of $|f_n(x)-f(x)|$ should be zero. $\endgroup$
    – zkutch
    Commented Sep 6, 2021 at 3:18
  • $\begingroup$ @zkutch could you please clarify your comment? $\endgroup$
    – SunRoad2
    Commented Sep 6, 2021 at 3:24
  • 1
    $\begingroup$ It's classical equivalent condition for uniform convergence and is used in José Carlos Santos answer below. $\endgroup$
    – zkutch
    Commented Sep 6, 2021 at 3:35

1 Answer 1


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 $n\gg1$.

  • $\begingroup$ Thanks for your answer. Having to prove this exercise seems silly to me because, intuitively, it is so obvious and the proof is so simple, but nonetheless, I appreciate your help. $\endgroup$
    – SunRoad2
    Commented Sep 6, 2021 at 3:36
  • 2
    $\begingroup$ I'm glad I could help. $\endgroup$ Commented Sep 6, 2021 at 3:45

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