$(f_n) $ is uniformly convergent iif $\exists f:D\subseteq \mathbb{R}\longrightarrow \mathbb{R}$ such that $\forall \epsilon > 0 \quad\exists N=N(\epsilon) \in \mathbb{N}$ such that $\forall n \geq N, \forall x\in D, |f_n(x)-f(x)|<\epsilon$

So the negation is:

$(f_n)$ is not uniformly convergent iff $\exists \epsilon_0 >0$ such that $\forall N\in\mathbb{N}, \exists k \in \mathbb{N}, k \geq N$ and $\exists x_k\in D$ such that $|f_k(x_k)-f(x_k)|\geq\epsilon_0$

The negation with subsequences is:

$(f_n)$ is not uniformly convergent iff $\exists \epsilon_0 >0, \exists (f_{n_k})$ subsequence of $(f_n)$, $\exists (x_k)$ sequence in $D$ such that $\forall k\in\mathbb{N}$ $|f_{n_k}(x_k)-f(x_k)|\geq\epsilon_0$

Show the equivalence between the two negations.

I've been trying this proof without success. Any suggestions would be great!


1 Answer 1


Hint: Idea is to construct a subsequence by taking $N=1,2,3,\cdots $
From the first definition, for $N=1,\exists k_1\gt 1: x_{k_1}\in D$ etc.
Similarly, get $x_{k_2},x_{k_3}\cdots $ etc.

  • $\begingroup$ Thanks! For the other way what you do recomend? $\endgroup$ Aug 26, 2020 at 23:22
  • 1
    $\begingroup$ @SofíaContreras: For the other way, note the "$\forall k\in \mathbb N$" part and think about $n_k\ge k$. $\endgroup$
    – Koro
    Aug 26, 2020 at 23:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.