$(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

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .