# Sequential characterization of non-equicontinuous sequence

Assume $f_n$ is a sequence of uniformly continuous functions.

I was given the following criterion to determine if $f_n$ is equicontinuous.

If $\exists t_n,s_n.t_n-s_n \to 0 \land \|f_n(t_n) - f_n(s_n)\| \not\to 0$ then $f_n$ is not equicontinuous.

Proof (draft)

If $f_n$ is not equicontinuous then there exists $\epsilon_0$ such that $\forall \delta > 0,t,s \in I. |t-s| < \delta \implies |f_n(t)-f_n(s)| \geq \epsilon_0 \forall n$.

So we take $\delta = \frac 1 m$ and we get a sequence $|f_n(t_m)-f_n(s_m)|$. The double index is something we want to get rid of, here is where the hypothesis of uniformly bounded comes into play.

Question

I was told the converse does not hold. So I want a counterexample for:

If $f_n$ is not equicontinuous then $\exists t_n,s_n.t_n-s_n \to 0 \land \|f_n(t_n) - f_n(s_n)\| \not\to 0$

So what I want to prove is that for some $f_n$:

$f_n$ is not equicontinuous and $\forall t_n,s_n.t_n-s_n \to 0 \implies \|f_n(t_n) - f_n(s_n)\| \to 0$

• $\sqrt{t}$ is one function, not a sequence. Hint: take a periodic (nonconstant) continuous function on some $[0, a]$ (sine willl be a perfect choice), and squeeze its graph horizontally by factor $2, 3, 4, \dots, n, \dots$. – user539887 Mar 19 '18 at 8:07
• Yes, your proof below is O.K. (only in (1) there should be $f_{n_{\delta}}(t_\delta) - f_{n_{\delta}}(s_\delta)$, and $\delta^n = \min \{ \delta_1, \dots, \delta_n, \frac{1}{n} \}$). My "I do not know" referred (and still refers) to your last problem: finding a sequence $f_n$ that is not semicontinuous, however for any two sequences $t_n - s_n \to 0$ there holds $\lVert f_n(t_n) - f_n(s_n) \rVert \to 0$. – user539887 Mar 21 '18 at 20:02

The proposition seems to be right.

Proposition:

If $\forall n \in \mathbb{N}, f_n$ is uniform continuous but the sequence $f_n$ is not equicontinuous then $\exists t_n,s_n \in I.t_n-s_n \to 0$ but $\|f_n(t_n) -f_n(s_n)\| \not\to 0$.

Solution:

If $f_n$ is not equicontinuous then: $$\exists \epsilon_0 > 0.\forall \delta > 0. \exists t_{\delta},s_{\delta} \in I.|t_{\delta} - s_{\delta}| < \delta \land \exists n_{\delta} \in \mathbb{N}.\|f_{n_{\delta}}(t_{\delta}) -f_n(s_{\delta})\| \ge \epsilon_0 \;\;\;\; (1)$$

Given $n \in \mathbb{N}$, we have that $f_1,\ldots,f_n$ are uniformly continuous and therefore, for $k = 1,\ldots,n$: $$\exists \delta_1,\ldots,\delta_n \in \mathbb{R}^{+}.t,s \in I,|t-s| < \delta_k \implies \|f_k(t)-f_k(s)\| < \epsilon_0 \;\;\;\; (2)$$

We now take in $(1)$, $\delta^n = \{\delta_1,\ldots,\delta_n,\frac 1 n\}$ and this gives us that: $$\exists \epsilon_0 > 0.\text{ for this } \delta^n.\exists t_{\delta^n},s_{\delta^n} \in I,n_{\delta^n} \in \mathbb{N}.|t_{\delta^n}-s_{\delta^n}| < \delta^n \land \|f_{n_{\delta^n}}(t_{\delta^n}) -f_{n_{\delta^n}}(s_{\delta^n})\| \ge \epsilon_0$$ so that if we denote $t_n := t_{\delta^n}$,$s_n := s_{\delta^n}$ then we have that $t_n-s_n \to 0$ since $|t_n-s_n| < \delta \le \frac 1 n \to 0$.

By $(2)$, we have that for $k = 1, \ldots,n.\|f_{k}(t_{\delta^n}) -f_k(s_{\delta^n})\| < \epsilon_0$ and therefore, necessarily $n_{\delta^n} > n$. We may denote $\sigma(n) = n_{\delta^n}$. Clearly, $\sigma(n) > n$.

In conclusion, we have proved that: $$\exists \epsilon_0.\forall n_0 \in \mathbb{N}.\exists n = \sigma(n_0) \ge n_0.\|f_m(t_n)-f_n(s_n)\| \ge \epsilon_0$$ This means that $\|f_n(t_n)-f_n(s_n)\| \not\to 0$. As we wanted.