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.


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 $\forall t_n,s_n.t_n-s_n \to 0 \implies \|f_n(t_n) - f_n(s_n)\| \to 0$

  • $\begingroup$ $\sqrt{t}$ 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 $2, 3, 4, \dots, n, \dots$. $\endgroup$ – user539887 Mar 19 '18 at 8:07
  • $\begingroup$ 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 $\lVert f_n(t_n) - f_n(s_n) \rVert \to 0$. $\endgroup$ – user539887 Mar 21 '18 at 20:02

The proposition seems to be right.


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 $\|f_n(t_n) -f_n(s_n)\| \not\to 0$.


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.

| cite | improve this answer | |

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.