I have some problem about my homework.

Is the sequence of function $f_n:\mathbb R \to \Bbb R$ defined by $$f_n(x) = cos(n+x) + log(1+\frac{1}{\sqrt{n+2}}sin^2(n^nx))$$equicountinuous? prove or disprove.

But I have checked the textbook, which says

A sequence of functions ($f_n$) in $C^0$ is equicontinuous if $$\forall \epsilon >0, \exists\delta>0 $$ such that $$|s-t|<\delta, n\in \Bbb N \implies |f_n(s)-f_n(t)|<\epsilon$$For total clarity, the concept might better be labeled uniform equicontinuity, in contrast to pointwise equicontinuity, which requires $$\forall\epsilon>0\forall x\in [a,b],\exists\delta>0 $$such that $$|x-t|<\delta,n\in\Bbb N \implies|f_n(x)-f_n(t)|<\epsilon$$

Now my question is, what the difference between pointwise-equicontinuous and uniform-equicontinuous? Can I have some examples? And which does it usually means when only mentioned equicontinuous?

  • $\begingroup$ Pugh, Real Mathematical Analysis, Chapter 4? $\endgroup$ – Divide1918 Oct 19 '20 at 15:42

Let $I$ be an index set. Often $I=\mathbb{N}$, but in general $I$ needn't even be countable.

A family of functions $\{ f_\alpha \}_{\alpha \in I}$ are:

  • continuous if each $\varepsilon$,$x$, and $\alpha$ has an appropriate $\delta$.

  • uniformly continuous if each $\varepsilon$ and $\alpha$ has an appropriate $\delta$ that doesn't depend on $x$

  • pointwise equicontinuous if each $\varepsilon$ and $x$ has an appropriate $\delta$ which doesn't depend on $\alpha$

  • uniformly equicontinuous if each $\varepsilon$ has an appropriate $\delta$ which depends neither on $\alpha$ nor on $x$.

In a lot of applications of the concept of equicontinuity, we are working on a compact metric space. On a compact metric space, essentially the exact same proof that works for the Heine-Cantor theorem lets us see that pointwise equicontinuity and uniform equicontinuity are equivalent. I think this causes some authors to define "equicontinuous" to mean what I am calling "uniformly equicontinuous". Your author seems to be adopting this convention.

In your problem you are not on a compact metric space. So while each function in your family is certainly pointwise continuous, you need to worry both about $n$ being a problem and about $x$ being a problem. The first term and the log aren't too big of a deal. However, you should be worried about $\sin^2(n^n x)$: that $n^n$ is creating a lot of oscillation on short length scales when $n$ is large. You need to see whether the division by $\sqrt{n+2}$ is enough to mitigate that.

  • $\begingroup$ actually, I can "prove" the question in the begining is not uniform equicontinuous. But as you just said, in compact set uniform is equivalent to not uniform(equicontinuous). $\endgroup$ – Vito Chou Mar 18 '15 at 18:25
  • $\begingroup$ @VitoChou I'm not sure I understand where you're going with your comment. $\endgroup$ – Ian Mar 18 '15 at 18:28
  • $\begingroup$ @VitoChou It is important to note in your problem that $\mathbb{R}$ is not compact. $\endgroup$ – Ian Mar 18 '15 at 18:37
  • $\begingroup$ Is that means when I prove f is pointwise equicontinuous, I shall : Given $\epsilon$,$x$, and then find $\delta (\epsilon ,a)$ s.t. for all $\alpha$... and when I prove f is uniform equicontinuous, I shall : Given $\epsilon$, and then find $\delta (\epsilon)$s.t. for all $x$ and $\alpha$ .... Is that right? $\endgroup$ – Vito Chou Mar 19 '15 at 12:50
  • $\begingroup$ In this case, seems I only need to take $\delta = 2\pi$, then whenever what $\epsilon$is taken, $f(x+\delta)-f(x)=0<\epsilon$, so it's uniformly equicontinuous? $\endgroup$ – Vito Chou Mar 19 '15 at 13:22

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