If a family of functions has an unbounded derivative, is it necessarily not uniformly equicontinuous?

I'm having trouble wrapping my head around uniform equicontinuity for a family of functions. I can often spot if a family is uniformly equicontinuous, but have trouble telling when they're not, and I was wondering if there are any signs which would allow me to spot the non-uniformly equicontinuous families. Intuitively, it seems to me that if the derivative of a family of functions is unbounded then it can't be uniformly equicontinuous.

For example, if the family is given by $F=\{f_n:n\in\mathbb N\}$ and $\forall n \in \mathbb N$, $|f_n'(x)|=|e^{n}|$, then it seems to me that the family cannot be uniformly equicontinuous.

Since this isn't a common theorem, I'm guessing my intuitiong is wrong, but a proof or a counter example would be very helpful, thanks!

Note: my definition of uniform equicontinuity is the following:

A family of functions, $F$, consisting of functions $f:(X,d)\to (Y,\rho)$ is uniformly equicontinuous on $X$ if for every $\epsilon >0$ there exists $\delta >0$ such that $$d(x,y)<\delta \implies \rho(f(x),f(y))<\epsilon$$ for all $x\in X$ and $f \in F$. The key here is that $\delta$ depends on $\epsilon$, but not on $x$ or $f$.

• do you mean like $\sup|f'_n(x)|$ is increasing and going to infinity? May 18, 2017 at 1:07
• @clark, yes that's what I mean May 18, 2017 at 2:39
• $F$ could have just one member $f.$ There exists a uniformly continuous, differentiable $f:\mathbb R\to \mathbb R$ with an unbounded derivative, May 18, 2017 at 5:13

Take for instance $$f_n (x) = \dfrac{\sin n^2x}{n}$$ then $$f_n$$ is an equicontinuous family. For a proof see at the end.

However, $$\sup|f'_n(x)| = n$$.

So, the point is imagine a family where the values increasingly oscillate faster and faster (making the sup of the derivative increase) however the values are staying close to each other.

I think the moral of the story is, when $$\sup|f'_n(x)| \leq M,~\forall n$$ then the values of the family of functions can be simultaneously controlled, however even when the derivatives behave radically differently the values can still be not very different.

Proof for equicontinuity:

Observe $$|f_n(x)-f_n(y) | \leq \frac{1}{n}|\sin(n^2x) -\sin (n^2y) | \leq \frac{2}{n}.$$

Now, let $$\epsilon >0$$ and find $$N$$ such that for $$n\geq N$$ we have $$\dfrac{2}{n}<\epsilon$$.

Also, for $$1\leq i\leq N-1$$ we can find $$\delta _i$$ such that $$|f_i(x) -f_i(y)| < \epsilon$$ whenever $$|x-y| < \delta _i$$.

From here we choose $$\delta = \min_{1\leq i\leq N-1} \delta _i$$, and we obtain $$|f_m(x) -f_m(y)| < \epsilon$$ whenever $$|x-y| < \delta$$ for all $$m$$.

• Thanks very much, I can see what you mean by plotting this function for a few different n. Would you mind giving a proof that this family is equicontinuous though? May 18, 2017 at 2:57
• @user428487 I have included a proof for the equicontinuity, please tell me if you need any further details May 18, 2017 at 3:12

Hint: Consider the family of functions consisting of just one function, namely $f(x)$ on $[0,1],$ where $f(x)=x^2\sin (1/x^2),x\in (0,1],$ $f(0)=0.$

• If we want an infinite family of functions, I guess we can take $f$ as in your answer, and $\mathcal{F} = \{f+n | n \in \mathbb{N}\}$
– Ovi
May 3, 2020 at 18:59
• @Ovi Yes, but it seemed to me a stronger result to show boundedness of $f'$ need not hold even for just one function.
– zhw.
May 3, 2020 at 19:26
• Ah I see ${}{}{}$
– Ovi
May 3, 2020 at 19:36