# Showing that a sequence has bounded derivatives at a point but is not equicontinuous at that point

Problem Statement: The sequence of $f_{n}(x) = nx^{2}$ ($n\geq 1$) has bounded derivatives at $0$ but is not equicontinuous at $0$.

I am a little confused how to approach this problem, but I don't think it should be as difficult as I am making it.

So to show $f_{n}$ has bounded derivatives at $0$, we must show that $$\lvert f(x)-f(y)\rvert \leq M\lvert x-y\rvert,$$ where $M> \lvert f'(0)\rvert=0$, correct?

But do I need to show this for any $x,y$ (I assume in $\mathbb{R}$)? Or just for $x$ and $y$ such that $\lvert f_{n}(x)-f_{n}(y)\rvert=f'(0)\lvert x-y\rvert=0$? In this case, would it not just be any $x,y$ such that $x=-y$, since $\lvert f_{n}(x)-f_{n}(y)\rvert=\lvert f_{n}(-y)-f_{n}(y)\rvert=\lvert n(-y)^{2}-ny^{2}\rvert=\lvert ny^{2}-ny^{2}\rvert=\lvert 0\rvert=0$. But then wouldn't the inequality hold for any $M$?

Then for the second part, by definition of equicontinuous, $f_{n}$ is equicontinuous at $x=0$ if $\forall \epsilon >0$, $\exists \delta=\delta(\epsilon,0)>0$ such that $$\lvert x-0\rvert=\lvert x\rvert <\delta\ \Rightarrow\ \lvert f_{n}(x)-f_{n}(0)\rvert =\lvert f_{n}(x)\rvert <\epsilon.$$ Thus, to show not equicontinuous, I must show that $\exists \epsilon>0$ such that $\forall \delta >0$ $$\lvert x\rvert <\delta\ \nRightarrow\ \lvert f_{n}(x)\rvert <\epsilon.$$ Is this the right idea, or should I take a different approach to show that the sequence is not equicontinuous at $0$? I have been keeping up in my Analysis class (for the most part), but this latest section has me feeling very lost. Any tips to help me understand this problem are appreciated!

• To show that $f_n(x)$ has bounded derivative at zero, all you need to do is to compute the derivative at zero. On the other hand, equicontinuity requires that you examine a neighbourhood of $0$. – астон вілла олоф мэллбэрг Oct 8 '16 at 2:56
• $\{f_n\}$ has bounded derivatives at $0$ if there is $M>0$ so that $|f'_n(0)|\le M$ for all $n$. – user99914 Oct 8 '16 at 3:00
• So basically, if I take $0<\epsilon<1$ and $x=\frac{1}{\sqrt{n}}$, then $\lvert x\rvert=\lvert \frac{1}{\sqrt{n}}\rvert<\delta$ for some $\delta$, but $\lvert f_{n}(\frac{1}{\sqrt{n}})\rvert=\lvert n\big(\frac{1}{\sqrt{n}}\big)^{2}\rvert=\lvert n\frac{1}{n}\rvert =\lvert 1\rvert \geq 1$. Thus, the sequence is not equicontinuous at $0$? – yung_Pabs Oct 8 '16 at 4:26
• @yung_Pabs Yes. – user99914 Oct 8 '16 at 6:54