# Sequence uniform convergence but the derivatives are not.

Give a sequence $(f_n)_{n\in \mathbb{N}}$ of differentiable functions which uniformly converge to $0$, but for which the seqeunce $(f_n')_{n\in \mathbb{N}}$ of the derivatives isn't even pointwise convergent.

I found this one in my textbook marked as "Fun things to solve" and although it may be fun, it's kind of hard to do.

To be honest I already failed at the first hurdle. I couldn't even find a sequence of functions which uniformly converges to $0$.

Is there a certain way of dealing with this kind of problem? Because it seems kind of hard for me to come up with sequences without a mathematical of way of doing so (or maybe there is a mathematical way I just don't know yet).

Consider

$\displaystyle f_n:\mathbb R \to \mathbb R, x\to \frac{\sin(nx)}{\sqrt{n}}$

Each $f_n$ is $C^1$, $(f_n)$ converges uniformly to $0$ over $\mathbb R$, but $(f_n')$ fails to converge anywhere.

The rationale behind coming up with this sequence is the following: you need functions that go uniformly to $0$ and sines with decreasing magnitudes fit the bill. Furthermore you want $f_n'$ to behave wildly. This suggests each $f_n$ does not vary in the same way: sines with increasing oscillations achieve this requirement.

• nice function +1 – idm Jul 5 '15 at 7:32

Consider:

$$f_n:(0,1]\to\mathbb{R}$$, where $$f_n(x)=\frac{1}{\sqrt{n}}x^n$$. Clearly, $$f_n$$ is differentiable for each $$n$$ and $$f_n\to\hat{0}$$ (uniformly). As $$f_n'(x)=\sqrt{n}x^{n-1}$$, we immediately see that the sequence is unbounded for $$x=1$$ and hence not pointwise convergent.