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).



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

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. enter image description here

  • $\begingroup$ nice function +1 $\endgroup$ – idm Jul 5 '15 at 7:32


$f_n:(0,1]\to\mathbb{R}$, 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.


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.