function differentiable in no point

Hello,

I would like to ask the following question:

How can I prove that the continuous function given as

$\displaystyle f(x)=\sum_{k=1}^{\infty} \frac{\sin{(kx)}}{k^{p}}$

is not differentiable at any real x?

there is given $\displaystyle p>1$ ..

I thought I could try for arbitrary, but fixed $\displaystyle x \in \mathbb{R}$ to find the sequence $\displaystyle \{x_n\}_{n=1}^{\infty} \subset \mathbb{R} $

such that $\displaystyle x_n>x\,,\,\forall n\in\mathbb{N}$ and $\displaystyle \lim_{n\to\infty}x_n=x $ and for which there holds

$\displaystyle \lim_{n\to \infty}\frac{f(x_n)-f(x)}{x_n-x}=\pm\infty $

but it looks I am not able to do it ..

thank you very much for any ideas

Re: function differentiable in no point

I am sorry, I forgot the upper limit of $\displaystyle p$

the condition says, that $\displaystyle 2>p>1 $

Re: function differentiable in no point

I'll start out with some hints and if you are still stuck please ask again.

What happens if you differentiate $\displaystyle f(x)$ and consider the power of k in the denominator ?

Hint: the upper limit of p is important.

Re: function differentiable in no point

Hi, I guess I know what you mean

if it was possible to differentiate the series term-by-term, then I have

$\displaystyle f'(x)=\sum_{k=1}^{\infty}\frac{\cos{(kx)}}{k^{p-1}}$

but I didn't find a reason why it should be possible to do it term-by-term ..

I thought for that I need at least uniform convergence of the series $\displaystyle \sum_{k=1}^{\infty}\frac{\cos{(kx)}}{k^{p-1}}$,

which doesn't even converge pointwise at some points

on the other side,

if there was true, that $\displaystyle f'(x)=\sum_{k=1}^{\infty}\frac{\cos{(kx)}}{k^{p-1}}$,

then there are points, in which the derivative exists if you choose any $\displaystyle p$

Re: function differentiable in no point

You can consider each term independently,

$\displaystyle f(x) = f_1(x) + f_2(x)_+ ...$

And thus differentiate term-by-term.

And as you mention for $\displaystyle f(x)$ to be differentiable you do need convergence of the series formed.

This is where the limits on $\displaystyle p$ is needed.

Do you need to show that $\displaystyle f(x)$ is convergent ?

Re: function differentiable in no point

It is not the problem to show that the series $\displaystyle \sum_{k=1}^{\infty}\frac{\sin{(kx)}}{k^{p}}$ for given limits of $\displaystyle p$ converges

uniformly to its sum $\displaystyle f(x)$

what I am trying to say is, that this fact is not enough to say, that the statement

$\displaystyle f'(x)=\sum_{k=1}^{\infty}\frac{\mathrm{d}}{\mathrm {d}x}\Big(\frac{\sin{(kx)}}{k^p}\Big)$

is true, and in fact it doesn't hold,

so anything you do after this wrong assumption, can be true or also false,

so it's not the way how to do that

Re: function differentiable in no point

Yes, I did realize that it is in general not true that you can differentiate term by term ( that your last statement holds ).

My analysis isn't really that strong ( or even good ).

I believe that the argument needed, that you can differentiate term by term ( btw. you probably should be using partial derivative ), is that the series $\displaystyle f'_k(x)$ converges uniformly.

Maybe writing out the limits makes it easier to provide the proper arguments along the way, similar to your original thoughts,

$\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h \to 0} \frac{\sum_{k=1}^{\infty} \frac{sin(k\cdot(x+h))}{k^p} - \sum_{k=1}^{\infty} \frac{sin(kx)}{k^p}}{h} $

Because of uniform convergence the limit can be taken under the summation sign and you can continue from there.

I don't know if that is what you are looking for - and as noted my analysis is not really that strong..

I just spotted ( after you added the upper limit on $\displaystyle p$ ) that taking the derivative term by term would yield a series that is not convergent, and got myself messed up into something that I probably shouldn't have :-)

Hope it helps anyway :-)

Edit:

You can take a look here:

http://www.math.ku.edu/~lerner/m500f...onvergence.pdf

On page 7 is described what you need ..

Re: function differentiable in no point

Hi Alteraus,

I assume you know you're trying to prove the given function is a Weierstrass function. The only proof that I'm familiar with is his original function. This itself is pretty involved. I tried briefly to solve your problem, but with no success. If you Google Weierstrass function, you'll get lots of references. I didn't find one which is exactly your problem, but Eric Weisstein's World of Mathematics has quite a bibliography about such functions. Not much help, I know.

Re: function differentiable in no point

Yes, I realize that although each term $\displaystyle f'_k(x)$ is continuous the sum does not converge uniformly ( unless $\displaystyle p>2$ ) .. I got in over my head there :-)

Re: function differentiable in no point

If you google Riemann function nowhere differentiable ( or permutations of those words ) you will find lots of articles mentioning the function

$\displaystyle R(x) = \sum\limits_{k=1}^{\infty}\frac{1}{k^2} \sin{k^2x}$

Which does have derivatives at isolated points.

That's the most similar function I have been able to find.

But it does suggest that there may be real $\displaystyle x$ where the function you mention is differentiable.

Good luck - and please do keep up updated, I am curious to learn something here!

Re: function differentiable in no point

Thanks for the interest to both of you guys ...

I found the Weierstrass function and corresponding proof before,

then I found more examples of such a function and I started thinking about the differences among them and the given function in my problem,

and from there I got an idea that there's something wrong in the given problem ..

This function probably has derivatives at least in some points because there is no chance that it could oscillate as extremely as the pathological functions I mentioned ..

If my deductions are right, there is at least derivative at points of the set $\displaystyle \{(2n+1)\pi\,,\,n\in\mathbb{Z}\}$

and there holds

$\displaystyle f'((2n+1)\pi)=(2^{2-p}-1)\zeta (p)$

I'll probably ask the lecturer if the given problem is right

Re: function differentiable in no point

oh, the relation I mentioned should be

$\displaystyle f'((2n+1)\pi)=(2^{2-p}-1)\zeta (p-1)$

I made a mistake in typing

Re: function differentiable in no point

so that's true, it was messed up and the given function belongs to another problem,

not this one ..