1
$\begingroup$

I have a homework problem which consists of two parts, the first of which I have been staring at for several days with very little (constructive) progress.

I need to show that the function $$f(t) = \sum_{n=1}^{\infty}\frac{\cos(3^{n}t)}{3^{n\alpha}}\in\Lambda^{\alpha}$$ when $0 < \alpha \leq 1$. The second part is to show that if $\alpha < \beta < 1$, then $f\notin\Lambda^{\beta}$, but I'll worry about the second part later.

I tried considering $ \begin{eqnarray*} |f(t+h) - f(t)| &=& \left|\sum_{n=1}^{\infty}\frac{\cos(3^{n}(t + h))}{3^{n\alpha}} - \sum_{n=1}^{\infty}\frac{\cos(3^{n}t)}{3^{n\alpha}}\right|\\ &=& \left|\sum_{n=1}^{\infty}\frac{\cos(3^{n}t + 3^{n}h)}{3^{n\alpha}} - \sum_{n=1}^{\infty}\frac{\cos(3^{n}t)}{3^{n\alpha}}\right|\\ &=& \left|\sum_{n=1}^{\infty}\frac{\cos(3^{n}t)\cos(3^{n}h) - \sin(3^{n}t)\sin(3^{n}h)}{3^{n\alpha}} - \sum_{n=1}^{\infty}\frac{\cos(3^{n}t)}{3^{n\alpha}}\right|\\ &=& \left|\sum_{n=1}^{\infty}\frac{\cos(3^{n}t)\cos(3^{n}h) - \sin(3^{n}t)\sin(3^{n}h)}{3^{n\alpha}} - \sum_{n=1}^{\infty}\frac{\cos(3^{n}t)}{3^{n\alpha}}\right|\\ &=& \left|\sum_{n=1}^{\infty}\frac{\cos(3^{n}t)\cos(3^{n}h) - \sin(3^{n}t)\sin(3^{n}h) - \cos(3^{n}t)}{3^{n\alpha}}\right|\\ &\leq& \sum_{n=1}^{\infty}\left|\frac{\cos(3^{n}t)\cos(3^{n}h) - \sin(3^{n}t)\sin(3^{n}h) - \cos(3^{n}t)}{3^{n\alpha}}\right|\\ \end{eqnarray*} $

EDIT: Removed the last half - dozen lines which turned out to be completely non-constructive.

Now I'm not sure if I'm even remotely close to going down the right path, but if I could get this manipulated into something of the form $C^{\alpha}$ I'd be done. But I just can't seem to go any further. Any suggestions?

  • 0
    Since your geometric series starts at $n=1$, the sum is$$\frac{\left(\frac{1}{3}\right)^\alpha}{1-\left(\frac{1}{3}\right)^\alpha}$$ but you've still lost your $h$.2011-10-05
  • 0
    Right. In fact the $h$ was lost a long time ago. So taking the $sin/cos$ functions and bounding them by $1$ is going too far I guess.2011-10-05

1 Answers 1

4

Use the Mean Value Theorem on the terms with $n

To be more precise, let $N=\lfloor\log_3(\frac{1}{h})\rfloor$. Then $|h|3^{N(1-\alpha)}\le|h|^\alpha$ and $\frac{1}{3^{N\alpha}}\le3|h|^\alpha$. Thus, $$ |f(t+h)-f(t)|\le\left(\frac{1}{3^{1-\alpha}-1}+\frac{6}{1-\frac{1}{3^{\alpha}}}\right)|h|^\alpha\tag{2} $$ Note that the $\Lambda_\alpha$-norm in $(2)$ blows up near $\alpha=0$ and $\alpha=1$.

  • 2
    Very nice! [For the sake of completeness...](http://en.wikipedia.org/wiki/Weierstrass_function)2011-10-05
  • 0
    I think I see where you're going - thank you for this. Can you explain what you mean by $\sim$?2011-10-05
  • 0
    Instead of $\frac{3^{N(1 - \alpha)} - 3^{1-\alpha}}{3^{1-\alpha} - 1}$, I get $\frac{3^{N(1-\alpha)} - 1}{3^{1-\alpha} - 1}$.2011-10-06
  • 1
    @Kyle: are you summing from $n=0$ or $n=1$?2011-10-07
  • 0
    Ahh.. I was doing $0$; but should have be summing from $1$. Thank you.2011-10-07
  • 0
    OK I FINALLY have worked through the details. I will mention on my assignment that this was not my idea. Thanks again. I never would have figured this out.2011-10-07
  • 0
    @robjohn: How do you deal with the $3^{1-\alpha}$ term in the first fraction?2011-10-07
  • 1
    @Kyle: if something is less than $\displaystyle\frac{3^{N(1-\alpha)}-3^{1-\alpha}}{3^{1-\alpha}-1}$ it is certainly less than $\displaystyle\frac{3^{N(1-\alpha)}}{3^{1-\alpha}-1}$, so there is really nothing with which to deal.2011-10-07
  • 0
    Well when you put it like that, how can I disagree? :P I guess I'm just so used to working with things between $|\cdot|$ signs that I imagine them there when they aren't. Thanks again.2011-10-07