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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?

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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 $h$. –  robjohn Oct 5 '11 at 2:26
    
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. –  Kyle Schlitt Oct 5 '11 at 2:30

1 Answer 1

up vote 4 down vote accepted

Use the Mean Value Theorem on the terms with $n<N$: $$ \begin{align} |f(t+h)-f(t)|&\le\sum_{n=1}^\infty\left|\frac{\cos(3^n(t+h))-\cos(3^nt)}{3^{n\alpha}}\right|\\ &=\sum_{n=1}^{N-1}\left|\frac{3^nh\sin(3^n(t+\eta_n))}{3^{n\alpha}}\right|+\sum_{n=N}^\infty\left|\frac{\cos(3^n(t+h))-\cos(3^nt)}{3^{n\alpha}}\right|\\ &\le|h|\frac{3^{N(1-\alpha)}-3^{1-\alpha}}{3^{1-\alpha}-1}+2\frac{\frac{1}{3^{N\alpha}}}{1-\frac{1}{3^{\alpha}}}\tag{1} \end{align} $$ Choose $N$ so that $|h|\sim3^{-N}$. Then $|h|3^{N(1-\alpha)}\sim|h|^\alpha$ and $\frac{1}{3^{N\alpha}}\sim|h|^\alpha$. Thus, the right side of $(1)\sim|h|^\alpha$.

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 $\alpha=1$.

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Very nice! For the sake of completeness... –  t.b. Oct 5 '11 at 3:00
    
I think I see where you're going - thank you for this. Can you explain what you mean by $\sim$? –  Kyle Schlitt Oct 5 '11 at 3:16
    
Instead of $\frac{3^{N(1 - \alpha)} - 3^{1-\alpha}}{3^{1-\alpha} - 1}$, I get $\frac{3^{N(1-\alpha)} - 1}{3^{1-\alpha} - 1}$. –  Kyle Schlitt Oct 6 '11 at 22:41
1  
@Kyle: are you summing from $n=0$ or $n=1$? –  robjohn Oct 7 '11 at 0:13
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. –  robjohn Oct 7 '11 at 1:45

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