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Show that for each $\alpha \in (0,1)$ there exists a constant $C_\alpha$ such that $$ |F_\alpha(x)| \leq C_\alpha |x|^\alpha $$ for all $x \in \mathbf{R}$ where $F_\alpha$ is given as $$ F_\alpha(x) = \sum_{n=0}^\infty 2^{\alpha n} \sin(\frac{x}{2^n} ) $$

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  • $\begingroup$ Recall that $|\sin(\theta)|\le|\theta|$. Given $x$, since $2^n>|x|$ if $n$ is large enough, we have that for large $n$, the $n$th term in the series is at most $2^{-(1-\alpha)n}|x|$. This should get you started. $\endgroup$ – Andrés E. Caicedo Apr 27 '13 at 23:00
  • $\begingroup$ To get $|x|^\alpha$ in the inequality is where I am having trouble. For $|x|<1$ it is easy enough because $|x|<|x|^\alpha$. However, for $|x|>1$ that inequality does not work. So I get the inequality you get, but how to get the $\alpha$ power? $\endgroup$ – Eager Student Apr 28 '13 at 1:35
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We divide the argument into two cases, according to whether $|x|\le 1$ or not.

Suppose first that $|x|\le 1$. Since $|\sin(\theta)|\le|\theta|$, we have $$ \sum_n 2^{\alpha n}|\sin(x/2^n)|\le|x|\sum_n2^{-(1-\alpha)n}=|x|K_\alpha, $$ where $K_\alpha$ depends only on $\alpha$. Now, since $|x|\le1$, we have $|x|\le|x|^\alpha$, and we are done.

Suppose now that $|x|>1$. Fix $n_0$ least such that $|x/2^{n_0}|\le1$, so $n_0>0$ and $2^{n_0-1}\le |x|$. For $m\ge0$, we have $$|2^{\alpha (n_0+m)}\sin(x/2^{n_0+m})|\le2^{-(1-\alpha)m}|2^{-(1-\alpha)n_0}x|\le2^{-(1-\alpha)m}(2|x|)^\alpha,$$ using that $|\sin(\theta)|\le|\theta|$, that $|x/2^{n_0}|\le 1$, and that $2^{n_0}\le 2|x|$.

The series $\sum_{m\ge0}2^{-(1-\alpha)m}$ converges to $K_\alpha$ as before.

Now you need to deal with a few terms at the beginning: $$\sum_{n<n_0}2^{\alpha n}|\sin(x/2^n)|\le\sum_{n<n_0}2^{\alpha n}=\frac{2^{\alpha n_0}-1}{2^\alpha -1}<\frac{2^{\alpha n_0}}{2^\alpha-1}\le T_\alpha |x|^\alpha $$ for some constant $T_\alpha$, using again that $2^{n_0}\le 2|x|$.

Putting this together, we get that $|F_\alpha(x)|\le (2^\alpha K_\alpha +T_\alpha)|x|^\alpha$, and we are done also in this case.

Finally, we let $C_\alpha=2^\alpha K_\alpha+T_\alpha$, and note that $C_\alpha\ge K_\alpha$ so, for all $x$, regardless of whether $|x|\le 1$ or not, we have $$ |F_\alpha(x)|\le C_\alpha|x|^{\alpha}, $$ which is precisely what we wanted to prove.

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