Prove Holder-continuity for $\mu_\lambda (x) = \sum\limits_{n=1}^\infty \frac{ \cos(2^n x)}{2^{n \lambda} }$

Assume

$$2^{-k-1}\le |x-y|\le 2^{-k},  \text{for some}   k \in \mathbb{Z}.$$

Then by mean value theorem, there exists $c$ between $x$ and $y$ such that 
$$|\cos(2^nx)-\cos(2^ny)|=|\sin (c)(2^nx-2^ny)| \le 2^{n}|x-y|\le 2^{n-k}.$$
For $0<\alpha<\lambda =1$,  we have
$$|f(x)-f(y)|\le \sum_{n=1}^k\frac{|\cos(2^nx)-\cos(2^ny)|}{2^{n\lambda}}+\sum_{n=k+1}^\infty\frac{ |\cos(2^nx)-\cos(2^ny)|}{2^{n\lambda}}$$$$\le \sum_{n=1}^k2^{n-k-\lambda n}+2\sum_{n=k+1}^\infty 2^{-\lambda n}$$ $$ \le \sum_{n=1}^k2^{n-k-\alpha n}+2\sum_{n=k+1}^\infty 2^{-\alpha n}$$ $$ \le {2^{-k}2^{(1-\alpha)(k+1)}\over 2^{1-\alpha}-1}+2\cdot {2^{-\alpha(k+1)}\over 1-2^{-\alpha}}$$ $$ ={2\over 2^{1-\alpha}-1}2^{-\alpha(k+1)}+{2\over 1-2^{-\lambda}}2^{-\alpha(k+1)}$$ $$ =({2\over 2^{1-\alpha}-1}+{2\over 1-2^{-\alpha}})2^{-\alpha(k+1)}$$ $$ \le c_\alpha|x-y|^\alpha.$$

Thus $f$ is Holder continuous for any $0<\alpha <\lambda \leq 1.$