This is a theorem of Bernstein, found, for example, in Katznelson's Introduction to Harmonic Analysis.
A simple-minded approach would be to shift the variable in $$c_n=\frac{1}{2\pi}\int_0^{2\pi} f(x)e^{-inx}\,dx\tag1$$ to $y=x+\pi/n$, thus obtaining
$$c_n=\frac{1}{2\pi}\int_{0}^{2\pi} f_{\pi/n}(y)e^{-iny+\pi i}\,dy\tag2$$
where I write $f_h(x)=f(x-h)$. Then add (1) and (2):
$$2c_n=\frac{1}{2\pi}\int_0^{2\pi} (f(x)-f_{\pi/n}(x))e^{-inx}\,dx\tag3$$
and weep in despair: the estimate is only $|c_n|=O(n^{-\alpha})$...
The thing is, estimating $|c_n|$ by maximum of integrand in (3) and then summing the estimates is just too crude. Parseval's identity is more precise.
So, let's try Parseval. Since the $n$th Fourier coefficient of $f_h-f$ is $(e^{-inh}-1)\widehat{f}(n)$, we have
$$ \sum_n |e^{-inh}-1|^2|\widehat{f}(n)|^2 \le Ch^{2\alpha} \tag4$$
The inequality (4) isn't worth much unless we can bound $ |e^{-inh}-1|$ from below. There is no uniform bound for all $n$, but if we focus on some dyadic range $2^k\le |n|< 2^{k+1}$, then choosing $h=\frac{2\pi}{3}\cdot 2^{-k}$ gives good result: $\frac{2\pi}{3}\le |nh|< \frac{4\pi}{3}$, which keeps $e^{-inh}$ far away from $1$.
So,
$$ \sum_{n=2^k}^{2^{k+1}-1} |\widehat{f}(n)|^2 \le \widetilde{C} 2^{-2k\alpha} \tag5$$
and the rest is a mopping-up operation: Cauchy-Schwarz turns (5) into
$$ \sum_{n=2^k}^{2^{k+1}-1} |\widehat{f}(n)| \le 2^{k/2}\cdot\widetilde{C} 2^{-k\alpha} \tag6$$
which sums up.
Observations:
- $f$ is odd
- $f$ is continuous on $\mathbb R$, because the series converges uniformly. Together with 1, this implies $f(\pi)=0$.
- $f$ is a cubic polynomial on $(-\pi,\pi)$, because the Fourier coefficients of $x^k$ involve $1/n^k$ (integration by parts happens $k$ times).
Odd cubic polynomials vanishing at $\pi$ are of the form $A(x^3-\pi^2 x)$. There are various ways to find $A$, including boring integration. A less boring way is to observe that
$$ f'(x)= \sum_{n=1}^\infty \frac{(-1)^{n}}{n^2}\cos nx$$
is continuous on $\mathbb R$ and
$$f'(\pi)= \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi ^2}{6}$$
Since $A(x^3-\pi^2 x)'=A(3x^2-\pi^2)$ evaluates to $2\pi^2A$ at $x=\pi$, we have $A=1/12$.
Best Answer
The lacunary series is kind of self-similar. Namely,
$$ f(x) = e^{ix} + \sum_{k=1}^\infty 2^{-k\alpha}e^{i2^{k }x} = e^{ix} + \sum_{k=0}^\infty 2^{-(k+1)\alpha}e^{i2^{k+1 }x} \\ =e^{ix} + 2^{-\alpha} f(2x) \tag{1} $$ Take distinct $x,y\in \mathbb R$. If $|x-y|\ge 1$, we have the Hölder bound simply because $f$ is bounded. Suppose $|x-y|<1$. Apply (1) to get $$ |f(x)-f(y)| \le |x-y| + 2^{-\alpha} | f(2x) -f(2y)| \tag{2}$$ Iterate this $n$ times, where $n$ is the smallest integer such that $2^n|x-y|\ge 1$. The result is $$ |f(x)-f(y)| \le |x-y|\sum_{k=0}^{n-1} 2^{(1-\alpha) k} + 2^{-\alpha n} | f(2^n x) -f(2^n y)| \tag{3} $$ The geometric sum $ \sum_{k=0}^{n-1} 2^{(1-\alpha) k}$ is dominated by its largest term. The difference $| f(2^n x) -f(2^n y)|$ is bounded because $f$ is. Thus, $$ |f(x)-f(y)| \le C ( |x-y|\, 2^{(1-\alpha) n} + 2^{-\alpha n} ) \tag{4} $$ It remains to note that $2^{-n}$ is comparable to $|x-y|$, and the desired bound $$ |f(x)-f(y)| \le C |x-y|^\alpha \tag{5} $$ follows.