Hölder Condition for Fourier Series

Question

So I'm trying to prove that the function (as represented by a Fourier series) $ f(x) = \sum_{k=0}^\infty 2^{-k\alpha}e^{i2^kx}$ satisfies the Hölder Condition: $|f(x+h)-f(x)| \le C|h|^\alpha$, with $0 < \alpha < 1$. I've tried finding the function $f$ as well as using the series directly and have gotten nowhere...any help would be appreciated.

Answer 1

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.