# Differentiability of Fourier series

Consider the function defined by the Fourier series

$$f(x;\alpha) = \sum_{n=1}^\infty \frac{1}{n^\alpha} \exp(i n^2 x ) ,$$

where $$\alpha >1$$.

For what values of $$\alpha$$ is $$f$$ differentiable? Based on numerics, it is conjectured that $$\alpha = 2$$ is a critical value. For $$\alpha <2$$, the function is nowhere differentiable; while for $$\alpha >2$$, the function is differentiable almost everywhere.

• The $\alpha$th power of the popcorn function has similar differentiability properties, essentially by Thue–Siegel–Roth. I can imagine your result having something to do with rational approximation (although that's very far from a proof). – LSpice Nov 5 '18 at 2:59

Edit: fedja has pointed out the previous result (first theorem) stated was weaker than the present purposes require. This has now been addressed with the addition of references and the result of Luther below.

The fact that $$f(x)$$ is everywhere non-differentiable for $$f$$ with lacunary Fourier series follows by Theorem 2.1 here with earlier versions known to Freud and Hardy

Theorem [Hardy]: An integrable periodic function $$f$$ with Fourier series $$\sum a_k \sin(n_k x)$$, satisfying $$\frac{n_k+1}{n_k} >\lambda > 1$$ is differentiable at a point only if $$\lim\limits_{k\rightarrow\infty} a_{k}n_k = 0$$.

The particular function that is provided (known as Weirstrass' function) is not lacunary, but many results are known. For $$\alpha=2$$ one has a mix of non-differentiability and differentiability results known, one interesting result being the following of Gerver

Theorem [Gerver, 1970]: $$f(x)=\sum\limits_{n=1}^{\infty} \frac{\sin{n^2 x}}{n^2}$$ is differentiable at points $$k\pi$$ if $$k=\frac{2p+1}{2q+1}$$, $$p,q\in\mathbb{Z}$$.

A literature search shows that the question for varying $$\alpha$$ was considered even earlier by Hardy here (see second paragraph on the fifth page) who showed that this function has no derivative for $$x$$ irrational and $$\alpha<\frac{5}{2}$$. A more recent result of Luther (Theorem 2, here) shows more, namely that the above series is nowhere differentiable for $$1\leq \alpha\leq \frac{3}{2}$$ and that for $$\frac{3}{2}<\alpha<3$$ the function is differentiable at $$x=\frac{p}{q}$$, $$(p,q)=1$$, only for $$pq \equiv\ 1\ (\text{mod }2)$$.

This last result shows that the mix between differentiability and non-differentibility happens for $$\alpha$$ smaller than two ($$\alpha=\frac{3}{2}$$) while for $$f(x;\alpha)$$ to be differentiable it is necessary that $$\alpha \geq 3$$.

• Erm... It is $n^2$, not $2^n$, so $\inf_k \frac{n_{k+1}}{n_k}=1$. – fedja Nov 5 '18 at 15:00
• @fedja Yes, the edits above now reference proper results for the non-differentiability in the region of interest. – Josiah Park Nov 5 '18 at 15:36
• But Gerver's results refers only to the imaginary part of my function – pie Nov 5 '18 at 23:36
• For statements about the cosine and sine series see theorem 4.31 in Hardy's paper ( perso-math.univ-mlv.fr/users/jaffard.stephane/pdf/Hardy.pdf ). The proof explains that Hardy's result applies to the real part as well. – Josiah Park Nov 6 '18 at 0:11

The critical value is $$a= 5/2$$.

Chamizo and Cordoba showed that the fractal dimension of $$f(x,a)$$ is $$2 + \frac{1}{2}(1/2 - a), \ \ \ 1 \leq a \leq \frac52$$.Thus $$f(x,a)$$ is not differentiable for $$a < 5/2$$ (differentiable functions have dimension 1). On the other hand, for $$\alpha > 5/2$$, $$f(x,a)$$ is differentiable almost everywhere. They also include pictures for the $a = 3/2 case. The point is that $$\sum_{N \leq n \leq 2N} e^{itn^2}$$ typically has size around $$\sqrt{N}$$ (for instance Weyl-differencing) and so at scale $$N$$ on the Fourier side, we see a decay of about $$N^{1/2 - a}$$. For differentiability, we want decay of $$N^{-2}$$ and so the critical value is $$5/2$$. See Jaffard's "The spectrum of singularities of Riemann's function" for precise results on the case $$\alpha = 2$$. He shows the set of points for which $$f(x,2)$$ is differentiable has Hausdorff dimension zero and gives a very precise formulation of the regularity at each point in terms of the continued fraction expansion of $$x$$. • "On the other hand, for α>5/2, it is easy to see that f(x,a) is differentiable". This is incompatible with the result of Luther (for$\frac{3}{2}<\alpha<3$, the only rational values for which the (sin) series is differentiable are of the form$\frac{p}{q}$,$(p,q)=1$with$pq \equiv 1\ (\text{mod } 2)\$). – Josiah Park Nov 6 '18 at 3:32
• Fair enough, I had in mind almost everywhere. I edited accordingly – George Shakan Nov 6 '18 at 3:38