Is the function $\sum_{k=1}^{+\infty}\frac{\cos kx}{k^{6/5}}$ differentiable at $x=0$?

Question

I am considering the series $\sum_{k=1}^{+\infty}\frac{\cos kx}{k^{6/5}}$. Since $\frac{|\cos kx|}{k^{6/5}}\leq \frac{1}{k^{6/5}}$ for all $x\in\mathbb{R}$, the function $f(x)=\sum_{k=1}^{+\infty}\frac{\cos kx}{k^{6/5}}$ is well-defined and continuous on $\mathbb{R}$. Now I want to find the differentiability of $f(x)$. Using the formula $$\sum_{k=1}^{n} \sin(kx)=\frac{\sin\frac{nx}{2}\sin\frac{(n+1)x}{2}}{\sin\frac{x}{2}}$$ and Dirichlet test, I shown that the series $\sum_{k=1}^{+\infty}(\frac{\cos kx}{k^{6/5}})'=\sum_{k=1}^{+\infty}\frac{\sin kx}{k^{1/5}}$ uniformly converge on $[0+\epsilon,2\pi-\epsilon]$ for all small $\epsilon>0$. Therefore $f(x)$ is differentiable for all $x\in\mathbb{R}\backslash (2\pi\mathbb{Z})$.

My question: what is the differentiability of $f(x)$ when $x\in 2\pi\mathbb{Z}$?

I guess that $f(x)$ is differentiable at $x=0$. Because if we replace the power $6/5$ by $2$, it can be shown that $f(x)$ is a Bernoulli polynomial. (How to do $\sum_{k=1}^{\infty} \frac{\cos(kx)}{k^2}$?)

Answer 1

Not differentiable. Here's a way to prove it that doesn't rely on computing the sum.

$$\begin{split} f(0)-f(x)&=\sum_{k=0}^{+\infty}\frac{1-\cos(kx)}{k^{\frac 6 5}}\\ &= 2\sum_{k=0}^{+\infty}\frac{\sin^2\left(\frac {kx}2\right)}{k^{\frac 6 5}}\\ &\geq 2\sum_{k=0}^{M(x)}\frac{\sin^2\left(\frac {kx}2\right)}{k^{\frac 6 5}}\\ &\geq \frac{2}{\pi}x \sum_{k=0}^{M(x)}\frac{1}{k^{\frac 1 6}} \end{split} $$

where we have set $M(x)=\lfloor \frac \pi x\rfloor$ and we have used the fact that if $0\leq u\leq\frac \pi 2$, then $\frac 2 \pi u\leq \sin u$.

Thus, if you set $x_n=\frac \pi {2n}$, $$\frac{f(x_n)-f(0)}{x_n}\leq -\frac 2 \pi\cdot\sum_{k=0}^{M(x_n)}\frac{1}{k^{\frac 1 6}}$$ Now the right-hand side explodes to $-\infty$ as $n\rightarrow+\infty$, so the derivative doesn't exist at $x=0$. The slope of the function's graph is vertical.