This paper claims that the function $\sum_{n=1}^\infty \frac{e^{2\pi in^kx}}{n^\alpha}$ is absolutely continuous if $\alpha > k+\frac{1}{2}$. Whilst it makes that claim, it never further shows/elaborates as to why this is the case.
Now I am trying to replicate this result but am failing miserably - I'm specifically interested in the case where $k=2$ and $\alpha=3$ - so showing absolute continuity for those two values would be sufficient for me.
NOTE: There is one thing that I really don't understand here, and it would be absolutely amazing if somebody could clear this up: My understanding is that if a complex function, in this case $\sum_{n=1}^\infty \frac{e^{2\pi in^kx}}{n^\alpha}$ is absolutely continuous, then the real part of this function, in this case $\sum_{n=1}^\infty \frac{\cos(2\pi n^kx)}{n^\alpha}$, should also be absolutely continuous.
For my case where $k=2$ and $\alpha=3$, the term-wise derivative of the real part equals $\sum_{n=1}^\infty \frac{\sin(2\pi n^2x)}{n}$. I previously hoped to show that this derivative is well defined, which led to this question I asked on this platform: Is $\sum_{n=1}^{\infty} \frac{\sin(n^2x)}{n}$ uniformly convergent on $(0,\pi)$?
For this series, metamorphy was able to show that it diverges on the dense subset $(0,\pi)$. Now as I understand a function that is absolutely continuous must have a derivative almost everywhere, meaning that those two statements:
- $\sum_{n=1}^\infty \frac{e^{i2\pi n^2x}}{n^3}$ is absolutely continuous
- $\frac{d}{dx}\left[\Re \left( \sum_{n=1}^\infty \frac{e^{in^2x}}{n^3}\right)\right] $ diverges on the dense subset $(0,\pi)$
Should contradict each other.
I would be more than thankful if someone could help me clear up my misconceptions here/show me how it can be shown that $\sum_{n=1}^\infty \frac{e^{i2\pi n^2x}}{n^3}$ is actually absolutely continuous! Thanks so much in advance already!