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@RiverLi I do not assume that $x_n=n f(1/n)$, I assume $x_{2n}=nf(1/n)$ and $x:_{2n+1}=n g(1/n)$. My power series exist. If they have positive radius of convergence, and I think this is not very difficult to prove, then my assumptions about $x_{2n}$ and $x_{2n+1}$ are proved and all is justified.
@Jun This remind me of Salem criterium for the Riemann hypothesis in: Salem, R., Sur une proposition equivalente à l'hypothèse de Riemann, C. R. Acad. Sci. Paris vol 236 (1953) pp. 1127--1128.
@GHfromMO RH could be unproved by a complicated argument, for example by a very careful study of the behaviour of $\pi(x)$ something as the elementary proof of the prime number theorem but more complicated. In this case the prize will be well deserved I think.
Nice start, I wrote similar things but I do not recall to use Fourier coefficients here. Jan van de Lune is the real proponent of this problem, he asked me to propose it in MathOverflow. I have sent mails to him about your answer, but his health is not good and I have not received any answer until now. I am sure he will like it, his PhD-thesis is full of trapezoidal sums for $z^x$.
@KungYao Your use of the word boundary is not clear to me. Are you taking the elements in $[0,1]^2$ mod 1? It is your first condition, for example, equivalent to $f(0, x_2)=e^{-\pi i x_2} f(1,x_2)$? This is what I will have called a boundary condition. Or can we use $x_1=1/3$? In this case what is the meaning of $x_1+1$?
@GerryMyerson I was confused. It is true that there is an unproved assertion of Ramanujan at the point I referred but it is unrelated to the $2^x$, $3^x$ question.
@Wojowu In the paper "Highly composite numbers" point 36 in page 114 of Ramanujan's Collected Papers, it is said that the quotient of two consecutive superior highly composite numbers is a prime. This is not proved, but follows it the problem have a positive answer. I think this is not known.
@AymanMoussa Consider the case $d=1$, and let $A$ be the set of function of $\mathcal{C}^\infty$ such that $$\Vert f\Vert=\sum_{n=0}^\infty \frac{\Vert f^{(n)}\Vert_\infty}{n!}<+\infty.$$ It is not this an example of what you want?
Yes, there are Borel sets with projection non Borel. Lebesgue, in one of his paper pretended to proof that the projection of a Borel set is a Borel set. Lusin detected the error. This started the Theory of analytic sets. But the examples, I think, are always difficult.