Assume $\{a_k\}_{k\ge1}$ is a real sequence such that $u(x) = \sum_{k\ge 1}a_k\sin(kx)$ is a smooth function, and for every $x \in [\pi, \pi]$ $$\left(\sum_{k\ge 1}\frac{a_k}{k}\sin(kx)\right)\left(\sum_{k\ge1}a_k\cdot k\cos(kx)\right) = \left(\sum_{k\ge1}a_k\sin(kx)\right)\left(\sum_{k\ge1}a_k\cos(kx)\right).$$ What can we say about $\{a_k\}$? Can we show that there exists at most one term of $a_k$ that is nonzero? Note that if we set $w(x) = \sum_{k\ge 1}\frac{a_k}{k}\sin(kx)$, we have the relation $u_xw = uw_x$ and $w_x = Hu$. Here $H$ denotes the standard Hilbert transform.

6$\begingroup$ Yes, this says that $fg'=gf'$ (defining $f,g$ as the two series with the sines), so $(f/g)'=0$ and thus also $a_k/k=ca_k$, so $a_k=0$ for $k>1$. (There are probably other ways of doing it too.) $\endgroup$ – Christian Remling Feb 13 at 4:32

1$\begingroup$ Thanks for your comment, Christian. This is the first intuition to this question, but the problem is that $g$ may have strange zero sets even the function is smooth. The fraction $f/g$ is not always defined. So we cannot use this to prove the argument rigorously. $\endgroup$ – Cohen Lu Feb 13 at 4:56

2$\begingroup$ I think we need to be careful about whether $f/g$ is defined. For example, let $f = x\exp(\frac{1}{x^2})$. Define the function $g$ as $g = f$when $x\ge 0$, $g = f$ when $x < 0$. One can check that both $f$ and $g$ are smooth and $f'g = fg'$. But $f/g$ is not a constant in this situation. The problem is that $x = 0$ is a zero point of infinite order. Thus if $g$ has weird zero set, $f/g$ may have strange behaviour. $\endgroup$ – Cohen Lu Feb 13 at 6:30

1$\begingroup$ Yes, you are right, if both $f$ and $g$ are analytic, then everything should be ok. The question is what happens if we only assume smoothness of $f$ and $g$. I think I should modify the question to make it more clear. $\endgroup$ – Cohen Lu Feb 14 at 3:18

2$\begingroup$ This ($f,g$ smooth, but not more) seems an interesting question. I guess (assuming the statement is true) one would have to use that $f,g$ are not just any two functions, but $Hf=g'$, with $H$ denoting the Hilbert transform . I have no clear idea how to proceed though $\endgroup$ – Christian Remling Feb 14 at 18:54
In order to perform the division which is needed to prove
$f_{tt} Hf=f_tHf_t \Rightarrow f_t=\lambda Hf$ one can use first Plemelij and write as usual $f=F^+F^$. Extend $f$ as a $0$homogeneous function to
$\Bbb C^\ast$. The function
$F^\varepsilon (z)=F^+(\frac{z}{1+\varepsilon})F^(\frac{z}{1\varepsilon})$
is holomorphic in the anulus
$L_\varepsilon=\{1\varepsilon<z<1+\varepsilon\}$ and
$F^\varepsilonf\leq C \varepsilon$ in $L_\varepsilon$. Angular derivation $D_t$ and Hilbert transform $H$ commute with the operator $f\mapsto F^\varepsilon$ and are continuous in the natural topologies. We have
$F^\varepsilon_{tt}\cdot HF^\varepsilonF^\varepsilon_t\cdot HF^\varepsilon_t\leq C\varepsilon$. Topologize meromorphic functions in $L_\varepsilon$ as maps to the Riemann sphere. We have
$D_t(\frac{F^\varepsilon_t}{HF^\varepsilon})\leq C \varepsilon$. This gives
$F^\varepsilon_t\lambda HF^\varepsilon =O(\varepsilon)$.