# A question on trig series

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.

• 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.) – Christian Remling Feb 13 '20 at 4:32
• 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. – Jacob Lu Feb 13 '20 at 4:56
• 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. – Jacob Lu Feb 13 '20 at 6:30
• 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. – Jacob Lu Feb 14 '20 at 3:18
• 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 – Christian Remling Feb 14 '20 at 18:54