1
$\begingroup$

I've been working on an exercise and have gotten stuck:

Suppose that $T(x)=\sum_{n=0}^N a_n\cos(2\pi nx)+b_n\sin(2\pi nx)$ is non-negative on $[0,1]$. Show that there exist $c_0,...,c_N\in \mathbb{C}$ such that $$ T(x)=\left|\sum_{n=0}^{N}c_n \exp(2\pi inx) \right|^2. $$

My strategy was to expand the argument using $|z|^2=z\bar{z}$. Equating coefficients and working from the largest value $N$ first, I obtained $$ a_N \cos(2N\pi x)+b_N\sin(2N\pi x)=c_N\bar{c_0}\exp(2N\pi ix)+\overline{c_N\bar{c_0}}\exp(-2N\pi ix) $$ If we call $c_N\bar{c_0}=u-iv$, for instance, then we get equality if $u=a_N/2$ and $v=b_N/2$. Thus, upon choosing $c_0$, for instance, one can solve for $c_N$ to get these terms to match up. Now, looking at the $N-1$ term, we get $$ a_N \cos(2N\pi x)+b_N\sin(2N\pi x)=(c_N\bar{c_1}+c_{N-1}\bar{c_0})\exp(2N\pi ix)+\overline{(c_N\bar{c_1}+c_{N-1}\bar{c_0})}\exp(-2N\pi ix) $$ Again, calling the coefficient $c_N\bar{c_1}+c_{N-1}\bar{c_0}=u-iv$, we get equality if $u=a_{N-1}/2,v=b_{N-1}/2$. Since $c_N$ and $c_{N-1}$, are known, this comes down to solving $$ c_N\bar{c_1}+c_{N-1}\bar{c_0}=a_{N-1}/2-ib_{N-1}/2 $$ Equating real and imaginary parts, this gives us an inhomogeneous system of 2 linear equations to solve. This is where I'm stuck; I haven't used the (clearly necessary) hypothesis that $T\geq 0$, and I'm assuming it will somehow imply that the system has a solution (in which case we could iterate this method and get systems of 2 equations for smaller values of $n$ as well), but I can't see how. I also suspect that the freedom to choose $c_0$ might be important. I'd appreciate any advice on if this method seems viable, or if there is an easier way to do this exercise.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.