 Solution

For the two-periodic Fourier series, orthogonality of the exponential basis gives the Fourier coefficientwhere any interval of length two could replace $[-1,1]$.

Let $C_{nm}={\hat{c}}_{n-m}$ for $-d\le n,m\le d$. Since $c$ is real,so $C_{mn}=\overline{C_{nm}}$ and $C$ is a Hermitian matrix. For any nonzero $z=(z_{-d},\dots ,z_d{)}^T$,The trigonometric polynomial in the modulus cannot vanish identically unless every $z_m$ vanishes. Since $c(x)>0$, the integral is strictly positive. Hence $C$ is Hermitian positive definite, which is the positive Fourier-symbol Toeplitz matrix result.