= Solution
For the two-periodic <Fourier series>, orthogonality of the exponential basis gives the <Fourier coefficient>
$$
\boxed{\widehat c_n=\frac12\int_{-1}^{1}c(x)e^{-i\pi nx}\,dx},
$$
where any interval of length two could replace $[-1,1]$.
Let $C_{nm}=\widehat c_{n-m}$ for $-d\leq n,m\leq d$. Since $c$ is real,
$$
\widehat c_{-k}=\overline{\widehat c_k},
$$
so $C_{mn}=\overline{C_{nm}}$ and $C$ is a <Hermitian positive-definite matrix>[Hermitian matrix]. For any nonzero $z=(z_{-d},\ldots,z_d)^T$,
$$
\begin{aligned}
z^*Cz
&=\sum_{n,m=-d}^d\overline{z_n}\widehat c_{n-m}z_m\\
&=\frac12\int_{-1}^{1}c(x)
\left|\sum_{m=-d}^dz_me^{i\pi mx}\right|^2dx.
\end{aligned}
$$
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 matrix>[Hermitian positive definite], which is the <positive Fourier-symbol Toeplitz matrix> result.
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