Let $f : [0,2\pi] \to \mathbb{R}$ be some function. Then the discrete Fourier transform of $f$ when sampled at $2\pi i/N$ is then given by $$ X_n := \sum_{i=0}^{N-1}\cos\left(\frac{2\pi n i}{N}\right)f\left(\frac{2\pi i}{N}\right), \quad n = 1,\ldots,N-1. $$

Question: What conditions are necessary on $f$ such that there exists a convex function $G : [0,1] \to \mathbb{R}$ with $G(n/N) = X_n$? That is, when is the DFT a discrete convex function with respect to $n$?

• Would it be sufficient for $f$ to be smooth and convex on $(0,2\pi)$?
• In the references post above it was shown that $f(x) = |\pi-x|$ defies this claim.

Notes: This question is related to the recent post: Convexity of discrete Fourier transform

Let $f : [0,2\pi] \to \mathbb{R}$ be some function. Then the discrete Fourier transform of $f$ when sampled at $2\pi i/N$ is then given by $$ X_n := \sum_{i=0}^{N-1}\cos\left(\frac{2\pi n i}{N}\right)f\left(\frac{2\pi i}{N}\right), \quad n = 1,\ldots,N-1. $$

Question: What conditions are sufficient on $f$ such that there exists a convex function $G : [0,1] \to \mathbb{R}$ with $G(n/N) = X_n$? That is, when is the DFT a discrete convex function with respect to $n$?

• In the references post above it was shown that $f(x) = |\pi-x|$ defies this claim.

Notes: This question is related to the recent post: Convexity of discrete Fourier transform