When are Fourier cosine coefficients convex? In the question When are Fourier coefficients monotonic it was determined that, if a function $f$ is (the restriction to $[0,2\pi]$) of a completely monotone function, then its Fourier coefficients, defined as
$$
\hat{f}(n) := \int_{0}^{2\pi}f(x)\cos(nx) dx, \quad n = 1,2,\ldots,
$$
are monotonically decreasing (decreasing due to the decay of Fourier coefficients).
An interesting extension to this result would be, what further conditions on $f$ would be sufficient for $\hat{f}$ to be convex, i.e. in the sense that
$$
\hat{f}(n+1) + \hat{f}(n-1) - 2\hat{f}(n) \geq 0 \quad \text{for } n \geq 1.
$$
Note: Such a function does exist, since it is known that if there is a convex sequence of numbers $(a_n)_{n\in\mathbb{N}}$, in the above sense, that tends to zero, then there exists an $f \in L^1(\mathbb{T})$ with $f \geq 0$ such that $\hat{f}(n) = a_n$. See Lemma 1.12 of Classical and Multilinear Harmonic Analysis by C.Muscalu and W. Schlag, page 16.
 A: Let us describe all the functions $f\in L^2[0,2\pi]$ such that $\hat f$ is convex on the set $\{0,1,\dots\}$. Since, by the Riemann–Lebesgue lemma, $\hat f(n)\to0$ as $n\to\infty$, for all such $f$ we have
$$\hat f(n)=\sum_{m=1}^\infty c_m(1-n/m)_+ \tag{1}$$
for some nonnegative $c_m$'s such that $\sum_{m=1}^\infty c_m<\infty$ and all $n\in\{0,1,\dots\}$, where $u_+:=\max(0,u)$. See details on representation (1) at the end of this answer.
Therefore,

a function $f\in L^2[0,2\pi]$ is such that $\hat f$ is convex on the set $\{0,1,\dots\}$ if and only if
$$f(x)=\sum_{m=1}^\infty c_m f_m(x)+\sum_{j=1}^\infty b_j\sin(jx)\tag{2}$$
for some nonnegative $c_m$'s with $\sum_{m=1}^\infty c_m<\infty$ and some real $b_j$'s with $\sum_j b_j^2<\infty$, where
$$f_m(x):=\frac12+\sum_{j=1}^m(1-j/m)\cos(j x)=\frac{\sin ^2\left(mx/2\right)}{2 m \sin ^2\left(x/2\right)}.$$
The convergence in (2) is in $L^2[0,2\pi]$.


Here are the graphs of the first five "basic" functions $f_1$ (red), $f_2$ (orange), $f_3$ (green), $f_4$ (blue), and $f_5$ (magenta).


Details on representation (1): For $g_n:=\hat f(n)$, let
$$d_n:=g_{n+1}+g_{n-1}-2g_n=(g_{n+1}-g_n)-(g_n-g_{n-1}),$$
so that $d_n\ge0$ for all $n=1,2,\dots$. Then
$$D_k:=\sum_{m=k}^\infty d_m=g_{k-1}-g_k$$
for all $k=1,2,\dots$ and hence
$$g_n=\sum_{k=n+1}^\infty D_k=\sum_{k=n+1}^\infty\sum_{m=k}^\infty d_m
=\sum_{m=n+1}^\infty d_m\sum_{k=n+1}^m 1
=\sum_{m=n+1}^\infty d_m(m-n)
=\sum_{m=1}^\infty d_m(m-n)_+
=\sum_{m=1}^\infty md_m(1-n/m)_+,
$$
so that we have (1) with $c_m=md_m\ge0$. Also, we have $\sum_{m=1}^\infty c_m=\hat f(0)<\infty$. Vice versa, if (1) holds for some nonnegative $c_m$'s such that $\sum_{m=1}^\infty c_m<\infty$ and all $n\in\{0,1,\dots\}$, then $\hat f$ is convex on $\{0,1,\dots\}$, because $(1-n/m)_+$ is convex in $n\in\{0,1,\dots\}$ for each natural $m$.
