Given some sufficiently smooth function $f$ what conditions would be sufficient for its Fourier coefficients, as defined by
$$
\hat{f}(n) := \int\_{0}^{2\pi}\cos(nx)f(x)\ dx, \quad \text{for } n = 1,2,\ldots,
$$
to be monotonic? Given the decay properties of Fourier coefficients, the monotonicity result would translate to
$$
|\hat{f}(n)| \geq |\hat{f}(n+1)|, \quad n = 1,2,\ldots.
$$
I haven't been able to find any literature regarding this and a result of this nature would be very interesting.



When are Fourier coefficients monotonic?
It suffices that $f$ be (the restriction to $[0,2\pi]$ of) a [completely monotone](https://en.wikipedia.org/wiki/Bernstein%27s_theorem_on_monotone_functions) real-valued function defined on $[0,\infty)$. Indeed, then for some finite measure $\mu$ on $[0,\infty)$ and all real $x\ge0$ we have
$$f(x)=\int\_0^\infty\mu(da) e^{-a x},$$
whence for natural $n$
$$\hat f(n)=\int\_0^\infty\mu(da) \int\_0^{2\pi}dx\,\cos(nx)e^{-a x}
=\int\_0^\infty\mu(da) \frac{a \left(1-e^{-2 \pi a}\right)}{a^2+n^2},$$
which is obviously decreasing in $n$ (to $0$, by dominated convergence or by the Riemann--Lebesgue lemma).


Note that, if $f(x)\equiv1$ or $f(x)\equiv x$, then $\hat f(n)=0$ for all natural $n$. So, if $f$ has the desired property, then the function $[0,2\pi]\ni x\mapsto a+bx+f(x)$ also has it for any real $a$ and $b$. Also, clearly, if $f$ has the desired property, then do does the function
$$[0,2\pi]\ni x\mapsto f^-(x):=f(2\pi-x)$$
-- because $\widehat{f^-}(n)=\hat f(n)$ for all natural $n$. It follows that, if $f$ and $g$ have the desired property, then the function
$$[0,2\pi]\ni x\mapsto a+bx+f(x)+g(2\pi-x)$$
also has it for any real $a$ and $b$.


**Added:**


1. As noted in a comment by Fedor Petrov, if $f(x)=h(\pi-x)$ for some odd function $h$ and all $x\in[0,2\pi]$, then $\hat f(n)=0$ for all natural $n$.
2. It follows from [this answer by fedja](https://mathoverflow.net/a/379776/36721) that, if
$$f(x)=\int\_1^\infty[\mu(dp) x^p+\nu(dp)(2\pi-x)^p]<\infty$$
for some measures $\mu$ and $\nu$ on $[1,\infty)$ and all $x\in[0,2\pi]$, then $f$ has the desired property.


