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.

9$\begingroup$ I doubt there will be such a condition in terms of smoothness. Given any smooth function $f$ and a positive integer $n$, the function $f \hat{f}(n) cos(nx)$ has the same Fourier coefficients as $f$, besides the $n$th, which would be 0. $\endgroup$– ItayDec 24, 2020 at 19:11

$\begingroup$ Hi @Itay, I'm not necessarily looking for a smoothness condition, but just another general condition for this to hold, e.g. a sufficient decay, has to satisfy condition (X) etc. type of condition. The smoothness assumption was put in place to simplify matters. $\endgroup$– spacemanDec 24, 2020 at 19:40

4$\begingroup$ @Itay but it's higher derivatives are quite large in a sense. $\endgroup$– Fedor PetrovDec 24, 2020 at 19:41

$\begingroup$ Are you only interested in even functions $f$, i.e. those satisfying $f(x)=f(2\pi x)$? If not then your definition of "Fourier coefficient" might not be appropriate $\endgroup$– Yemon ChoiJan 6, 2021 at 11:55

$\begingroup$ Can I just point out that this definition of "Fourier coefficient" looks a bit strange for $f(x)=\sin(x)$, which last time I looked was a perfectly reasonable $2\pi$periodic smooth function... $\endgroup$– Yemon ChoiJan 6, 2021 at 18:15
2 Answers
It suffices that $f$ be (the restriction to $[0,2\pi]$ of) a completely monotone realvalued 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 $n$ $$\hat f(n)=\int_0^\infty\mu(da) \int_0^{2\pi}dx\,\cos(nx)e^{a x} =\int_0^\infty\mu(da) \frac{a \left(1e^{2 \pi a}\right)}{a^2+n^2},$$ which is obviously decreasing in $n$ (to $0$, by dominated convergence or by the RiemannLebesgue 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 $$[0,2\pi]\ni x\mapsto f^(x):=f(2\pix)$$  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 $$[0,2\pi]\ni x\mapsto a+bx+f(x)+g(2\pix)$$ also has it for any real $a$ and $b$.
Added:
As noted in a comment by Fedor Petrov, if $f(x)=h(\pix)$ for some odd function $h$ and all $x\in[0,2\pi]$, then $\hat f(n)=0$ for all natural $n$.
It follows from this answer by fedja that, if $$f(x)=\int_1^\infty[\mu(dp) x^p+\nu(dp)(2\pix)^p]<\infty$$ for some measures $\mu$ and $\nu$ on $[1,\infty)$ and all $x\in[0,2\pi]$, then $f$ has the desired property.

$\begingroup$ Why does $bx+f(x)$ also have this property? $\endgroup$ Dec 25, 2020 at 14:49

$\begingroup$ @FedorPetrov : I have added a detail on this. $\endgroup$ Dec 25, 2020 at 15:00

$\begingroup$ ah, I see, actually any function of the form $h(x\pi)$, where $h$ is odd, works $\endgroup$ Dec 25, 2020 at 20:39

$\begingroup$ @IosifPinelis Thank you for a wonderful answer, I'm very happy that this led to some interesting discussions and even an extended question. $\endgroup$– spacemanDec 28, 2020 at 16:47
A comment on this problem would be that if $\hat f(n)$ are monotone ( here $f$ any continuous function, not necessarily odd or even, also I assume that $\hat f(n)$ is monotone not $\hat f(n)$ ) then one can assume that they are positive. And if Fourier coefficients are real and positive then they must be absolutely convergent, that is $\{\hat f(n)\} \in l_1$. This follows easily from property of Fejer's kernel, i.e. that it is positive operator with integral 1:
$$ \sum_k (1k/n)\hat f(k)exp(ikt) = \int F(ts)f(s) \le \supf$$ so $1/2 \sum_{k \in (n/2, n/2)} \hat f(k)) \le \supf$.