TITLE: When are Fourier coefficients monotonic?
QUESTION [19 upvotes]: 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.

REPLY [3 votes]: 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:
$$ \sum_k (1-|k|/n)\hat f(k)exp(ikt) = \int F(t-s)f(s) \le \sup|f|$$
so
$1/2 \sum_{k \in (-n/2, n/2)} \hat f(k))  \le \sup|f|$.