Vtyjkyuk

This question arises as a follow-up to MSE2874264.
Let us assume to have $fin L^2(-pi,pi)$ defined by
$$ f(x) = sum_ngeq 1 c_n cos(nx) $$
where the coefficients $c_n$ are non-negative, decreasing to zero and such that $c_n=Oleft(frac1nright)$.

Is it true that for any $ain(1,+infty)$ the quantity
$ c_a stackreltextdef=frac1piint_-pi^pif(x)cos(ax),dx $ belongs to the interval $left[c_lfloor arfloor+1,c_lfloor a rfloorright]$? In such a case, what conditions on $f$ ensure that $c_a$ is a decreasing function?

We may easily check that
$$ c_a = sum_ngeq 1frac2a(-1)^n sin(pi a)pi(a^2-n^2)c_n $$
so if $ainmathbbN^+$ we have that $c_a$ agrees with the coefficient $c_lfloor arfloor$ from the Fourier cosine series of $f$.
If that is not the case, an efficient strategy for bounding the previous series seems to invoke the Poisson summation formula and the fact that the Fourier transform of the $operatornamesinc$ is well-known. Besides that I have not been able to produce sharp inequalities for $c_a$ in terms of $c_lfloor arfloor$ and $c_lfloor arfloor+1$ only.

The answer relies in Gibbs phenomenon. As pointed out in the comments below the main question it is not granted that if $ain(n,n+1)$ then $c_ain(c_n+1,c_n)$, but due to the following lemma, the given constraints ensure that $c_a$ cannot lie too far from such interval:

For any $ngeq 4$ and any $xin[0,n]$ we have that
$$ f_n(x)stackreltextdef=sum_k=0^ntextsinc(pi(x-k)) in left[frac2pitextSi(2pi),frac2pitextSi(pi)right]subset[0.902,1.179]$$
where $textsinc(x)=left{beginarraycl1&textif x=0\ fracsin xx&textotherwiseendarrayright.$ and $textSi(x)=int_0^xtextsinc(t),dt.$