# When are Fourier coefficients monotonic?

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.

• 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. – Itay Dec 24 '20 at 19:11
• 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. – Spaceman Dec 24 '20 at 19:40
• @Itay but it's higher derivatives are quite large in a sense. – Fedor Petrov Dec 24 '20 at 19:41
• 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 – Yemon Choi Jan 6 at 11:55
• 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... – Yemon Choi Jan 6 at 18:15

It suffices that $$f$$ be (the restriction to $$[0,2\pi]$$ of) a completely monotone 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 $$b$$.

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 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.
• Why does $bx+f(x)$ also have this property? – Fedor Petrov Dec 25 '20 at 14:49
• ah, I see, actually any function of the form $h(x-\pi)$, where $h$ is odd, works – Fedor Petrov Dec 25 '20 at 20:39
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 $$1/2 \sum_{k \in (-n/2, n/2)} \hat f(k)) \le \sup|f|$$.