# Do flat functions with flat Fourier transform exist?

Call a smooth function flat (in zero) if $f^{(k)}(0) = 0$ for all $k\ge 1$. Does there exist a flat Schwartz function $f\in\mathcal S(\mathbb R)$ such that its Fourier transform $\hat f = \mathcal F(f)$ is flat, too?

The construction of such a function has proven difficult to me. The simplest example of flat Schwartz functions are Bump functions which are constant around zero. However since, by the Paley-Wiener theorem, the Fourier transform of a bump function is real analytic, it cannot be flat as the Taylor series in zero is a constant.

Also note that the important identity

$$\int_{-\infty}^{+\infty} f(x) x^k dx = \mu_k \hat f^{(k)}(0)$$

implies that all but the zeroth moment of the fourier transform of a flat function vanish. Typically the graph of a function whose moments are almost all zero looks similar to a sinc function: oscillating indefinitely and decaying as $|x|\to\infty$.

As a second remark, if such a function exists, it can w.l.o.g. be chosen as an Eigenfunction of the Fourier transform since for arbitrary $f$

$$g = f + \mathcal F(f) + \mathcal F^2(f) + \mathcal F^3(f)$$

is an Eigenfunction of the Fourier transform, and the property of the derivatives being $0$ in the origin carries over to $g$ (since $\mathcal F^2(f)(t) = f(-t)$). This leads to a slight reformulation

Question: Does there exist an (even) Schwartz function $f\in\mathcal S(\mathbb R)$ such that

• $f$ is a density function: $\int_{-\infty}^{+\infty} f(x) dx = 1$

• All higher moments of $f$ vanish: $\int_{-\infty}^{+\infty} f(x)x^k dx = 0$ for all $k\ge 1$

• $f$ is an Eigenfunction of the Fourier transform: $\mathcal F(f) = f$

Any ideas, remarks, or references to relevant literature are greatly appreciated!

EDIT: 7th JUL

I am still interested in this question and have started a bounty. My progress towards finding such a function has been minimal. Although it is easy to find Schwartz functions that are flat - take e.g. $e^{-x^2-1/x^2}$ - it seems very hard to impossible to control both the moments in time and frequency space simultaneously. It doesn't help that concrete examples like the above have seemingly no closed form Fourier transform. Moreover forcing $f(0)=1$ makes thing even harder. The simplest example of such a flat schwartz function I was able to find is $\exp(-\exp(x^2-1/x^2))$.

On the other hand there might be an avenue towards disproving the existence of such function. As pointed out above a flat function cannot be real analytic, but there are theorems connecting analyticity of the Fourier transform to properties like its decay rate. However the one I am familiar with (Schwartz-Payley-Wiener) deals with functions of compact support only. There seem to be generalizations with some interesting discussion found here https://mathoverflow.net/questions/23679/fourier-transform-of-analytic-functions

• In $L^2 \supset \mathcal{S}$, a basis of eigenfunctions of the Fourier transform is given by the Hermite functions ($H_n(x)e^{-\pi x^2}$ where $H_n$ are Hermite polynomials, or some appropriate scaling thereof). This may be useful. – Chappers Jun 22 '17 at 12:24
• I don't see why $g$ and hence $f \ast f$ stays flat – reuns Jun 22 '17 at 12:27
• You mean non-zero function, right? Because $f(x)=0$ is flat and is its own Fourier transform. – skyking Jun 22 '17 at 12:33
• @user1952009 I cannot find it in my notes right now and it might be that I made a mistake here. Anyway, there is a different construction that works and I edited my post. – Hyperplane Jun 22 '17 at 13:19
• @skyking Yes. In the second part this is implied. – Hyperplane Jun 22 '17 at 13:20

Not sure why you want the $0$th moment to be $1$, this construction makes $f$ Schwartz such that all the moments of $f,\hat{f}$ vanish :

Take $h \in C^\infty_c([1/4,3/4])$ and let $$c_n = \int_0^1 h(x) e^{-2i \pi n x}dx, \qquad h(x) = \sum_{n=-\infty}^\infty c_n e^{2i \pi nx}, \qquad \forall k, \ \ \sum_{n=-\infty}^\infty c_n n^k = 0$$

Then take another $g \in C^\infty_c([1/4,3/4])$ and let

$$f(x) = \sum_{n=-\infty}^\infty c_n g(x-n) \quad \in S(\mathbb{R})$$

With $\displaystyle d_m = \int_{-\infty}^\infty g(x) x^mdx$ we obtain

$$\int_{-\infty}^\infty f(x) x^k dx= \sum_{n=-\infty}^\infty c_n \int_{-\infty}^\infty g(x) (x+n)^k dx = \sum_{m=0}^k {k \choose m}d_m \sum_{n=-\infty}^\infty c_n n^{k-m} = 0$$

And the same holds for $\hat{f}$ since $f$ vanishes on $[-1/4,1/4]$.

• Thank you for this answer! First of all, the case $M_0(f)=1$ is interesting because it would make $f$ a density function. Secondly I am wondering at the moment whether your method can be adapted to construct a flat function that is 'as regular as possible' by which I mean that it is flat only in the origin. Obviously your example leaves us with a function that is constant on some open sets, even even if we take $g,h\in C^\infty_c([0,1])$ such that they are flat at the boundaries, the resulting function is still flat at a countable number of points. – Hyperplane Jul 10 '17 at 15:17
• To this end it might make sense to consider a continuous version of your construction, i.e. start out with even, flat functions $g,h$ and consider the convolution $f=\hat h * g$. It is easily shown that $f$ has zero moments: $$\int f(x) x^k dx \sim D^k [\mathcal F(f)](0) = D^k [h\hat g](0) = 0$$ However here we are left with the problem that $f$ itself might not be flat. – Hyperplane Jul 10 '17 at 15:25
• That is very elegant (sign error in the first equation, one of the exponents should be negative)! In fact, you could also just compute $\widehat{f}=\sum c_n e^{-2\pi in\xi}\widehat{g}(\xi)=h(\xi)\widehat{g}(\xi)$. This means that you also answer the follow-up question I asked on MO: feel free to post your answer there also, and I'll accept it: mathoverflow.net/questions/275072/… – user138530 Jul 10 '17 at 16:42

I think a simple inductive construction will work, though the details are somewhat tedious. My function $f(x)=\sum g_j(x-a_j)$ will consist of (widely) separated bumps located at $0<a_1<a_2<\ldots$.

I must show that these can be chosen such that $\int x^n f(x)\, dx=0$ and $f\in\mathcal S$. I start out with an odd $g_1$, so $\int f=0$, which is good, but unfortunately $\int xf\not=0$, which isn't. To address this issue, I use $g_2$, which will be chosen orthogonal to $1$, and such that $\int x(g_1+g_2)=0$ also. In the next step, I use $g_3$, which will be chosen orthogonal to $1,x$, and such that $\int x^2(g_1+g_2+g_3)=0$ also.

The key observation is that the orthogonality that I impose to preserve earlier achievements is translation invariant in the sense that $g(x)$ is orthogonal to $1,x,\ldots , x^N$ precisely if $g(x-a)$ is. This means that by taking $a$ large, I can make $g_n$ as small as I want to in a given step, and this allows me to keep $f\in\mathcal S$ overall. (This requirement is really the list of conditions $|f^{(k)}(x^2+1)^m|\le 1$, $x\ge x_{N,k}$, and I satisfy these by paying attention to the first $N$ of these at a given step, and $N$ is increased gingerly.)

Remark: I'm no expert on these things, but I think it's known that there are measures (and functions, I would think) $f$ such that both $f$ and $\widehat{f}$ vanish on an interval, but I don't know if there are such examples in $\mathcal S$.

• I think you meant take $g \in C^\infty_c([0,1])$ and let $f_0 = g(x), f_{n+1}(x) = f_n(x)- c_n f_n(b_n x-a_n)$ with $a_n$ arbitrary and $b_n = c_n^{-1/(1-n)}$ so that by induction $\int_{-\infty}^\infty f_{n+1}(x) x^{n} dx =0$. – reuns Jul 10 '17 at 4:15
• See my answer for a special non-recursive case – reuns Jul 10 '17 at 4:34
• @user1952009: Yes, of course my $g_n$'s are smooth and compactly supported, but I meant what I wrote (which contains what you suggest as a special case anyway, but that's not the procedure I had in mind). – user138530 Jul 10 '17 at 16:23

Here is some numerical evidence to the affirmative.

Let $g(x)=\exp(-x^2/2)$ be the Gaussian function such that $\mathcal Fg=g$. Let us define our function of interest as a linear combination of copies of $g$ stretched by powers of $2$ (factors in blue are mathematically superfluous but improve the numerical conditioning of the system we will form): $$f_n(x) = \sum_{i=-n}^n \color{blue}{2^{-i^2}}a_ig(2^ix).$$ By choosing $a_{-i}=2^{-i}a_i$ for negative $i$, we satisfy $\mathcal Ff_n=f_n$. We are left with $n+1$ degrees of freedom $a_0,\dots,a_n$, so we may impose $n+1$ equations: \begin{align} \int_{\infty}^\infty f_n(x)\,\mathrm dx &= 1, \\ \color{blue}{\frac12}\,f''(0) &= 0, \\ &\vdots \\ \color{blue}{\frac{2^{-n^2}}{(2n - 1)!!}}\,f^{(2n)}(0) &= 0. \end{align} (Note that $f_n$ is an even function so all odd derivatives are automatically zero.) These equations form a linear system in the coefficients $a_i$, in fact a nearly tridiagonal one with diagonal entries converging to $\pm1$, which can be easily solved numerically. Let's look at the resulting functions:

Sure seems to be converging to something.

Here's a table of the values of $a_i$ for different $n$:

$$\begin{array}{rcccccc} n & a_0 & a_1 & a_2 & a_3 & a_4 & a_5 \\ 0 & 0.398942 & & & & & \\ 1 & 0.626909 & -0.303956 & & & & \\ 10 & 0.729176 & -0.465597 & 0.246237 & -0.124822 & 0.0626214 & -0.0313298 \\ 100 & 0.729177 & -0.465598 & 0.246238 & -0.124825 & 0.0626265 & -0.0313400 \\ \end{array}$$

• Interesting idea! However there is one possible problem here: as I recall when I tried an expansion by means of Hermite function I got a function that looked similar to yours, but for large $n$ it degraded towards a constant function. With your Ansatz it however seems not easy to consider large $n$ since the matrices are incredibly bad conditioned. So I'm unsure whether or not the same problem will arise here. – Hyperplane Jul 10 '17 at 16:38
• You're right, a numerical solution is uselessly inaccurate even for $n=10$; I've been computing exact solutions using a CAS. Maybe there is some structure in the equations that can be exploited to improve the conditioning, but I haven't put any thought into it. – Rahul Jul 10 '17 at 17:55
• I fixed the numerical conditioning problem. We still get a dense matrix of course, but it has decent conditioning ($\kappa$ is only $27.7$ for $n=100$, roughly the same as for $n=10$). – Rahul Jul 11 '17 at 4:42