# Non-zero, bounded, continuous, differentiable at the origin, compactly supported functions with everywhere non-negative Fourier transforms

Do there exist functions $$F(x) \! : \, \mathbb R \to \mathbb R$$ which are non-zero and bounded: $$\mathrm {Range} (F) = [l, u] \, , \quad \mathrm {where} \quad l, u \in \mathbb R \land u > l \, ; \tag 1$$ continuous; differentiable at the origin; and compactly supported: $$\mathrm {supp} (F) = (a, b) \, , \quad \mathrm {where} \quad a, b \in \mathbb R \, ; \tag 2$$ such that the Fourier transform, $$\tilde F(t) \! : \, \mathbb R \to \mathbb C \, , \,$$ defined as $$\tilde F(t) = \int_{-\infty}^\infty \! e^{i t x} F(x) \, \mathrm d x \, ; \tag 3$$ exists and is everywhere real and non-negative: $$\mathrm {Range} \! \left ( \tilde F(t) \right ) \subseteq \mathbb R_{{\ge}0} \, ? \tag 4 \label {Condition}$$

I believe one can easily show that $$\tilde F(t)$$ must be bounded and non-zero: $$\mathrm {Range} \! \left ( \tilde F(t) \right ) = [0, c] \, , \quad \mathrm {where} \quad c \in \mathbb R_{{>}0} \, ; \tag 5$$ and converge to zero:$$~~\tilde F(t \to \infty) \to 0^+ \, .$$

In order to have a real Fourier transform: $$\mathrm {Range} \! \left ( \tilde F(t) \right ) \subseteq \mathbb R \, , \tag 6$$ $$F(x)$$ must be even: $$\forall x \in \mathbb R \! : \, F(x) = F(-x) \, , \tag 7$$ which implies $$b > 0 \, , \,$$ $$a = -b \, , \,$$ $$F' \! (0) = 0 \, , \,$$ and that $$\tilde F(t)$$ is also even: $$\forall t \in \mathbb R \! : \, \tilde F(t) = \tilde F(-t) \, . \tag 8$$ So, without loss of generality, we can ask the same question of the cosine transform, $$\tilde F^c \! (t) \! : \, \mathbb R_{{\ge}0} \to \mathbb R \, , \,$$ defined as $$\tilde F^c \! (t) = \int_0^b \! \cos{(t x)} \, F(x) \, \mathrm d x \, ; \tag 9$$ namely, $$\mathrm {Range} \! \left ( \tilde F^c \! (t) \right ) \subseteq \mathbb R_{{\ge}0} \, ? \tag {10}$$ Furthermore, $$\tilde F^c \! (t)$$ should obey the same conditions as $$\tilde F(t)$$ laid out in the previous paragraph.

I understand that condition$$~\eqref {Condition}$$ is equivalent to requiring that $$F(x)$$ be a positive-definite function. Also, I am under the impression that this paper shows that if $$F(x)$$ is “convex”, $$\forall x > 0 \! : \, F'' \! (x) > 0 \, , \tag {11}$$ then it is positive-definite. I am doubtful, however, that such a convex $$F(x)$$ can satisfy the requirements laid out in the first paragraph. The Paley–Wiener theorem also seems potentially relevant. I have thusfar neither been able to use these results to construct an $$F(x)$$ satisfying those requirements nor to prove their non-existence.

Two functions which come close are $$F(x) = (|x| - 1)^2 \, \mathbf 1_{[-1, 1]} (x) \, , \tag {12}$$ and $$F(x) = -\ln{|x|} \, \mathbf 1_{[-1, 1]} (x) \, , \tag {13}$$ where $$\mathbf 1_S (x)$$ is the indicator function. Both are non-differentiable at $$x = 0 \, , \,$$ and the latter is unbounded:$$~~F(x \to 0) \to \infty \, .$$

I am also interested in the generalization of this question to $$D > 1$$-dimensional isotropic Fourier transforms, $$t^{1 - D/2} \! \int_0^b \! J_{D/2 - 1} (t x) \, F(x) \, x^{D/2} \, \mathrm d x \, , \tag {14}$$ where $$J_\alpha$$ is a Bessel function.

Thanks!

• As many as you want: just take any smooth even real-valued compactly supported function and convolve with itself. Aug 30, 2020 at 12:37
• @fedja That was remarkably fast. Neat trick. I get why it works. I also confirmed it numerically using the standard bump function. Thanks! Additionally, I see that by adding two incommensurately scaled copies of the convolution, the Fourier transform can be made everywhere strictly positive. Aug 30, 2020 at 14:30

I confirmed this numerically using the standard bump function $$e^{\frac 1 {r^2 - 1}} \, \mathbf 1_{(-1, 1)}$$ and a version horizontally stretched by $$\sqrt 2$$.