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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!

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    $\begingroup$ As many as you want: just take any smooth even real-valued compactly supported function and convolve with itself. $\endgroup$ – fedja Aug 30 at 12:37
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    $\begingroup$ @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. $\endgroup$ – OzoneNerd Aug 30 at 14:30
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Credit to @fedja who answered my question in a comment ten minutes after I posted it. Thanks!

As many as you want: just take any smooth even real-valued compactly supported function and convolve with itself.

This works because convolving the function with itself squares the Fourier transform. We get minima of zero wherever the original Fourier transform changed signs.

By adding the self-convolution of two functions with no shared Fourier roots, we can in fact strengthen the original requirement to an everywhere strictly positive Fourier transform, instead of just non-negative.

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 $\sqrt 2$.

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