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:
$$ \mathrm {Range} \! \left ( \tilde F(t) \right ) \subseteq \mathbb R\_{{\ge}0} \, ? \tag 4 \label {Condition} $$




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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:$~~\tilde F(t \to \infty) \to 0^+ \, .$




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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.




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I understand that condition$~\eqref {Condition}$ is equivalent to requiring that $F(x)$ be a [positive-definite function](https://en.wikipedia.org/wiki/Positive-definite_function). Also, I am under the impression that [this paper](http://www.maths.adelaide.edu.au/yvonne.stokes/Tuck/pdfiles/posFT5_06.pdf) 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](https://en.wikipedia.org/wiki/Paley%E2%80%93Wiener_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.




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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](https://en.wikipedia.org/wiki/Indicator_function).
Both are non-differentiable at $x = 0 \, , \,$ and the latter is unbounded:$~~F(x \to 0) \to \infty \, .$




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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.




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Thanks!



Non-zero, bounded, continuous, differentiable at the origin, compactly supported functions with everywhere non-negative Fourier transforms
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 $\sqrt 2$.


