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!