Return to Answer

A simple argument that the Fourier transform cannot be nonnegative for any $m>1$, integer or not, is given in my 1991 paper with Odlyzko and Rush:

Noam D. Elkies, Andrew M. Odlyzko, and Jason A. Rush: On the packing densities of superballs and other bodies, Invent. Math. 105 (1991), 613-639.

The Fourier transform $f$ of a nonnegative even function must satisfy $4f(\xi) \leq 3 f(0) + f(2\xi)$ for all $\xi$. This fails for any $m>1$ once $\xi$ is close enough to (but not equal) zero.

Postscripts:

0. The inequality $4f(\xi) \leq 3 f(0) + f(2\xi)$ holds because the difference is $\int_{-\infty}^\infty \hat f(\eta) \, w(\eta) \, d\eta$ with $w(\eta) \geq 0$ for all $\eta$ because $3 - 4\cos \theta + \cos 2\theta = 8 \sin^4 (\theta/2) \geq 0$.

1. The nonnegativity of $3 - 4\cos \theta + \cos 2\theta$ is also a key ingredient in the proof of thenonvanishing of the Riemann zeta function $\zeta(s)$ on the edge $s = 1+it$ of the critical strip, and thus of the Prime Number Theorem. See the last exercise of this chapter my notes on analytic number theory.

2. Sergei noted that it is also known that for $0 < m \leq 1$ the Fourier transform is positive. As I wrote earlier this year on Math Stackexchange, the same paper with Odlyzko and Rush also proves that fact as follows, crediting the argument to B.F. Logan (see Lemma 5 on page 626).
   We first prove this for $m=1$, then reduce $0<m<1$ to this case. For $m=1$, the function $\exp(-t|\xi|^{2m})$ is a Gaussian, and thus so is its Fourier transform; Gaussians are positive everywhere, so the case $m=1$ is done. For $0<m<1$, the function $\exp(-t|\xi|^{2m})$ is a nonnegative mixture of Gaussians $\exp(-c\xi^2)$ by Bernstein's theorem. Therefore the Fourier transform is also a nonnegative mixture of Gaussians, so again positive everywhere. QED

A simple argument that the Fourier transform cannot be nonnegative for any $m>1$, integer or not, is given in my 1991 paper with Odlyzko and Rush:

Noam D. Elkies, Andrew M. Odlyzko, and Jason A. Rush: On the packing densities of superballs and other bodies, Invent. Math. 105 (1991), 613-639.

The Fourier transform $f$ of a nonnegative even function must satisfy $4f(\xi) \leq 3 f(0) + f(2\xi)$ for all $\xi$. This fails for any $m>1$ once $\xi$ is close enough to (but not equal) zero.

Postscripts:

0. The inequality $4f(\xi) \leq 3 f(0) + f(2\xi)$ holds because the difference is $\int_{-\infty}^\infty \hat f(\eta) \, w(\eta) \, d\eta$ with $w(\eta) \geq 0$ for all $\eta$ because $3 - 4\cos \theta + \cos 2\theta = 8 \sin^4 (\theta/2) \geq 0$.

1. The nonnegativity of $3 - 4\cos \theta + \cos 2\theta$ is also a key ingredient in the proof of thenonvanishing of the Riemann zeta function $\zeta(s)$ on the edge $s = 1+it$ of the critical strip, and thus of the Prime Number Theorem. See the last exercise of this chapter my notes on analytic number theory.

2. Sergei noted that it is also known that for $0 < m \leq 1$ the Fourier transform is positive. As I wrote earlier this year on Math Stackexchange, the same paper with Odlyzko and Rush also proves that fact as follows, crediting the argument to B.F. Logan (see Lemma 5 on page 626).
   We first prove this for $m=1$, then reduce $0<m<1$ to this case. For $m=1$, the function $\exp(-t|\xi|^{2m})$ is a Gaussian, and thus so is its Fourier transform; Gaussians are positive everywhere, so the case $m=1$ is done. For $0<m<1$, the function $\exp(-t|\xi|^{2m})$ is a nonnegative mixture of Gaussians $\exp(-c\xi^2)$ by Bernstein's theorem. Therefore the Fourier transform is also a nonnegative mixture of Gaussians, so again positive everywhere. QED

A simple argument that the Fourier transform cannot be nonnegative for any $m>1$, integer or not, is given in my 1991 paper with Odlyzko and Rush:

Noam D. Elkies, Andrew M. Odlyzko, and Jason A. Rush: On the packing densities of superballs and other bodies, Invent. Math. 105 (1991), 613-639.

The Fourier transform $f$ of a nonnegative even function must satisfy $4f(\xi) \leq 3 f(0) + f(2\xi)$ for all $\xi$. This fails for any $m>1$ once $\xi$ is close enough to (but not equal) zero.

A simple argument that the Fourier transform cannot be nonnegative for any $m>1$, integer or not, is given in my 1991 paper with Odlyzko and Rush:

Noam D. Elkies, Andrew M. Odlyzko, and Jason A. Rush: On the packing densities of superballs and other bodies, Invent. Math. 105 (1991), 613-639.

The Fourier transform $f$ of a nonnegative even function must satisfy $4f(\xi) \leq 3 f(0) + f(2\xi)$ for all $\xi$. This fails for any $m>1$ once $\xi$ is close enough to (but not equal) zero.

Postscripts:

0. The inequality $4f(\xi) \leq 3 f(0) + f(2\xi)$ holds because the difference is $\int_{-\infty}^\infty \hat f(\eta) \, w(\eta) \, d\eta$ with $w(\eta) \geq 0$ for all $\eta$ because $3 - 4\cos \theta + \cos 2\theta = 8 \sin^4 (\theta/2) \geq 0$.

1. The nonnegativity of $3 - 4\cos \theta + \cos 2\theta$ is also a key ingredient in the proof of thenonvanishing of the Riemann zeta function $\zeta(s)$ on the edge $s = 1+it$ of the critical strip, and thus of the Prime Number Theorem. See the last exercise of this chapter my notes on analytic number theory.

2. Sergei noted that it is also known that for $0 < m \leq 1$ the Fourier transform is positive. As I wrote earlier this year on Math Stackexchange, the same paper with Odlyzko and Rush also proves that fact as follows, crediting the argument to B.F. Logan (see Lemma 5 on page 626).
   We first prove this for $m=1$, then reduce $0<m<1$ to this case. For $m=1$, the function $\exp(-t|\xi|^{2m})$ is a Gaussian, and thus so is its Fourier transform; Gaussians are positive everywhere, so the case $m=1$ is done. For $0<m<1$, the function $\exp(-t|\xi|^{2m})$ is a nonnegative mixture of Gaussians $\exp(-c\xi^2)$ by Bernstein's theorem. Therefore the Fourier transform is also a nonnegative mixture of Gaussians, so again positive everywhere. QED

A simple argument that the Fourier transform cannot be nonnegative for any $m>1$, integer or not, is given in my 1991 paper with Odlyzko and Rush:

Noam D. Elkies, Andrew M. Odlyzko, and Jason A. Rush: On the packing densities of superballs and other bodies, Invent. Math. 105 (1991), 613-639.

The Fourier transform $f$ of a nonnegative even function must satisfy $4f(\xi) \leq 3 f(0) + f(2\xi)$ for all $\xi$. This fails for any $m>1$ once $\xi$ is close enough to (but not equal) zero.

A simple argument that the Fourier transform cannot be nonnegative for any $m>1$, integer or not, is given in my 1991 paper with Odlyzko and Rush:

Noam D. Elkies, Andrew M. Odlyzko, and Jason A. Rush: On the packing densities of superballs and other bodies, Invent. Math. 105 (1991), 613-639.

The Fourier transform $f$ of a nonnegative even function must satisfy $4f(\xi) \leq 3 f(0) + f(2\xi)$ for all $\xi$. This fails for any $m>1$ once $\xi$ is close enough to (but not equal) zero.