# Fourier cosine transforms of Schwartz functions and the Fejer-Riesz theorem

This question spanned from a previous interesting one.
Let $k$ be a real number greater than $2$ and
$$\varphi_k(\xi) = \int_{0}^{+\infty}\cos(\xi x) e^{-x^k}\,dx$$
the Fourier cosine transform of a function in the Schwartz space.

Is is possible to use the Fejér-Riesz theorem or some variation of it,
to prove that $\varphi_k(\xi)<0$ for some $\xi\in\mathbb{R}^+$?

#### Solutions Collecting From Web of "Fourier cosine transforms of Schwartz functions and the Fejer-Riesz theorem"

One proof is given in the paper

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.

Instead of Fejér-Riesz, we use the Fourier inversion formula directly.
Assume for contradiction that $\phi_k(\xi) \geq 0$ for all $\xi$.
Then for all $x$ we would have
$$0 \leq \int_0^\infty \phi_k(\xi) \, (1 – \cos (\xi x))^2 \, dx = \frac12 \int_0^\infty \phi_k(\xi) \, (3 – 4 \cos (\xi x) + \cos (2\xi x)) \, dx,$$
which means
$$3 – 4 e^{-x^k} + e^{-(2x)^k} \geq 0$$
for all $x$. But for $x$ near $0$ we may write
$e^{-x^k} = 1 – \epsilon$ and
$e^{-(2x)^k} = (1 – \epsilon)^{2^k} = 1 – 2^k \epsilon + O(\epsilon^2)$,
so
$$3 – 4 e^{-x^k} + e^{-(2x)^k} = 3 – 4(1-\epsilon) + (1 – 2^k \epsilon + O(\epsilon^2)) = (4 – 2^k) \epsilon + O(\epsilon^2).$$
Thus $3 – 4 e^{-x^k} + e^{-(2x)^k}$ becomes negative for small $x$
once $2^k > 4$, which is equivalent to $k=2$,
and we have our desired contradiction.