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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 $\xi\in\mathbb{R}^+$?

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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.

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