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Suppose we define a function \begin{align} f(k ;a,b) =\frac{ \int_0^\infty \cos(a x) e^{-x^k} \,dx}{ \int_0^\infty \cos(b x) e^{-x^k} \,dx} \end{align}

can we show that \begin{align} |f(k ;a,b)| \le 1 \end{align} for $ 0<k \le 2$ and $a\ge b$?

This question was motivated by the discussion here.

Note that for $k=1$ and $k=2$ this can be done, since

\begin{align} \int_0^\infty \cos(a x) e^{-x^1} \,dx=\frac{1}{1+a^2}\\ \int_0^\infty \cos(a x) e^{-x^2} \,dx=\frac{\sqrt{\pi}}{2}e^{-a^2/4}\\ \end{align}

So, we have that \begin{align} f(1;a,b)&=\frac{1+b^2}{1+a^2} \\ f(2;a,b)&=e^{ \frac{b^2-a^2}{4}} \end{align}

In which case, we have that the conjectured bound is true.

Edit: The bounty was posted specifically to address this question and a question raised by Jack D'Aurizio in the comments.
The question is:

Let \begin{align} g_k(z)=\int_0^\infty \cos(zx) e^{-x^k} dx \end{align}

What is the largest value of $k$ such that $g_k(z)$ is non-negative and decreasing for $z\in \mathbb{R}^{+}$?

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Yes, the Fourier transform of $\exp(-|x|^k)$ is positive and decreasing for all $k$ such that $0 < k \leq 2$.

This follows from the known case of $k=2$ (Gaussians) via an argument of B.F.Logan cited 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.

See Lemma 5 on page 626 (with $k=\sigma$; in that paper we needed only positivity, not that the Fourier transform is decreasing). The key is that $\exp(-t^{k/2})$ is a "totally monotone" function of $t>0$ (its $n$-th derivative has sign $(-1)^n$ for all $t>0$), and decays to zero as $t \to \infty$, whence it is a nonnegative mixture of decreasing exponentials $e^{-ct}$ $(c>0)$ by Bernstein's theorem. Taking $t=x^2$ we deduce that $\exp(-|x|^k)$ is a nonnegative mixture of Gaussians $\exp(-cx^2)$ $(c>0)$. Since the Fourier transform of $\exp(-cx^2)$ is positive and decreasing for all $c>0$, the same is true of the Fourier transform of $\exp(-|x|^k)$, QED.

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We have to prove that the Fourier cosine transform of $e^{-x^k}$ is non-negative and decreasing on $\mathbb{R}^+$.

We have: $$ \int_{0}^{+\infty}\cos(\xi x)e^{-x}\,dx = \frac{1}{1+\xi^2},\qquad \int_{0}^{+\infty}\cos(\xi x)e^{-x^2}\,dx = \frac{\sqrt{\pi}}{2}\,e^{-\xi^2/4} \tag{1} $$ hence the claim is expected to hold by some kind of interpolation argument. Let $f_k(x)=e^{-x^k}$ and $$ g_k(\xi)=\int_{0}^{+\infty}\cos(\xi x)\,e^{-x^k}\,dx.\tag{2} $$ We have: $$ g_k'(\xi) = -\int_{0}^{+\infty}\sin(\xi x) x e^{-x^k}\,dx =-\frac{1}{\xi}\int_{0}^{+\infty}\cos(\xi x)\left(e^{-x^k}-kx^k e^{-x^k}\right)\,dx\tag{3}$$ hence: $$ g_k(\xi)+\xi g_k'(\xi) = \frac{k}{\xi}\int_{0}^{+\infty}\cos(\xi x) x^k e^{-x^k}\,dx \tag{4} $$ and as soon as we prove that the RHS of $(4)$ cannot be too large (for instance, through $w e^{-w}\in\left[0,\frac{1}{e}\right]$ for any $w\in\mathbb{R}^+$) we have that $g_k(\xi)$ is close to $g_1(\xi)$ or $g_2(\xi)$ by a Gronwall-like inequality.

As already shown in your other question, $g_k(\xi)$ does not preserve its non-negativity and monotonicity for every $k$. That is not surprising, since $e^{-x^k}$, regarded as a distribution over $\mathbb{R}^+$, weakly converges to $c_k\cdot\mathbb{1}_{(0,1)}(x)$ as $k\to +\infty$, and the Fourier transform of $\mathbb{1}_{(0,1)}(x)$ is extremely regular (as any Fourier transform of a compact-supported function, by the Paley-Wiener theorem) but has plenty of real zeroes. Another interesting (but probably very hard) question arises:

What is $$ \sup\left\{k\in\mathbb{R}^+: g_k(\xi)\text{ is non-negative and decreasing over } \mathbb{R}^+\right\}$$ ?

Numerical simulations seem to suggest that such a supremum is exactly $\color{red}{2}$.

I strongly suspect that the Fejer-Riesz theorem is involved.

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  • 2
    $\begingroup$ Thank you. Great answer and indeed a very interesting conjecture. $\endgroup$ – Boby Jul 1 '16 at 20:05
  • $\begingroup$ I think once. I will be able to post a bounty I will do that. To attract more people. $\endgroup$ – Boby Jul 2 '16 at 2:21
  • $\begingroup$ Jack Thanks for reference (+1) $\endgroup$ – Behrouz Maleki Jul 11 '16 at 14:26
  • $\begingroup$ I was wondering if you could add a little more details on your Gronwall-like argument? $\endgroup$ – Boby Dec 4 '16 at 1:24

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