An inequality concerning the Fourier transform

Question

I want to know that the inequality $$ \left| \int_{-\infty}^\infty\frac{f(\xi) e^{ix\xi}}{\sqrt{1+\xi^2}} d\xi\right| \le C \left| \int_{-\infty}^\infty f(\xi) e^{ix\xi}d\xi\right| $$ holds or not with some constant $C>0$, if the right hand side is finite. Here $f(\xi)$ is a complex-valued function.

Answer 1

Here's what I've come up with so far (incomplete):

Rewriting the LHS and using the convolution theorem: $$\left\| \int_{-\infty}^{+\infty} f(\xi)g(\xi)e^{ix\xi}d\xi \right\|= \left\|F(x)*G(x) \right\|$$

Where $F(x)$ and $G(x)$ are the inverse Fourier transforms of $f(\xi)$ and $g(\xi)$ respectively. Then applying Young's inequality for convolution where $p=q=r=1$: $$\left\|F(x)*G(x) \right\| \leqslant \left\|F(x) \right\|\left\|G(x) \right\|$$

$G(x)$ turns out to be $2K_0(|x|)$ where $K_n(x)$ is the modified Bessel function of the second kind that only diverges at $x=0$.

Answer 2

Switching to Fourier transforms, we want to compare the absolute values of: $$ \int_{-\infty}^{+\infty}\widehat{f}(s)\,\frac{K_0(|x-s|)}{\pi}\,ds \qquad \text{and}\qquad \widehat{f}(x), $$ where $\frac{1}{\pi}\,K_0(|s-x|)$ is a modified Bessel function of the second kind that acts like an approximate Dirac delta $\delta(s-x)$. In particular: $$\min(0,-\log|x|)\leq K_0(|x|)\leq -\log\left(\frac{|x|}{|x|+1}\right),$$

hence if $\widehat{f}$ is well-behaved your inequality is very likely to hold. Also notice that: $$ \left\|\widehat{f}*\frac{K_0}{\pi}\right\|_2 = \left\|\widehat{f}\right\|_2 \cdot \left\|\frac{K_0}{\pi}\right\|_2 = \frac{\|f\|_2}{\sqrt{2}}$$ and by Young's inequality: $$ \left\|\widehat{f}*\frac{K_0}{\pi}\right\|_1 \leq \left\|\widehat{f}\right\|_1.$$