For the following function: $$ u(x,y)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{-\hat{f}(\omega)e^{-|\omega|y}}{1+|\omega|}e^{i\omega x}d\omega $$
can we conclude that $u\rightarrow 0$ as $|x|\rightarrow \infty$? Here, $y>0$ and $f$ is a compactly supported smooth function.
If the terms multiplying $\hat{f}$ were smooth, I think I could argue that the integrand is Schwartz because $\hat{f}$ is Schwartz and the FT preserves that property, and then use the same reasoning to conclude that $u$ decayed at infinity. But unless I'm mistaken, the terms with $|\omega|$ aren't differentiable at $\omega=0$.
The Fourier transform of any $L^1$ function decays at infinity. Is the function $$g(\omega) = \frac{-\hat{f}(\omega)e^{-|\omega|y}}{1+|\omega|}$$ in $L^1$? Yes, it is, for a number of reasons. Even the boundedness of $\hat f$ is enough here, thanks to the exponential term $e^{-|\omega|y}$. Or, even neglecting the exponential term, just having $f\in L^2$ is enough, since the product $$\hat{f}(\omega) \frac{1}{1+|\omega|}$$ is integrable by the Cauchy-Schwarz inequality (both terms are in $L^2$).