Consider the distributional equation $$\Delta \omega-\omega=\mu$$ Then it is easy to verify by Fourier transform that $$\omega=-\mathcal{F}^{-1}\left(\frac{1}{|\cdot|^2+1}\hat{\mu}\right)$$ is the only solution of the above equation. Now let $\mu\in\mathcal{S}'(\mathbb{R}^2)$ be defined by $$\mu(\varphi)=\int_\mathbb{R}\varphi(0,y)\ dt,\forall \varphi\in\mathcal{S}(\mathbb{R^2})$$ The problem asks to show that $\omega$ is a bounded function on $\mathbb{R}^2$.
Intuitively we can calculate as follows: \begin{equation*} \begin{split} \omega(\varphi)&=(-\mathcal{F}^{-1}\left(\frac{1}{|\cdot|^2+1}\hat{\mu}\right),\varphi)\\ &=(\hat{\mu},-\frac{1}{|\cdot|^2+1}\mathcal{F}^{-1}\varphi)\\ &={-1 \over {2\pi}}(\mu,f*\varphi)\\ \end{split} \end{equation*} where $f=\mathcal{F}\left(\frac{1}{|\cdot|^2+1}\right) \in L^2$ since $\frac{1}{|\cdot|^2+1} \in L^2$. Now we compute formally with \begin{equation*} \begin{split} \omega(\varphi)&={-1 \over {2\pi}}(\mu,f*\varphi)\\ &={-1 \over {2\pi}}\int_{\mathbb{R}_y}f*\varphi(0,y)\ dy\\ &={-1 \over {2\pi}}\int_{\mathbb{R}_y}\ dy\int_{\mathbb{R}^2_\eta}f(x-\eta)\varphi(\eta)|_{x=(0,y)}\ d\eta\\ &={-1 \over {2\pi}}\int_{\mathbb{R}^2_\eta} \varphi(\eta)\ d\eta\int_{\mathbb{R}_y}f(-\eta_1,y)\ dy \end{split} \end{equation*} Now formally define $$\omega(x,y)={-1 \over {2\pi}}\int_{\mathbb{R}}f(-x,z)\ dz$$ and I want to do estimation to it but have no idea of how to proceed. Am I doing wrong here?
We are only left with showing that $\omega$ is bounded. Writing out the Fourier transform: $$ \omega(x) = \int_{\mathbb{R}} e^{+2\pi i 0 z}\int_{\mathbb{R}}e^{-2\pi i z \xi_2} \int_{\mathbb{R}} e^{+2\pi i x \xi_1} \frac{1}{1+\xi_1^2+\xi_2^2} d\xi_1 \, d\xi_2 \, dz. $$
By Fourier inversion we can cancel the two outer integrals substituting $\xi_2 = 0$, so we arrive at $$ \omega(x) = \int_{\mathbb{R}}e^{+2\pi i x \xi_1} \frac{1}{1+\xi_1^2} \, d\xi_1. $$ And this is bounded since $1/(1+x^2)$ is in $L^1(\mathbb{R})$.