On showing a distribution is a function

On showing a distribution is a function

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?