If $f\in H^1(\mathbb{R}^2)$, does $\Delta^{-1}\partial_xf(x,y)\in L^\infty(\mathbb{R}^2)$?

Question

I am trying to understand the integrability properties of a function defined in terms of the $\Delta^{-1}$ operator. Let's consider $f:\mathbb{R}^2\to\mathbb{R}$ belonging to the Sobolev space $f\in H^1(\mathbb{R}^2)$. Now, define $$ g(x,y)=\Delta^{-1}\partial_xf(x,y), $$ where $\Delta^{-1}$ is defines in Fourier variables as $$ \Delta^{-1}h(x,y)=\mathcal{F}^{-1}\Big(\dfrac{-\hat{h}(\xi,\mu)}{\xi^2+\mu^2}\Big). $$ Now, from Plancharel Theorem we can see that $g_x,g_y\in H^1(\mathbb{R}^2)$. However, I was wondering if we can say something about $g$ itself and not about its derivatives. It is clear to me that we cannot expect that $g\in L^2(\mathbb{R})$ since, again from a Fourier point of view, the multiplier $1/(\xi^2+\mu^2)$ has a non-integrable singularity at the origin. Although, I have the feeling that the following should holds $$ g\in L^\infty(\mathbb{R}^2), $$ but I am not sure how to prove it, since $L^\infty$ does not get along with the Fourier transform. Is there any trick one can perform to prove such type of property? Or maybe that is immediately false? Or maybe $g$ might belong to other $L^p$ spaces with $p\neq 2$?