Bounded solutions of Poisson's equation in $n = 2$ I want to prove is the following statement;

Let $f \in C_c^2(\mathbb R^2)$ ($f$ is twice continuously differentiable function with compact support)  then $u(x) = \int_{\mathbb R^2} \phi (x-y) f(y) dy$ is bounded. ($\phi$ here is the fundamental solution for $n = 2$)  

I don't know how to use the compactness of support of $f$ to make $u(x)$ bounded as the integral will blow-up when $x = y$ as $\phi(0)$ is unbounded. 
 A: This is not exactly true; $u$ will not be globally bounded in general, because 
$$
\phi(x-y) = \frac{1}{2\pi} \log|x|  + O(1) \quad \text{as } |x|\to \infty, \ y\in\operatorname{supp} f
$$
which yields
$$u(x) = \frac{1}{2\pi} \log|x| \int_{\mathbb{R}^2} f(y)\,dy + O(1), \quad |x|\to\infty
$$
What we can prove is that $u$ is locally bounded. The problem is when $y$ is near $x$, so let's focus on that: 
$$ u(x) = \int_{|y-x|<1} \phi (x-y) f(y)\, dy +  \int_{|y-x|\ge 1} \phi (x-y) f(y)\, dy$$
The second integral is bounded as long as $x$ is bounded. 
In the first integral, counter the singularity of logarithm with the continuity of $f$: 
$$
\int_{|y-x|<1} \phi (x-y) f(y)\, dy
 = \int_{|y-x|<1} \phi (x-y) (f(y)-f(x))\, dy 
+  f(x) \int_{|y-x|<1} \phi (x-y)\, dy
$$
Both integrals on the right converge. Indeed, $f(y)-f(x) = O(|y-x|)$ as $y\ to x$, which makes $\phi (x-y) (f(y)-f(x))$ a continuous function of $y$, being integrated on a bounded set. The second integral converges by a direct calculation in polar coordinates.
