I have  a real-valued function $f$ defined on a ball $B$ of  $\mathbb{R}^{N}$, $N\geq2$.  I have found a constant $M>0$ such that for all $x\in B$ and $B(x,R)$ (ball of center $x$ and radius $R>0$) such that its closure  $\overline{B(x,R)}$ is in $B$, we have 
$$|f(x)|\leq M\max_{B}|\Delta\phi|,$$
where $\phi$ is a non-negative test function with support in $B(x,R)$ such that $\phi=1$ on  $\overline{B(x,r)}$ with $0<r<R$.  Here, $r$ is arbitrary and $ \phi $ is the function given by Urysohn's lemma. My question is: Can I conclude that $f$ is locally bounded on the ball $B$? It seems to me that since the right side of the above inequality depends on $ \phi $, the answer is no. Is it possible to get rid of $ \phi $?