Suppose $u(x)$ is in $W^{1,1}(R^2)$ (Sobolev space on $R^2$ with 1 weak derivative and $L_1$ norm) and is locally bounded. Must $u(x)$ be continuous?
Motivation: In $R^1$, a weakly differentiable function is continuous, but that is not the case for $R^2$ because of $f(x) = |x|^{0.5}$. However, this function is not bounded. So I am looking for a function that is bounded, weakly differentiable and not continuous.