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.
As Jose27 mentioned, we look for functions of the form: $$f(x) = \sin(|x|^{-\alpha}), \alpha > 0$$
$$\displaystyle f_{x_i} = -\alpha \cos(|x|^{-\alpha})|x|^{-\alpha-1}\frac{x_i}{|x|}$$
Similar to the argument on Evan's page 260, we can prove $f$ is weakly differentiable.
For any $\phi \in C^{\infty}_c(R^2)$ $$\int\int_{R^2 - B(0,\epsilon)} f\phi_{x_i} dx = -\int\int_{R^2 - B(0,\epsilon)} f_{x_i}\phi dx + \int_{\partial B(0,\epsilon)}f\phi \nu_i dS$$
As $\epsilon \rightarrow 0$, the last term goes to zero. The remaining equation is what is required by the weak derivative. The weak derivative also needs to be $L_1$ local. We have $$|Df| = |\cos(|x|^{-\alpha})| \alpha |x|^{-\alpha-1}$$
As long as $\alpha + 1 < 2$, $|Df|$ is $L_1$-local. Then $\alpha = 0.5$ should work.