0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Think $\sin(|x|^{-\alpha})$. $\endgroup$ – Jose27 Apr 2 '17 at 5:31
  • $\begingroup$ $\alpha=0.5$ should work. I posted more detail below. $\endgroup$ – Arman Tavakoli Apr 12 '17 at 21:26
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.