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Not necessarily- let $\Omega = B_1 \cap \{x_3 > 0\}.$ Then $u(x) := (1-|x|^2)\frac{x_3}{|x|}$ is in $H^1_0(\Omega) \cap C^{\infty}(\Omega),$ but $u$ is discontinuous at the origin.

Not always. Consider the case $n \geq 5$, $K = B_1$, and $u = |x|^{\frac{4-n}{2}} - 1$. Then $u \in H^1_0(B_1)$ but $u \notin H^2(B_1)$, and $$\Delta u = \frac{n(4-n)}{4}(u+1)^{\frac{n}{n-4}} := c(u), …

The answer is indeed "no." To see this, instead of considering translations of radial bump functions, one needs to use bump functions whose level sets are deformed e.g. to ellipsoids. More precisely, …

This response is closely related to my answer Here. For the case $W^{2,n}$ we automatically get Holder regularity by applying Sobolev and then Morrey. However, we don't get Lipschitz. The following e …

I'll take a stab. In the following we consider the case $W^{1,n}$ in $\mathbb{R}^n$. My short answer is that under rescaling by factor $\lambda$, derivatives scale by $\lambda$ and volumes by $\lambda …