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A Sobolev space is a vector space of functions equipped with a norm that is a combination of Lp-norms of the function itself and its derivatives up to a given order.

3 votes
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Sharpness of the Sobolev embedding theorem

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 …
Connor Mooney's user avatar
3 votes
Accepted

improved Sobolev embedding

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, …
Connor Mooney's user avatar
11 votes
Accepted

Is an $H_0^1$ function continuous to the boundary if it is continuous in the interior?

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.
Connor Mooney's user avatar
6 votes

Regularity of the quasi-linear PDE $-\Delta u + c(u) = f$

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), …
Connor Mooney's user avatar
20 votes
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What goes wrong for the Sobolev embeddings at $k=n/p$?

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 …
Connor Mooney's user avatar