# Higher integrability for Sobolev functions - part 2

This is a follow-up to the question asked in Higher integrability for Sobolev functions

Updated question: Given the very helpful counterexamples and the ideas, I have the following question: Suppose for each $$r>0$$, there holds $$\int_{B_r(x)} |\nabla u|^2 dy \leq \frac{r}{c}\int_{\partial B_r(x)} |\nabla u|^2 d\sigma$$ for some $$c \in (0,2)$$ which is a uniform constant, then does this imply some form of higher integrability?

Clearly, when $$c=2$$, it is easy to see that $$u \in W^{1,\infty}_{loc}(\mathbb{R}^2)$$, so does $$c<2$$ still retain some form of higher integrability?

• Perhaps it is worth looking at an intermediate version that is between what you have here and your previous post: for every $x$, every $r \in (0,1]$ and every $\lambda \in (0,1]$ you have $$\int_{B_{\lambda r}(x)} |\nabla u|^2 \leq \lambda^c \int_{B_r(x)} |\nabla u|^2$$ (this is implied by what you posed above, and is strictly stronger than what you had in the last question. Feb 10 at 14:10
• Come to think of it, the spaces defined by the inequality I wrote in the previous comment are precisely the Morrey spaces. And so the answer there is also negative: the only embeddings between the Morrey spaces on bounded sets are those given by Holder's inequality. Feb 10 at 14:50
• You are right, the Morrey embedding is false, but for solutions, maybe this or the inequality you wrote might imply something better in the Sobolev sense, for solutions. This is what I'm trying to understand
Feb 10 at 17:27
• Or the inequality in the question plus something additional information from the equation could give some better Sobolev regularity, if yes, what could that look like? And if no, what's the obstruction?
I noticed derivative of volume integral is the surface one. Then $$\varphi(r) = \int_{B_r}$$ satisfies the differential inequality $$r\varphi'(r) \leq \varphi(r)$$ You could deduce properties from that.