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I am reading through "A GEOMETRIC APPROACH TO THE CALDERON–ZYGMUND ESTIMATES" by Lihe Wang and I am perplexed by an assertion in Lemma 7. The claim is that whenever $\Delta u = f$:

$$\frac{1}{|B_{4}|} \int_{B_{4}} |D^{2}u|^2 \leq 2^{n}$$ if $$\frac{1}{|B_{r}(x)|}\int_{B_{r}(x)} |D^{2}u|^2 \leq 1$$ for some $x \in B_{1},$ all $B_{r}(x) \subset \Omega$

There is an assumption that $B_{4} \subset \Omega,$ and clearly $B_{4} \subset B_{5}(x)$ but what if $B_{5}(x) \not \subset \Omega$?

I have tried to show $\int_{B_{4}(x)}|D^{2}u|^2 \leq \int_{B_{6}(x)}|D^{2}u|^2 \leq 2^{n}\int_{B_{3}(x)}|D^{2}u|^2$ but haven't figured it out. Is this the right approach?

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    $\begingroup$ In many cases like this there is a typo or other minor oversight that can be corrected with small changes to the text. For instance, here, perhaps all that is needed is that the assumption $B_4 \subset \Omega$ be replaced by $B_5 \subset \Omega$. I would continue reading and see if this sort of fix doesn't already resolve all issues. (See also my essay terrytao.wordpress.com/advice-on-writing-papers/… ) $\endgroup$
    – Terry Tao
    Commented Sep 30, 2023 at 0:00
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    $\begingroup$ Provide a link to the paper you are talking about. It is not my job to find it. If you want help, help those you want to help you. $\endgroup$
    – Piotr Hajlasz
    Commented Sep 30, 2023 at 1:58
  • $\begingroup$ The statement you are trying to prove is not true (even for $f=0$) if $\Omega =B_4$, so there is for sure some typo in the paper, as other pointed out. Note also that the main theorem in the paper is an interior estimate (solution in $B_2$ implies estimate in $B_1$), so you don't really care. My advice with interior estimate would be to allow yourself the space you need to forget all this empty-information stuff. Even if you start from $B_2$ and obtain the estimate in $B_{1/100}$, you can recover the same result for $B_1$ covering just at the end. $\endgroup$
    – Michele Caselli
    Commented Oct 2, 2023 at 8:38

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