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The folowing theorem

Let $u$ be a solution to $$ \left\{ \begin{array}{ccccc} \Delta u &=& f \chi_{\{u\neq 0\}} &\mbox{in}& B_1,\\ u &=& g &\mbox{on}& \partial B_1, \end{array} \right. $$ in a suitable weak sense and assume furthermore that $f=\Delta v$ where $v\in C^{1,1}(B_1)$ and that $g \in C( \partial B_1)$. Then $u \in C^{1,1}(B_{1/2})$ and $$\|D^{2}u\|_{L^{\infty}(B_{1/2})} \le C( \|u\|_{L^{1}(B_1)} + \|D^{2}v\|_{L^{\infty}(B_1)}),$$ where $C$ depends on the dimension.

This theorem suggests that there is a relation like $\|u\|_{C^{1,1}(B_{1/2})} \le C_n\|D^{2}u\|_{L^{\infty}(B_{1/2})}$. Is this what happens? You can find the details in reference 1 and I found a more simple case in the proof of theorem 1.1 on the page 11 in reference 2. Thank you.

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up vote 2 down vote accepted

Your last formula is missing a function. I'm going to assume you meant $$\|u\|_{C^{1,1}(B_{1/2})} \le C_n\|D^{2}u\|_{L^{\infty}(B_{1/2})}$$ This is a true statement. Indeed, the norm on the left is the Lipschitz norm of $\nabla u$. The norm on the right gives an upper bound for the 1st derivatives of $\nabla u$. A function with bounded partial derivatives on a convex set is Lipschitz, by the Mean Value Theorem (applied to restrictions to line segments).

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Ok, But. Do we need some hypothesis like $\nabla u \in C^1$ to apply the Mean Value theorem? If yes. Can we conclude that $\nabla u \in C^1$. – user29999 Jul 18 '12 at 23:59
@user29999 OK, this is not quite the MVT on line segments. The derivatives are taken in the weak sense, so $D^2u\in L^{\infty}$ really means $u\in W^{2,\infty}$. And since $W^{1,\infty}=\mathrm{C^{0,1}}$ on convex domains (standard fact), we also have $W^{2,\infty}=\mathrm{C^{1,1}}$. To show $W^{1,\infty}=\mathrm{C^{0,1}}$ one can still use the line segments, but not every one. The point is that a Sobolev function is absolutely continuous on a.e. line, which means it's the integral of its (bounded) derivative, hence Lipschitz. And "Lipschitz on a dense set" => "Lipschitz". – user31373 Jul 19 '12 at 0:07
Cool. Thank you. – user29999 Jul 19 '12 at 0:12

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