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Observe that if $u$ is homogeneous of degree $\alpha$, then $N(r) \equiv \alpha$. (After integrating by parts, the numerator becomes $r \int_{\partial B_r} u u_{\nu} = \alpha \int_{\partial B_r} u^2$. …

This is a nice question. In my view, the choice of test function is motivated by the idea of taking some positive quantity that is a supersolution to an elliptic equation, and looking at the equation …

A linear uniformly elliptic equation of the form $F(D^2u) = 0$ can only be $\Delta u = 0$ (up to an affine transformation of the solution) since the only inputs are the second derivatives. Equations l …

One approach is to observe that $$\|u_0\|_{L^{\infty}(B_1^+)} \geq \frac{1}{16n}|f(0)|.$$ It suffices by linearity to prove this when $f(0) = 1$. Consider the barrier $$b(x) = \frac{1}{2n}\left(\left| …

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), …

Consider $u_{\epsilon} = \rho_{\epsilon} \ast \log$ in $B_{1/2} \subset \mathbb{R}^2$, where $\rho_{\epsilon}$ are standard mollifiers. Then $|u_{\epsilon}(0)|$ blows up as $\epsilon \rightarrow 0$, a …

Yes, I think the estimate you propose is true. As a simple case take $f = 0$ and $M = B_1 \subset \mathbb{R}^n$. By adding a constant assume that $\inf_{B_1} u = 0$ and $\sup_{B_1}u = K,$ and note tha …

Another argument perhaps worth mentioning uses the maximum principle. To illustrate the idea we assume that $\Omega$ is a smooth bounded domain, that $L = \Delta$, and that we are dealing with Dirichl …

It is true and well-known (assuming $F$ is e.g. uniformly elliptic). The idea is sketched in ch. 9 of the book by Caffarelli-Cabre, but I am not sure of a precise reference at this level of generality …

Here is a counterexample: let $$u(x,y) = e^{-x^2}\sinh(y).$$ Then $$\det D^2u = -2e^{-2x^2}(\sinh^2(y) + 2x^2) < 0 \text{ ơn } \mathbb{R}^2 \backslash \{0\},$$ and the equation $$u_{xx} + (2-4x^2)u_{y …

Yes, this inequality is true, and in fact the assumption about the smallness of $D^2u$ in $L^{2p}$ is not needed. Using that $|\nabla u|$ is $\epsilon$-close to one in $L^1(B_2)$, we can find a point …

It depends on the geometry of the domain. Consider e.g. the function $f(x,\,y) = xy$ on $\{|xy| < 1\}$, which is bounded with constant Hessian on this domain but has linearly growing gradient. Or take …

The ABP estimate indeed holds in your setting. The key is that the concave envelope of $u$ is in $C^{1,\,1}$, so the area formula is valid for its gradient. Assuming for simplicity that $L = \Delta$, …

The proposition is true. One can argue as follows: for $x \in \Omega$, let $L$ be a supporting linear function to $u$ at $x$. Then $L < u$ on $\partial \Omega$ by (b). For $h > 0$ small we conclude th …

$U = 2x^2-y^2$ and $u = x^2-2y^2$ solve constant-coefficient elliptic equations, but $U - u = x^2 + y^2$ has an interior minimum and thus cannot solve an elliptic equation.