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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 …

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 …

One more perspective: after performing a diffeomorphism that locally flattens the boundary, we get an equation of the form $$a_{ij}(x)w_{ij} + b_i(x)w_i = f(x_n) \text{ in } B_1,$$ where the coefficie …

Yes, this is possible. Consider the intersection of the helicoid $z = \tan^{-1}(y/x)$ with its tangent plane $z = y$ at $(1,0,0)$. The projection of the intersection curves to $\{z = 0\}$ consists of …

One cannot weaken the regularity hypothesis on $A$ to $C^{\alpha}$. Consider for example the 1D case: when $f = 0$ the solution $u$ satisfies $$u'(x) = \frac{const.}{A(x)},$$ which is not in $H^1$ for …

For the second, rewrite $F(t) \leq \lambda$ as $$t - \lambda \leq \left(\int_{\mathbb{R}} a(s,\,t)\phi(s)\,ds\right)^q,$$ which reduces the problem to showing that $$\int_{\mathbb{R}} a(s,\,t)\phi(s)\ …

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 …

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 …

$U = 2x^2-y^2$ 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.

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 …

Below is a maximum principle-based alternative to the proof of Mateusz. We may assume that $r = 1$ by scaling. Let $w$ be the harmonic function on $B_1 \subset \mathbb{R}^n$ with boundary data $|\varp …

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), …

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 …

The discussion from Section 13.1 in the book of Gilbarg and Trudinger shows that $u \in C^{1,\,\alpha}\left(B_{3/4}^+\right)$. From here one can apply Schauder estimates for linear equations. For exam …

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| …