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I know this direct proof which works also for $\infty$. Assume $2/q=1/p+1/r$ and $q \geq 2$. Then $$ \int|D_ku|^q=\int D_k u D_k u |D_k u|^{q-2}=-(q-1)\int u|D_k u|^{q-2} D_{kk}u \leq (q-1)\int |u||D_ …

The sup-norm of a (normalized) spherical harmonic of degree $k$ is bounded by $C \sqrt d_k$, where $d_k$ is the dimension of all spherical harmonics of order $k$ and $C$ is also given. The estimate is …

Let $\psi$ be harmonic in $\Omega$, with $\psi=\phi$ on $\cup _{i \in S} \Gamma_i$, $\psi=m$ on $\cup_{i \not \in S}\Gamma_i$, where $m=\max_{i \not \in S} \max_{\Gamma_i} \phi$. By comparison, $\psi …

The first observation is that the $u$ above satisfies $\nabla u=0$ on $\partial \Omega$ if and only if $f$ is orthogonal to all harmonic functions $v$ in $\Omega$, continuous up the the boundary. In f …

This follows from the mean value theorem. Assume that (up to a subsequence) $w_n(x_n) \geq -B$ with $(x_n) \in K$ (a compact subset of $\Omega$). If $x_n \to x_0 \in K$ and $B(x_n,r) \in \Omega$ for e …

This is a direct proof which gives $V \in C^\infty (\bar \Omega)$ whenever $g \in C^\infty (\bar \Omega)$, $V$ being the Newtonian potential of $g$. As in the proof by @Terry Tao assume that locally $ …

There is no hope to gain summability without using boundary conditions. For example the function $\frac{1}{z \log z}$ is holomorphic, hence harmonic, and in $L^2$ in the disc (in the complex plane) c …

Let us assume Korn's inequality in the usual form $$\int_{\Omega} |\nabla u|^2 \leq C\left (Q(u)+\int_\Omega |u|^2 \right ),$$ with $Q(u)=\int_\Omega |\nabla u +\nabla u^{T}|^2$, for every $u \in H^1( …

Let us assume some regularity on $\partial D$ (bounded and $C^2$ suffices). Then the problem above has a unique solution $u \in W^{2,p}(\Omega) \cap W^{1,p}_0(\Omega)$ for every $p<\infty$ and taking …

This is a way to get an a-priori estimate, if I understood correctly the question. Multiply by $A$ and integrate by parts the left-hand-side. Then $$\int_{R^3}(A^2u^2+|\nabla A|^2)=-\int_{R^3}Au\nabla …

There is a trick that reduces the equation $u_t=Lu$, $L=\Delta-x \nabla$ to the heat equation $u_t-\Delta$. It is genuinely parabolic and gives the parabolic kernel in the whole space, from which the …

Let me give a positive answer perhaps omitting some details. Fact 1. Let $u'' \geq ku^\alpha$ in $[c,\ell[$ with $k>0, \alpha>1$ and $u,u' \geq 0$. Let $A=u(c)$, then $ \ell \to c$ as $A \to \infty$ …

The first is not true, and probably also the others. Take $L^2(0, \pi)$ and $u_1=\sin x$, $u_2=\sin (2x)$, so that $E_2=\{u=a\sin x+b \sin (2x)\}$ and $u \geq 0$ iff $a \geq 0$ and $2|b| \leq a$. If $ …

The result is true. Let $L=\sum_{ij}a_{ij}D_{ij}$ and consider $$L^{-1}: C^{2+\alpha}(\partial \Omega) \mapsto C^{2+\alpha}(\bar \Omega)$$ with $L^{-1}f=u$ is $Lu=0$ and $u=f$ at the boundary. $L^{-1} …

Yes, $u$ is strictly positive. Assume that $F=\{x:u(x)=0\}$ is non-empty. $F$ is clearly closed and I show that is open. Let $x_0 \in F$ and $r>0$ such that $u-\Delta u=u(1+Ku^{\frac{4}{n-2}}) \geq 0$ …