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Let us consider $\mathbb S^d$ the unit Euclidean sphere of $\mathbb R^{d+1}$ and let $\Delta_{\mathbb S^d}$ be the Laplace operator on $\mathbb S^d$. We have $$ -\Delta_{\mathbb S^d}=\sum_{k\in \mathb …

(1) Let me answer first to the last question: $\Delta \vert x\vert$ is homogeneous of degree $-1$ and radial. On $\mathbb R^d$ ($d\ge 2$) it is $$ (\partial_r^2+\frac{d-1}{r}\partial_r)(r)=\frac{d-1}{ …

$e^{it\Delta}$ is the Fourier multiplier $e^{-4it\pi^2\vert D\vert^2}$, i.e. the operator defined by $$ (e^{it\Delta} u)(x)=\int_{\mathbb R ^d} e^{2i\pi x \xi}e^{-4it\pi^2\vert \xi\vert^2}\hat u(\xi) …

Let me follow your notations with $\Delta=\sum_{1\le j\le 3}\partial_{x_j}^2$. You have with $r=\Vert x\Vert$ (the Euclidean norm) $$ r^2\Delta=(r\partial_r)^2+r\partial_r+\Delta_{\mathbb S^2},\quad\t …

Let $A$ be a selfadjoint (pseudo)differential operator of order 2 on $(M,g)$ with a nonnegative symbol. It is a consequence of the Fefferman-Phong inequality that $A$ is semi-bounded from below, i.e. …

The answer to the first question is negative on the Euclidean sphere $\mathbb S^2$. It is possible to prove that the Laplace operator on the sphere $\mathbb S^2$ is NOT the sum of two squares of smoot …

Isn't it $$ T(x,y)=\frac{1}{4π}\sum_{(k,l)\in \mathbb Z^2}\ln\bigl((x-k)^2+(y-l)^2\bigr)? $$ Reading the objections below, I note that $T$ is indeed well defined as a distribution on $\mathbb R^2$. Ta …