6 votes
Accepted

A constant ratio of integrals? Part II

In fact, this is true for any homogeneous polynomial $u$ (not identically $0$), be $u$ harmonic or not. Indeed, let $m\ge1$ be the degree of such a polynomial $u$. Then $$u(tx)=t^m u(x)$$ and $$v(tx)=...
user avatar
6 votes
Accepted

A constant ratio of integrals? Part I

No. E.g., if $n=2$ and $u(x,y)=x + x^2 - y^2$ for $(x,y)\in\mathbb R^2$, then $\dfrac{r^2A(r)}{B(r)}=2\dfrac{1+2r^2}{1+r^2}$.
user avatar
3 votes

Uniqueness of Kantorovich potentials?

Uniqueness of Kantorovich potentials (up to a constant shift) has been analyzed in a very general framework in this work of ours: "On the Uniqueness of Kantorovich Potentials" - https://...
user avatar
3 votes
Accepted

Bounding integral expression with total variation of integrand

$\newcommand{\ep}{\epsilon}\newcommand{\R}{\mathbb R}$Yes, this is true. Indeed, for $\ep\in(0,1/2]$ we have \begin{equation} \mathcal I\le2\|f\|_\infty^2\, J, \tag{1}\label{1} \end{equation} ...
user avatar
3 votes

Local differentiability of eigenvalues and eigenvectors of a real symmetric matrix

Theorem (1.1) of Perturbation theory for normal operators is likely what you are looking for. See also Differentiable perturbation of unbounded operators. If you would order the eigenvalues by their ...
user avatar
2 votes

A functional inequality which calculates the limitation of human eyes

Nice questions, with nice motivation. I think that $f_-(x) = x^2$ and $f_+(x) = 2x-x^2 = 1-(1-x)^2$ is a non-trivial solution. Indeed, for every $x$ and $y$ in $[0,1]$, $$0 \le x^2 \le x \le 2x-x^2 \...
user avatar
1 vote
Accepted

Sobolev estimates $\|\nabla\phi\|_{\infty}\leq C\|\phi\|_{H^2}$

In $d \geq 3$ the answer is no from scaling argument. WLOG we can assume $0\in \Omega$ (by translation) and that $B(0,r_0)\subset\Omega$. Take $\phi\in C^\infty_0(B(0,r_0))\subset C^\infty_0(\Omega)$. ...
user avatar
  • 31.3k

Only top scored, non community-wiki answers of a minimum length are eligible