4
$\begingroup$

Let $u(x)$ be a harmonic polynomial in the unit ball $B_1(0)\subset\mathbb{R}^n$ with $u(0)=0$.

For $0<r\leq1$, consider the average of its Dirichlet integral $$A(r):=\frac1{\vert B_r(0)\vert}\int_{B_r(0)}\vert\nabla u\vert^2dx,$$ and the average of the square function on the boundary $$B(r):=\frac1{\vert \partial B_r(0)\vert}\int_{\partial B_r(0)}u^2d\sigma.$$

I would like to ask:

QUESTION. Is this true? The ratio $\frac{r^2A(r)}{B(r)}$ is a constant in $r$.

Improve this question
$\endgroup$
2
  • $\begingroup$ Is $u$ real-valued, or did you mean $|u|^2$ in place of $u^2$? $\endgroup$
    – Iosif Pinelis
    Commented Jun 21, 2022 at 17:06
  • $\begingroup$ Yes, $u$ is real-valued. $\endgroup$
    – T. Amdeberhan
    Commented Jun 21, 2022 at 17:15

1 Answer 1

Reset to default
6
$\begingroup$

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

Improve this answer
$\endgroup$
4
  • $\begingroup$ Ouch! I know what I missed in the assumptions. $\endgroup$
    – T. Amdeberhan
    Commented Jun 21, 2022 at 21:52
  • 2
    $\begingroup$ @T.Amdeberhan : Despite the simplicity of the example, I spent quite a bit of time to find it. So, It would be great if you would reverse your latest edit. $\endgroup$
    – Iosif Pinelis
    Commented Jun 21, 2022 at 22:07
  • $\begingroup$ Done, Iosif Pinelis, now there is a Part II. $\endgroup$
    – T. Amdeberhan
    Commented Jun 21, 2022 at 22:16
  • $\begingroup$ Why the downvote? Can you explain? $\endgroup$
    – Iosif Pinelis
    Commented Jun 22, 2022 at 13:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .