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In a paper that I am reading the author quotes the following result about harmonic functions. According to him this should be "easy to show" but I don't seem to be able to do so.

Let $u:\overline{B^n}\to \mathbb{R}$ be harmonic, where $\overline{B^n}\subset\mathbb{R}^n$ is the closed unit ball. I would like to prove that $$ \int_{B_1} |Du|^2\leq \int_{\partial B_1} |\partial_{\tau}u|^2, $$ where $\partial_{\tau}u$ is the tangential derivative of $u$. I tried to use both the Gauss-Green theorem and to write the Laplacian in spherical coordinates, but I always get stuck.


Here are some correct but apparently useless computations.

Using the Gauss-Green formula we get $$ \begin{equation} 0=\int_{B_1} u\Delta u =-\int_{B_1}|Du|^2+\int_{\partial {B_1}}u\frac{\partial u}{\partial \nu}. \end{equation} $$ On the other hand, the expression of the Laplacian in spherical coordinates is $$ \begin{equation} \begin{aligned} 0=\Delta u=\frac{\partial^2u}{\partial r^2} +\frac{n-1}{r}\frac{\partial u}{\partial r}+ \frac{1}{r^2}\Delta_{\partial B_1} u\\ =\frac{1}{r^{n-1}}\frac{\partial}{\partial r}(r^{n-1}\frac{\partial u}{\partial r})+ \frac{1}{r^2}\Delta_{\partial B_1} u, \end{aligned} \end{equation} $$ and integrating this against $u$ and noting that $\partial_{\tau}u=1/r\partial_{\theta}u$ we find the same expression as before (obviously) $$ \begin{equation} \tag{2} \begin{aligned} 0=-\int_{\partial B_1} \int_0^1|\frac{\partial u}{\partial r}|^2 r^{n-1}+\int_{\partial B_1}u \frac{\partial u}{\partial \nu}-\int_0^1r^{n-3}\int_{\partial B_1}|\partial_{\theta}u|^2(r\theta)\\ =-\int_{B_1} |\frac{\partial u}{\partial r}|^2 +\int_{\partial B_1}u \frac{\partial u}{\partial \nu}-\int_{ B_1}|\partial_{\tau}u|^2\\ =-\int_{B_1}|Du|^2+\int_{\partial {B_1}}u\frac{\partial u}{\partial \nu}. \end{aligned} \end{equation} $$

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  • $\begingroup$ What about using the mean value property for each partial derivative, which are also harmonic? $\endgroup$
    – username
    Dec 3, 2022 at 10:37
  • $\begingroup$ Crossposted to math.stackexchange math.stackexchange.com/questions/4589982/… $\endgroup$
    – No-one
    Dec 3, 2022 at 18:41

1 Answer 1

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Consider first the $d=2$ case. Then, $u$ is a real part of an analytic function. We can write $$u(z)=\frac12\sum_{n=0}^{\infty}(a_nz^n+\overline{a}_n\overline{z}^n)$$ and $$\partial_\nu u(z)=\frac12\sum_{n=0}^{\infty}(a_nnz^n+\overline{a}_nn\overline{z}^n)$$ for $|z|=1$. When multiplying them out and integrating over the circle, we get terms of the form $c\cdot z^m=c\cdot e^{im\theta}$ for some $m\in\mathbb{Z}$, which integrage to zero unless $m=0$. This yields $$ \int_{\partial B_1} u\partial_\nu u=\pi\sum_{n=0}^\infty n|a_n|^2\leq \pi\sum_{n=0}^\infty n^2|a_n|^2=\int_{\partial B_1} |\partial_\nu u|^2. $$

The generalization to higher dimensions is straightforward, for one can still write an arbitrary harmonic function as $f(x)=\sum_{n,j}a_{n,j}r^n Y_{n,j}(x/r)$, where $Y_{n,j}$ are spherical harmonics (hence mutually orthogonal).

Update: I misread the question and the above proof proves a different inequality - with $\partial_\nu$ instead of $\partial_\tau$ in the right-hand side - I'm assuming $\partial_\tau$ stands for the spherical gradient.

However, the same idea works in the case of OP's inequality. Note that the spherical harmonics are an orthonormalgonal basis of $L^2(S_1)$, and also $Y_{n,j}$ is an eigenfunction of the Laplace-Beltrami operation on the sphere with eigenvalue $n(n+d-2)$. In particular, they are orthogonal with respect to the Dirichlet form: $$ \int_{S_1}\partial_\tau Y_{n,j}\cdot\partial_\tau Y_{\hat{n},\hat{j}}= \int_{S_1}\Delta_{S_1} Y_{n,j} Y_{\hat{n},\hat{j}}=n(n+d-2)\delta_{n,\hat{n}}\delta_{j,\hat{j}}. $$ This means that if $u(x)=\sum_{n,j}a_{n,j}r^n Y_{n,j}(x/r)$, then, as explained above, $$ \int_{S_1} u\partial_\nu u=\sum_{n,j} a^2_{n,j}n\leq\sum_{n,j} a^2_{n,j}n(n+d-2)=\int_{S_1} |\partial_\tau u|^2. $$

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  • $\begingroup$ You are right my comment was wrong. Now deleted. $\endgroup$
    – username
    Dec 3, 2022 at 10:34
  • $\begingroup$ Nice answer, thank you. To show the case $d=2$ using complex analysis I guess that one just have to further notice that $\partial_\tau u(z)=\frac{1}{2}\sum_{n=0}^\infty (e^{i\frac{\pi}{2}}a_n n z^n +e^{-i\frac{\pi}{2}}\overline{a}_n n \overline{z}^n )$, and hence that $\int_{\partial B} |\partial_\tau u|^2=\pi \sum_{n=0}^\infty n^2 |a_n|^2 = \int_{\partial B} |\partial_\nu u|^2$. $\endgroup$
    – No-one
    Dec 3, 2022 at 18:47
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    $\begingroup$ In general, when trying to prove a Hilbert space (or "$L^2$ type") identity or inequality that enjoys a group action invariance (in this case, rotational invariance), one typically is able to reduce to irreducible representations of that group action as a "free move". In this particular context, this suggests that the most important case to deal with is when the functions involved behave like a spherical harmonic in the angular variable (i.e., one has separation of variables), which can motivate the argument given in this answer. $\endgroup$
    – Terry Tao
    Dec 3, 2022 at 20:11
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    $\begingroup$ Incidentally, the above analysis also reveals the energy equipartition identity $\int_{\partial B_1} |\partial_\tau u|^2 = \int_{\partial B_1} |\partial_\nu u|^2 + (d-2) \int_{B_1} |Du|^2$, which provides an alternate proof of the inequality when $d \geq 3$. It is a nice exercise to figure out an integration by parts proof of this identity. $\endgroup$
    – Terry Tao
    Dec 3, 2022 at 20:16
  • $\begingroup$ @TerryTao Many thanks for the comments. The energy equipartition identity in this case is just a restatement of Pohozaev identity, right? $\endgroup$
    – No-one
    Dec 4, 2022 at 14:03

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