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)
$$
\begin{equation}
\tag{2}
\begin{aligned}
0=-\int\_{\partial B\_1} \int\_0^1|\frac{\partial u}{\partial r}|^2 r^{n-1}+\int\_{\partial B\_1}u \frac{\partial u}{\partial \nu}-\int\_0^1r^{n-3}\int\_{\partial B\_1}|\partial\_{\theta}u|^2(r\theta)\\
=-\int\_{B\_1} |\frac{\partial u}{\partial r}|^2 +\int\_{\partial B\_1}u \frac{\partial u}{\partial \nu}-\int\_{ B\_1}|\partial\_{\tau}u|^2\\
=-\int\_{B\_1}|Du|^2+\int\_{\partial {B\_1}}u\frac{\partial u}{\partial \nu}.
\end{aligned}
\end{equation}
$$



An inequality for harmonic functions
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,
$$
\int\_{S\_1} u\partial\_\nu u=\sum\_{n,j} a^2\_{n,j}n\leq\sum\_{n,j} a^2\_{n,j}n(n+d-2)=\int\_{S\_1} |\partial\_\tau u|^2.
$$


