# Dirichlet energy of a harmonic function bounded above by the energy of the boundary function?

$\Omega$ is a domain in $R^2$ with sufficient smooth boundary. Given an absolutely continuous function f difined on $S^1$($[0,2\pi]$). Then there exists an unique harmonic function u defined on $\overline \Omega$ with boundary value f, which is the function minimizing the Dirichlet energy.

In the case $\Omega$ is a disk, is it true that the energy of u is bounded above by the energy of f along the boundary? ie $$\int_\Omega |\nabla u|^2 dxdy \leq \int_0^{2\pi} (\frac{d f}{d \theta })^2 d\theta$$

This is correct (if $f$ is real, of course). Use the Green formula to transform the LHS: $$\int_\Omega|\nabla u|^2dxdy=R\int_0^{2\pi}uu_rd\theta,$$ where $R$ is the radius of the disc. Now expand $$u(r,\theta)=\sum_{-\infty}^\infty r^{|n|}c_ne^{in\theta},$$ where $c_n=\overline{c_{-n}}$, because $f$ is real. Now compute, using orthogonality: $$R\int_0^{2\pi}uu_rd\theta=2\pi\sum_{-\infty}^\infty|n|R^{2n}|c_n|^2,$$ and $$\int_0^{2\pi}u_\theta^2d\theta=2\pi\sum_{-\infty}^\infty n^2R^{2n}|c_n|^2.$$ We conclude that LHS$\leq$RHS, with equality only if $c_n=0$ for $|n|\geq 2$.

As suggested in the very nice answer of Prof. Eremenko, I think the key point of this kind of inequality is homogeneity, and it is related to some monotonicity formulae for harmonic functions. We expand briefly below.

If we take the boundary data of $u$ and extend inwards to have homogeneity $\alpha$, i.e. let $v = r^{\alpha}u(x/r)$ (with $r=|x|$) then it is easy to verify $$|\nabla v(x)|^2 = r^{2\alpha-2}(|\nabla_Tu(x/r)|^2 + \alpha^2u(x/r)^2)$$ where $\nabla_Tu$ is the tangential part of the gradient, giving (in dimension $n$) that $$(n-2+2\alpha)\int_{B_1} |\nabla v|^2 = \int_{\partial B_1} (|\nabla_Tu|^2 + \alpha^2u^2).$$ Furthermore, since $v$ is a competitor for $u$ we obtain $$(n-2+2\alpha)\int_{B_1}|\nabla u|^2 \leq \int_{\partial B_1} (|\nabla_Tu|^2 + \alpha^2u^2).$$ Thus, we get a similar inequality to the one you want in $n \geq 3$ by taking $\alpha = 0$. Furthermore, in $\mathbb{R}^2$ we may take $\alpha$ to be the "homogeneity detected at radius $1$", $$\alpha = \frac{\int_{B_1}|\nabla u|^2}{\int_{\partial B_1}u^2}$$ to get $$\alpha\int_{B_1}|\nabla u|^2 \leq \int_{\partial B_1} u_{\theta}^2.$$ A very important monotonicity formula of Almgren (very much related to the solution given above by Prof. Eremenko) says that for nonconstant harmonic functions $\alpha \geq 1$ (roughly, the homogeneity detected at radius $r$ increases with $r$ and starts off at least $1$ since $u$ blows up to a plane).

Some nice notes on Almgren's monotonicity are here.

One of its interesting applications is a unique continuation theorem due to Garofalo and Lin, which can be found here.

• Monney:for general domain in $R^n$(or $R^2$), is there a similar inequality? – jiangsaiyin Jan 11 '14 at 8:47
• @Connor Mooney, could you give a reference for the monotonicity formula of Almgren? – Alexandre Eremenko Jan 11 '14 at 15:40
• @AlexandreEremenko: References for Almgren's monotonicity and a nice application are now included in my answer. – Connor Mooney Jan 11 '14 at 16:05
• @Connor Mooney: thanks! Do you think your inequality for $n\geq 3$ is best possible? I suppose the equality must be attained on some spherical harmonics. On the other hand, if you compute $u_T$ for the spherical harmonics, probably my argument will also work for $n\geq 3$. – Alexandre Eremenko Jan 11 '14 at 16:07
• So for $n=2$ and $n=3$ the inequality holds exactly as written, and for $n>3$ we have a constant factor $n-2$ in front of energy? – Alexandre Eremenko Jan 11 '14 at 16:16