Integral inequality for harmonic functions: $\frac{\int_{|w|=1}u^2(Rw)dS(w)}{\int_{|w|=1}u^2(rw)dS(w)}\leq \left(\frac{R}{r}\right)^{2N(R)}$

Question

$u$ be a nonzero harmonic function on the ball of radius $1$ centred at origin, $B_1(0)$ in $\mathbb{R}^n$. Lets set for $0<r<1$ : $$N(r)=\frac{r\int_{B_r(0)}|\nabla u|^2dx}{\int_{\partial B_r(0)}u^2dS(x)}$$ Then

(1) $N(r)$ is non-decreasing in $r$.

(2) $\lim\limits_{r\to 0^+}N(r)=?$

(3) For $0<r<R<1$ we have $$\frac{\int_{|w|=1}u^2(Rw)dS(w)}{\int_{|w|=1}u^2(rw)dS(w)}\leq \left(\frac{R}{r}\right)^{2N(R)}$$

My Try : After change of variables and by Gauss-divergence theorem I arrived at $$N(r)=\frac{r\int_{|w|=1}u(rw)\nabla u(rw)\cdot wdS(w)}{\int_{|w|=1}u^2(rw)dS(w)}$$. Denoting $g(r)=\int_{|w|=1}u^2(rw)dS(w)$ we see since $u$ is harmonic its analytic in $B_1(0)$ thus $g$ is analytic in $(0,1)$. Now if we suppose $u(0)\neq 0$ then by Dominated Convergence Theorem, I get $\lim_{r\to 0^+}N(r)=0$. If $g^{(k)}(0)\neq 0$ for the first $k$ in the power series of $g$, then by L'Hospital I get $\lim_{r\to 0^+}N(r)=\frac{k}{2}$. I still could not manage to show $N(r)$ is non-decreasing. Any idea to approach (1)-(3)?

I arrived at the below inequality which would show (1) if the following is true $$\int_{B_1(0)}|\nabla u(rw)|^2\cdot\int_{\partial B_1(0)}u^2(rw)+2r\int_{\partial B_1(0)}\left[u(rw)\sum_{1\leq i,j\leq n}w_iw_ju_{x_ix_j}(rw)\right]\cdot\int_{\partial B_1(0)}u^2(rw)\geq \frac{r}{2}\left[\int_{B_1(0)}|\nabla u(rw)|^2\right]^2$$

Answer 1

Log-convexity

Let's focus on the quantity $$ f(r) = r^{1-n}\int_{\partial B_r } u^2 $$ which, up to normalization, is the average of $u^2$ on the sphere of radius $r$. The derivative of the average with respect to $r$ is just the average of normal derivative of $u^2$, that is $$ f'(r) = r^{1-n} \int_{\partial B_r} 2u \frac{\partial u}{\partial n} $$ Since $$ \int_{\partial B_r} u \frac{\partial u}{\partial n} = \int_{B_r}\operatorname{div}\left(u\nabla u\right) = \int_{B_r}\left(|\nabla u|^2 + u\Delta u\right) = \int_{B_r} |\nabla u|^2 $$ the conclusion is that $$ N(r) = \frac{1}{2} \frac{rf'(r)}{f(r)} = \frac12 \frac{d\log f}{d\log r} $$ The property that $N$ is non-decreasing is equivalent to $\log f$ being a convex function of $\log r$.

The part I don't like

A harmonic function in a ball can be expanded into homogeneous harmonic polynomials: $u= \sum_{k=0}^\infty p_k$ where each $p_k$ is a harmonic polynomial, homogeneous of degree $k$. Moreover, $p_k$ are mutually orthogonal when integrated over a sphere. All this tells us that $$ f(r) = \sum_{k=0}^\infty r^{1-n} \int_{\partial B_r } p_k^2 $$ But for each $k$, the function $r^{1-n} \int_{\partial B_r } p_k^2$ is just a constant multiple of $r^{2k}$ due to homogeneity. Its logarithm is $2k\log r + C$, which is a convex function of $\log r$. It remains to use the fact that the sum of log-convex functions is log-convex to conclude that $f$ is a log-convex function of $\log r$, as was claimed.

Limit as $r\to 0$

Let $j$ be the smallest index for which the homogeneous polynomial $p_j$ is not identically zero. Then $p_j$ dominates the integral over $\partial B_r$ when $r$ is small, which means that when computing $\lim_{r\to 0} N(r)$ we can focus on $p_j$. With $u$ replaced by $p_j$, the ratio $N(r)$ is exactly $j$; so that is the limit.

In other words, it's the order of vanishing of $u$ at the center of the ball.

Inequality (3)

Take logarithm of both sides: the inequality becomes $$ \log f(R) - \log f(r) \le 2N(R) (\log R-\log r ) $$ or equivalently $$ \frac{\log f(R) - \log f(r)}{ \log R-\log r } \le 2N(R) $$ Think of the graph of $\log f$ with $\log r$ as horizontal axis. The quantity on the left is the slope of the secant over the interval $[\log r,\log R]$. The quantity on the right is the slope of the tangent at $\log R$. The inequality holds by convexity.

Answer 2

We write $N(r)=\frac{f(r)}{g(r)}$ where $f(r)=r\int_{B_r}|\nabla u|^2 dx=r\int_{0}^{r}\int_{\partial B_t}|\nabla u|^2 dS dt$ and $g(r)=\int_{\partial B_r}u^2 dS$. Now $N'(r)=\frac{f'g-fg'}{g^2}$ therefore we compute $f'=\int_{B_r}|\nabla u|^2+r\int_{\partial B_r}|\nabla u|^2$ and $g'=\frac{n-1}{r}\int_{\partial B_r}u^2+2\int_{\partial B_r}uu_r$. Here $u_r=\frac{\partial u}{\partial r}$. Now arranging terms we get $$f'g-fg'=\int_{\partial B_r}u^2\cdot\left(\int_{\partial B_r}(r|\nabla u|^2-(n-2)uu_r)\right)-2r\left(\int_{\partial B_r}uu_r\right)^2$$

Claim : $$\int_{\partial B_r}\left[r|\nabla u|^2-2r(u_r)^2-(n-2)uu_r\right]=0$$ For the moment assuming this we get $$f'g-fg'=\int_{\partial B_r}u^2\cdot\left(\int_{\partial B_r}2r(u_r)^2\right)-2r\left(\int_{\partial B_r}uu_r\right)^2\geq 0$$ By Cauchy-Schwartz and $N'(r)\geq 0$ follows.

For the claim we see by taking $$F=|\nabla u|^2 x-2(\nabla u \cdot x)\nabla u-(n-2)u\nabla u$$ and using gauss divergence $$\int_{\partial B_r}\left[r|\nabla u|^2-2r(u_r)^2-(n-2)uu_r\right]=\int_{\partial B_r}F\cdot \nu=\int_{B_r}\nabla\cdot F=0$$ Since by a straight forward calculation we have $\nabla\cdot F=0$.