Showing harmonicity of $\frac{1}{|x|}H(\frac{x}{|x|^2})$ with $H$ harmonic on $\mathbb{R}^3$.

Question

In this year's first exam, my teacher gave the following problem:

State and prove the spherical average theorem for harmonic functions in open sets of $\mathbb{R}^n$. Let $H$ be a harmonic function on $\mathbb{R}^3$. Show that the function $x\mapsto|x|^{-1}H(\frac{x}{|x|^2})$ is harmonic in $\mathbb{R}^3\smallsetminus\{0\}$.

I am a bit at a loss as to how to do the last part (the rest is easy). I tried calculating the laplacian and stopped midway because it seemed crazy. I will now try to end the calculations here, noting $\frac{1}{|x|}$ is harmonic in $\mathbb{R}^3\smallsetminus\{0\}$ and has gradient $-\frac{x}{|x|^3}$:

\begin{align*} \Delta\left(\frac{1}{|x|}H\left(\frac{x}{|x|^2}\right)\right)={}&\Delta\left(\frac{1}{|x|}\right)H\left(\frac{x}{|x|^2}\right)+2\nabla\left(\frac{1}{|x|}\right)\cdot\nabla\left(H\left(\frac{x}{|x|^2}\right)\right)+\frac{1}{|x|}\Delta\left(H\left(\frac{x}{|x|^2}\right)\right)={} \\ {}={}&-2\frac{x}{|x|^3}\cdot\left(\nabla H\left(\frac{x}{|x|^2}\right)\cdot\frac{1}{|x|^2}-2\nabla H\left(\frac{x}{|x|^2}\right)\cdot x\cdot\frac{x}{|x|^4}\right)+{} \\ &{}+\frac{1}{|x|}\Delta\left(H\left(\frac{x}{|x|^2}\right)\right)={} \\ {}={}&-\frac{2}{|x|^3}\partial_\nu H\left(\frac{x}{|x|^2}\right)+\frac{1}{|x|}\frac{2}{|x|^2}\partial_\nu H\left(\frac{x}{|x|^2}\right), \end{align*}

where $\partial_\nu$ at the end denotes the normal derivative on an appropriate ball. So those cancel out. Let me now show the justifications for the expressions at the end:

\begin{align*} \partial_j\left(H\left(\frac{x}{|x|^2}\right)\right)={}&\sum_k\partial_kH\left(\frac{x}{|x|^2}\right)\partial_j\frac{x_k}{|x|^2}=\sum_k\partial_kH\left(\frac{x}{|x|^2}\right)\left(\frac{\delta_{jk}}{|x|^2}-\frac{2x_kx_j}{|x|^4}\right)={} \\ {}={}&\nabla H\left(\frac{x}{|x|^2}\right)\cdot\left(\frac{e_j}{|x|^2}-\frac{2x_jx}{|x|^4}\right). \end{align*}

So the gradient of our composite function should be:

$$\nabla\left(H\left(\frac{x}{|x|^2}\right)\right)=\frac{1}{|x|^2}\nabla H\left(\frac{x}{|x|^2}\right)-\frac{2x}{|x|^4}\nabla\left(\frac{x}{|x|^2}\right)\cdot x.$$

And now things get messy as we deal with horrid pure second partials:

\begin{align*} \partial_j^2\left(H\left(\frac{x}{|x|^2}\right)\right)={}&\partial_j\left(\sum_k\partial_kH\left(\frac{x}{|x|^2}\right)\left(\frac{\delta_{jk}}{|x|^2}-\frac{2x_jx_k}{|x|^4}\right)\right)={} \\ {}={}&\sum_k\partial_j\left(\partial_kH\left(\frac{x}{|x|^2}\right)\left(\frac{\delta_{jk}}{|x|^2}-\frac{2x_jx_k}{|x|^4}\right)\right)+{} \\ &{}+\sum_k\partial_kH\left(\frac{x}{|x|^2}\right)\left(\frac{-2\delta_{jk}x_j}{|x|^4}-\frac{2x_k}{|x|^4}-\frac{2x_j\delta_{jk}}{|x|^4}+\frac{8x_j^2x_k}{|x|^6}\right)={} \\ {}={}&\sum_k\sum_\ell\partial_\ell\partial_kH\left(\frac{x}{|x|^2}\right)\left(\frac{\delta_{\ell j}}{|x|^2}-\frac{2x_\ell x_j}{|x|^4}\right)\left(\frac{\delta_{jk}}{|x|^2}-\frac{2x_jx_k}{|x|^4}\right)+{} \\ &{}+\sum_k\partial_kH\left(\frac{x}{|x|^2}\right)\left(-\frac{2\delta_{jk}x_j}{|x|^4}-\frac{2x_k}{|x|^4}-\frac{2x_j\delta_{jk}}{|x|^4}+\frac{8x_j^2x_k}{|x|^6}\right)={} \\ {}={}&\sum_k\sum_\ell\partial_\ell\partial_kH\left(\frac{x}{|x|^2}\right)\left(\frac{\delta_{\ell j}}{|x|^2}-\frac{2x_\ell x_j}{|x|^4}\right)\left(\frac{\delta_{kj}}{|x|^2}-\frac{2x_kx_j}{|x|^4}\right)+{} \\ &{}-\frac{4x_j}{|x|^4}\partial_jH\left(\frac{x}{|x|^2}\right)-\frac{2}{|x|^4}\nabla H\left(\frac{x}{|x|^2}\right)\cdot x+\frac{8x_j}{|x|^6}\nabla H\left(\frac{x}{|x|^4}\right)\cdot x={} \\ {}={}&\partial_j^2H\left(\frac{x}{|x|^2}\right)\frac{1}{|x|^4}-\frac{2}{|x|^6}\sum_k\partial_j\partial_kH\left(\frac{x}{|x|^2}\right)x_jx_k+{} \\ &{}-\frac{2}{|x|^6}\sum_\ell\partial_\ell\partial_jH\left(\frac{x}{|x|^2}\right)x_\ell x_j+\frac{4}{|x|^8}x_j^2\sum_{\ell,k}\partial_\ell\partial_kH\left(\frac{x}{|x|^2}\right)x_\ell x_k+{} \\ &{}-\frac{4x_j}{|x|^4}\partial_jH\left(\frac{x}{|x|^2}\right)-\frac{2}{|x|^4}\nabla H\left(\frac{x}{|x|^2}\right)\cdot x+\frac{8x_j}{|x|^6}\nabla H\left(\frac{x}{|x|^4}\right)\cdot x={} \\ {}={}&\partial_j^2H\left(\frac{x}{|x|^2}\right)\frac{1}{|x|^4}-\frac{4}{|x|^6}\sum_k\partial_k\partial_jH\left(\frac{x}{|x|^2}\right)x_jx_k+{} \\ &{}+\frac{4}{|x|^8}x_j^2\sum_{\ell,k}\partial_\ell\partial_kH\left(\frac{x}{|x|^2}\right)x_\ell x_k-\frac{4x_j}{|x|^4}\partial_jH\left(\frac{x}{|x|^2}\right)-\frac{2}{|x|^4}\nabla H\left(\frac{x}{|x|^2}\right)+{} \\ &{}+\frac{8x_j^2}{|x|^6}\nabla H\left(\frac{x}{|x|^2}\right)\cdot x. \end{align*}

Oh my. Now, summing over $j$, I should get:

\begin{align*} \Delta\left(H\left(\frac{x}{|x|^2}\right)\right)={}&\underbrace{\Delta H\left(\frac{x}{|x|^2}\right)}_{0\text{ by harmonicity of }H}-\underline{\frac{4}{|x|^6}x\cdot\operatorname{Hess}_H\left(\frac{x}{|x|^2}\right)\cdot x}+\underline{\frac{4}{|x|^6}x\cdot\operatorname{Hess}_H\left(\frac{x}{|x|^2}\right)\cdot x}+{} \\ &{}-\frac{4}{|x|^4}\nabla H\left(\frac{x}{|x|^2}\right)\cdot x-\frac{2\cdot3}{|x|^4}\nabla H\left(\frac{x}{|x|^2}\right)+\frac{8}{|x|^4}\nabla H\left(\frac{x}{|x|^2}\right)\cdot x={} \\ {}={}&-\frac{2}{|x|^2}\nabla H\left(\frac{x}{|x|^2}\right)\cdot\frac{x}{|x|^2}=-\frac{2}{|x|^2}\partial_\nu H\left(\frac{x}{|x|^2}\right). \end{align*}

Are we done? No! Now we need the other term, the gradient of $\frac{1}{|x|}$ scalar product with the gradient of $H(\frac{x}{|x|^2})$. Recall we found:

$$\nabla\left(H\left(\frac{x}{|x|^2}\right)\right)=\frac{1}{|x|^2}\nabla H\left(\frac{x}{|x|^2}\right)-\frac{2x}{|x|^4}\nabla\left(\frac{x}{|x|^2}\right)\cdot x.$$

So the other missing term would be:

\begin{align*} 2\nabla\left(\frac{1}{|x|}\right)\cdot\nabla\left(H\left(\frac{x}{|x|^2}\right)\right){}={}&-\frac{2x}{|x|^3}\cdot\frac{1}{|x|^2}\nabla H\left(\frac{x}{|x|^2}\right)-\frac{2x}{|x|^4}\nabla\left(\frac{x}{|x|^2}\right)\cdot x={} \\ {}={}&-2\sum_j\partial_jH\left(\frac{x}{|x|^2}\right)\frac{2x_j}{|x|^5}+\frac{8}{|x|^7}|x|^2\nabla H\left(\frac{x}{|x|^2}\right)\cdot c={} \\ {}={}&-2\nabla H\left(\frac{x}{|x|^2}\right)\cdot\frac{2x}{|x|^5}+\frac{8}{|x|^5}\nabla H\left(\frac{x}{|x|^2}\right)\cdot x={} \\ {}={}&\frac{4}{|x|^3}\partial_\nu H\left(\frac{x}{|x|^2}\right). \end{align*}

Oh no. I had forgotten a 2 in the laplacian of product formula. So the things do not cancel out, and I am left with a laplacian of:

$$\Delta\left(\frac{1}{|x|}H\left(\frac{x}{|x|^2}\right)\right)=\frac{2}{|x|^2}\partial_\nu H\left(\frac{x}{|x|^2}\right).$$

So I probably need that $\partial_\nu$ to be zero. Now, if $H$ is bounded, by Liouville I know it is, since $H$ is constant, but if it isn't? Besides, I am sure there must be another much less calc-y way of doing this, or this is not an exercise fit for an exam. It is absolutely impossible to get this right on a test, IMHO. So firstly, am I doing something wrong in the mess up there? And secondly, how else can one do this?

Answer 1

I didn't find a less calc-y solution, but assuming the mess is correct, we have two cases:

1. $H$ is bounded; then, by Liouville, it is constant, hence $\partial_\nu H\equiv0$, and we are done;
2. $H$ is not bounded; then we have a general argument; we know that $H$ is smooth by harmonicity, so if $\partial_\nu H$ is nonzero at some point, it is nonzero (and constant in sign) on a whole ball $B_r(x_0)$ with $x_0:\partial_\nu H(x_0)\neq0$; integrate the derivative on $\partial B_{\frac r2}$; on one side, that derivative has constant sign, but on the other side, we know the integral is zero, since it equates to the integral of the laplacian on the ball and $H$ is harmonic so the laplacian is zero; but with constant sign, the integral is, modulo a sign, the $L^1(\partial B_{\frac r2}(x_0))$ norm of the derivative, which being zero gives $\partial_\nu H$ is a.e. zero on the spherical surface, in particular we have at least a point in $\partial B_{\frac r2}(x_0)$ where $\partial_\nu=0$, contradicting the choice of $r$; so $\partial_\nu H\equiv0$, hence the above messy laplacian is zero.

But of course, the question stays open whether I have a better trick than such a mess. I tried using averages, but I don't know how to handle the Jacobian if I integrate on a surface, and though averaging on balls doesn't, AFAIK, give me harmonicity (only surface averages do), I tried, and the Jacobian comes out to mess up, because the integrand after setting $z=\frac{x}{|x|^2}$ should be like $\frac{1}{|z|^4}H$, or the likes, which I can't use to conclude.