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Someone has asked exactly this question but the answer was preciselly the thing I cannot use, so I'll ask again:

Let $$K(x,y) = \frac{1}{n \alpha(n)}\frac{1-\|x\|^2}{\|x-y\|^n}$$ be the poisson kernel for the ball in $\mathbb{R}^n$. Here $\alpha(n)$ is the volume of the $n$ dimensional ball of radius $1$.

I'd like to show that $$\int_{\partial B_1(0)}K(x,y)dH^{n-1}(y) = 1$$

(here $dH^{n-1}$ is the n-1 dimensional Hausdorff measure)

The thing is, I CANNOT USE the fact that

$K[f](x) = \int_{\partial B_1(0)}K(x,y)f(y)dH^{n-1}(y)$ is the solution of the Dirichlet problem $$\left\{\begin{matrix}\Delta u = 0 \; \mbox{ in } B_1(0)\\u = f\; \mbox{ in } \partial B_1(0)\end{matrix}\right.$$

as we used the fact that the integral is $1$ in the first place to prove that $K[f](x)$ is a solution of the Dirichlet problem.

On Lawrence Evans' book "partial differential equations" it says simply "A direct calculation, the details which we omit, verifies [...]" So it should be possible to calculate this integral. I just don't know how.

Any tips, or book references that might help me are appreciated. Thanks in advance.

Matheus barros castro
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2 Answers2

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In this post there is a very elegant proof of the statement using the strong maximum principle for harmonic functions. Although it uses the fact that $$ \Delta K[f] (x) = 0,$$ you don't need to use the fact that $\int_{\partial B(0,1)} K(x,y) dy =1$ in first place to show that the Poisson kernel for the ball is harmonic. I hope it helps.

J. David Beltran
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I know this is old, but I'll answer it anyway.

Define $k(x,y)=\frac{1-|x|^2}{|x-y|^n}$ which allows us to write

$$u(x)=\boldsymbol{-\kern-11.2pt}\int_{\partial\mathbb B(0,1)} k(x,y)g(y)\mathrm d^{n-1}y$$

With this in mind let $$F(x)=\boldsymbol{-\kern-11.2pt}\int_{\partial\mathbb B(0,1)} k(x,y)\mathrm d^{n-1}y$$ We remark that $F$ has spherical symmetry, that is $F(x)=F(x')$ whenever $|x|=|x'|$. This means that $F$ is equal to its spherical means, i.e

$$F(x)=f(|x|)=\boldsymbol{-\kern-11.2pt}\int_{\partial\mathbb B(0,|x|)}F(z)\mathrm d^{n-1}z$$

As given in one of the formulas in my question, which you can easily prove as an exercise, we know that $$f'(r)=\frac{r}{n}\boldsymbol{-\kern-11.2pt}\int _{\mathbb B(0,r)}\Delta F(x)\mathrm d^{n-1}x$$ However, $$\Delta F(x)=\boldsymbol{-\kern-11.2pt}\int_{\partial\mathbb B(0,1)}\Delta_x k(x,y)\mathrm d^{n-1}y$$

But, $\Delta_x k(x,y)=\Delta k(\cdot,y)=0$ and therefore $f'(r)=0$. But $\nabla F(x)=f'(|x|)~\frac{x}{|x|}$ and therefore $\nabla F=0$ hence $F$ is constant. Finally , $$F(0)=\boldsymbol{-\kern-11.2pt}\int_{\partial\mathbb B(0,1)} k(0,y)\mathrm d^{n-1}y=\boldsymbol{-\kern-11.2pt}\int_{\partial\mathbb B(0,1)} 1~ \mathrm d\mu^{n-1}=1.$$

Therefore $F\equiv 1$.

Note: Verifying $\Delta_x k(x,y)=0$ turns out to be quite difficult, but I might attach this computation later when I get the chance.

K.defaoite
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