# Holder continuous for the Poisson integral to Laplace equation on Half space

From Evans, we see that if $$u$$ satisfies the Laplace equation on the half space with boundary $$g$$, $$u$$ is given by the Poisson integral: $$u = \int_{\partial\mathbb{R}^n_+} \frac{2x_n}{n\alpha(n)}\frac{g(y)}{|x-y|^n}dy$$

I would like to show that if $$g$$ is Holder continuous, then $$u$$ is.

My attempt:

Using triangle inequality to plug in a $$g(x_0)$$ gives $$|u(x')-u(x)| \leq \int_{\partial\mathbb{R}^n_+} |K(x,y)-K(x',y)||g(x_0)-g(y)|dy$$ From this expression I see that Holder continuity of $$g$$ can be used directly, but not giving what we desired. Say it works on the region where $$|x-x'|\leq |x_0 - y|$$. I wonder what could we do on the other region where when $$y$$ is close to $$x_0$$?

Probably we could use mean value formula for $$K$$ ie. $$|K(x,y)-K(x',y)|\leq |x-x'||\nabla K(z,y)|$$ but I have no clue what $$\nabla K$$.

Let $$g: \mathbb R^{n-1}\to \mathbb R$$ and assume that $$g\in C^\alpha (\mathbb R^{n-1})$$ with $$\alpha \in (0,1)$$ is such that $$[g]_{C^\alpha (\mathbb R^{n-1})}<+\infty$$. Given a point $$x\in \mathbb R^n$$, I will write $$x'=(x_1,\dots,x_{n-1} )\in \mathbb R^{n-1}$$ and $$x=(x',x_n)$$. Then the Poisson kernel representation of $$u$$ is \begin{align*} u(x) &= \frac{2x_n}{n\alpha(n)} \int_{\partial \mathbb R^n_+} \frac{g(y')}{\vert x - y\vert^n} \, d\mathcal H^{n-1}_y \qquad \text{for all } x\in \mathbb R^n_+. \end{align*} Note that this is written slightly different to Evans. I've written $$d\mathcal H^{n-1}_y$$ to denote the $$(n-1)$$-dimensional Hausdorff measure (with respect to $$y$$) instead of $$d y$$ since $$d y$$ usually denotes the $$n$$-dimensional Lebesgue (and, indeed, does in Evans), but, in this context, $$d\mathcal H^{n-1}_y$$ will be $$(n-1)$$-dimensional Lebesgue measure which I will denote as $$d y'.$$ Evans also identifies the points $$y=(y',0) \in \partial \mathbb R^n_+$$ and $$y' \in \mathbb R^{n-1}$$ in order to write $$g(y)$$ (instead of $$g(y')$$), which is pretty standard, I just prefer to this make this distinction abundantly clear.
Anyways, making the change of variables $$z'=\frac{y'-x'}{x_n}$$ gives
\begin{align*} u(x) &=\frac{2x_n}{n\alpha(n)} \int_{ \mathbb R^{n-1}} \frac{g(y')}{\big ( \vert x' - y'\vert^2+ x_n^2\big )^{\frac n2}} \, d y' \\ &=\frac{2x_n}{n\alpha(n)} \int_{ \mathbb R^{n-1}} \frac{g(x_nz'+x')}{\big ( x_n^2\vert z'\vert^2+ x_n^2\big )^{\frac n2}} x_n^{n-1}\, d z'\\ &= \frac{2}{n\alpha(n)} \int_{ \mathbb R^{n-1}} \frac{g(x_nz'+x')}{\big ( \vert z'\vert^2+ 1\big )^{\frac n2}} \, d z'. \end{align*} Hence, given $$x,y\in \mathbb R^n_+$$, \begin{align*} \vert u(x)-u(y)\vert &\leqslant C\int_{ \mathbb R^{n-1}} \frac{\big \vert g(x_nz'+x')-g(y_nz'+y')\big \vert }{\big ( \vert z'\vert^2+ 1\big )^{\frac n2}} \, d z' \\ &\leqslant C [g]_{C^\alpha(\mathbb R^{n-1})} \int_{ \mathbb R^{n-1}} \frac{\big \vert (x_n-y_n)z'+x'-y'\big \vert^\alpha }{\big ( \vert z'\vert^2+ 1\big )^{\frac n2}} \, d z' \end{align*} with $$C=C(n)>0$$. Then $$\big \vert (x_n-y_n)z'+x'-y'\big \vert \leqslant \vert x_n-y_n \vert \cdot \vert z' \vert + \vert x'-y' \vert \leqslant \vert x - y \vert \big ( \vert z'\vert +1\big ),$$ so $$\vert u(x)-u(y)\vert\leqslant C [g]_{C^\alpha(\mathbb R^{n-1})} \vert x-y\vert^\alpha \int_{ \mathbb R^{n-1}} \frac{\big ( \vert z'\vert +1\big )^\alpha}{\big ( \vert z'\vert^2+ 1\big )^{\frac n2}} \, d z' .$$ Since $$\alpha \in (0,1)$$, it follows that $$z'\mapsto {\big ( \vert z'\vert +1\big )^\alpha}/{\big ( \vert z'\vert^2+ 1\big )^{\frac n2}}$$ is in $$L^1(\mathbb R^{n-1})$$, so we conclude that $$[ u]_{C^\alpha(\mathbb R^n_+)} \leqslant C [g]_{C^\alpha(\mathbb R^{n-1})}$$ with $$C=C(n,\alpha)>0$$.