1
$\begingroup$

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 $\nabla K$.

$\endgroup$

1 Answer 1

2
$\begingroup$

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 $C=C(n,\alpha)>0$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .