# Dirichlet Problem for the Half space: differentiating under the integral sign.

Let us denote points in $\mathbb{R}^{n+1}$ by $(x,y)$ where $x\in\mathbb{R}^n, y\in\mathbb{R}$. What I'm aware of is that the Poisson kernel $P$ defined on $\mathbb{R}^n\times(0,\infty)$ by$$P(x,y)=\frac{2}{\omega_{n+1}}\frac{y}{(|x|^2+y^2)^{\frac{n+1}{2}}}$$ is a 'good' kernel, $P$ is a solution of Laplace's equation, and that the Fourier transfrom of $P(x,y)$ (with respect to $x$-variable) equals $e^{-|\xi|y}$.

My textbook shows that after some formal deductions using Fourier transform and its inverse, it is reasonable to 'guess' the formula $$u(x,y)=\int_{\mathbb{R} ^n}P(x-z,y)f(z)dz$$ as a solution for the Dirichlet problem on the half space, where $f$ is the boundary condition. ($u$ is defined on $\mathbb{R}^n\times(0,\infty)$.) For convenience, let us assume that $f\in L^{\infty}(\mathbb{R}^n)$.

My questions are:

1. Is $u$ harmonic on $\mathbb{R}^n\times(0,\infty)$ no matter which $f\in L^{\infty}(\mathbb{R}^n)$ was given? Is it okay to differentiate under the integral sign?
2. If $f$ is continuous at $x_0\in\mathbb{R}^n$, does it follow that $$\lim_{(x,y)\rightarrow(x_0,0)}u(x,y)=f(x_0)\quad?$$ This is much more that the mere statement $\lim_{y\rightarrow0}u(x_0,y)=f(x_0)$, and I'm not sure how to show this.

Any advice is welcome! Since these are really basic problems, any (not too advanced) references are also welcome. Please enlighten me. Thank you.

• Both are true. Differentiation under integral sign can be justified by considering that the derivatives of $P$ are well behaved (locally). Two references that come to mind are Harmonic Measure by Garnett and Marshall and Classical Potential Theory by Armitage and Gardiner.
– user357151
Jan 26 '17 at 4:42

As Zaq mentioned in a comment, the answer to both questions are yes. I will try to outline an explanation of the first. The second follows similarly.

Let $P_{y}(x) = P(x,y) = \frac{2}{\omega_{n+1}} \frac{y}{(|x|^{2} + y^{2})^{\frac{n+1}{2}}}$.

Note that $P_{y} \in C^{\infty}(\mathbb{R})$ for all $y \neq 0$. Also, for all $y \neq 0$ we have $P_{y}(x) = y^{-n} P_{1}(y^{-1}x)$. Consequently, it can be shown that $\int_{x \in \mathbb{R}^{n}} P_{y}(x) d x = 1$ by computing that $\int_{x \in \mathbb{R}^{n}}P_{1}(x) d x = 1$. A quick aside here, since you have seen the Fourier analysis already, then we have that $\widehat{P}_{y}(0) = \int_{\mathbb{R}^{n}} P_{y}(x) d x$. So, plugging into your formula, the claim is immediate.

The three proceeding properties I mentioned about $P_{(\cdot)}$ tells us that the family of functions $\{P_{y}\}_{y > 0}$ shares many of the same properties as a smooth approximation to the identity sometimes also called a mollifier. See Chapter 4 section 1 of Evans and Gariepy's book "Measure Theory and Fine Properties of Functions" if you are unfamiliar. This text includes a very detailed proof of the commonly used properties of mollifiers.

A key property that mollifiers have that $P_{y}$ does not, is that mollifiers typically have compact support. This is not necessary, but it is quite convenient. However, we can get by with instead observing that $P_{y} \in L^{1}(\mathbb{R}^{n}) \cap L^{\infty}(\mathbb{R}^{n})$ (since $y \neq 0$, we're $C^{\infty}$ and decay like $|x|^{n+1}$). Similarly $\Delta_{x}P_{y}, \partial_{y}^{2}P_{y} \in L^{1}(\mathbb{R}^{n}) \cap L^{\infty}(\mathbb{R}^{n})$.

Now, suppose $f \in L^{\infty}(\mathbb{R}^{n})$. We note that $$u(x,y) = \int_{\mathbb{R}^{n}} P(x-z,y) f(z) d z = P_{y}*f(z) \quad \forall y > 0, x \in \mathbb{R}^{n}$$

By Young's inequality, and $P_{y}, \partial_{y}^{2} P_{y}, \Delta_{x} P_{y} \in L^{1}$ it follows that the convolutions: $g*P_{t}, ~ g* \Delta_{x}P_{y}$, and $g* \partial_{y}^{2} P_{y}$ are each absolutely convergent for all $x$ and $t$.

Therefore, if $v = (v_{1}, \dots, v_{n+1})$ is a unit vector in $\mathbb{R}^{n+1}$ with $v_{n+1} = 0$, we can compute the derivative of $u(x,y)$ in the direction of $v$ as $$\lim_{h \downarrow 0} \frac{u ((x,y) + hv) - u(x,y)}{h} = \lim_{h \downarrow 0} \frac{P_{y} *g(x +hv) - P_{y}*g(x+hv)}{h} = \lim_{h \downarrow 0} \left(\frac{ P_{y}(\cdot + hv) - P_{y}(\cdot)}{h} \right)*g(x) = D_{v}P_{y}*g(x).$$

In particular, $\Delta_{x} u(x,y) = \Delta_{x}P_{y}*g(x)$. It can similarly be shown that $\partial_{y}^{2}u(x,y) = \partial_{y}^{2}P_{y}*g(x)$, so that $\Delta u(x,y) = \left( \Delta P_{y} \right)* g(x) = 0 * g(x) = 0$.

Hence, under the assumption $y > 0$, we have that $u$ is harmonic at $(x,y)$. Note, this sketch of a proof does not actually require $g \in L^{\infty}$. We used that if $g \in L^{p}$ then $P_{y} \in L^{q}$ for $p^{-1} + q^{-1} = 1$. But, $P_{y} \in L^{1} \cap L^{\infty}$ implies $P_{y} \in L^{q}$ for all $1 \le q \le \infty$, so $g \in L^{p}(\mathbb{R}^{n})$ for any $p$ suffices.

The remaining question is a local question, so you may as well assume that $g$ is continuous everywhere (and bounded!) instead of just continuous at the point $x$. The type of argument follows similarly. If you desire more details, I suggest looking at Folland's "Introduction to Partial Differential Equations" Chapter 2 Section G.