Our professor proved a certain theorem in our PDE class and I have troubles understanding the proof. Let me start with the relevant definitions.
Problem: Suppose $\Omega \subset \mathbb{C}$ is a bounded region with a smooth boundary. Suppose we have two continuous functions $f: \overline{\Omega} \to \mathbb{R}$ and $g: \partial \Omega \to \mathbb{R}$. We are looking for a function $u : \overline{\Omega} \to \mathbb{R}$, such that $\Delta u = f$ on $\Omega$ and $u = g$ on $\partial \Omega$.
Definition: Let $z \in \Omega$. Let $h_z$ be a function such that $\Delta h_z = 0$ on $\Omega$ and $h_z (\xi) = - \frac{\log | z - \xi |}{2 \pi}$ on $\partial \Omega$. Then define $$ G(z, \xi) := \frac{1}{2 \pi} \log | z - \xi | + h_z(\xi). $$ This is called the Green's function of domain $\Omega$.
Theorem: Let $G(z, \xi)$ be the Green's function of domain $\Omega$ with a singularity at $z$. Then the following representation formula hold for every $u \in C^2( \overline{\Omega})$: $$ u(z) = \int_{\partial \Omega } u(\xi) P(z, \xi) ds(\xi) + \iint_{\Omega} G(z, \xi) (\Delta u)(\xi) dA(\xi). $$
Up to this, I would understand what he did and how he showed it. Now comes the theorem I have troubles understanding the proof of.
Theorem: The problem described above has a unique solutions, given by the formula $$ u(z) = \int_{\partial \Omega} P(z, \xi) g(\xi) d \xi + \iint_{\Omega} G(z, \xi) f(\xi) dA(\xi). $$
It is clear (because of representation formula) that there can be no other solution. So the following lemma remains to be proven.
Lemma: The function $F$ defined below is $C^2$ and satisfies $\Delta F = f$; $$ F(z) = \frac{1}{2 \pi} \iint_{\Omega} \log |z - \xi | f(\xi) dA(\xi). $$
Let me now write the proof that I have in my notes, and after that the questions I have regarding it.
Proof: Recall the Fourier transform $\mathcal{F}: f \mapsto \hat{f}$. Let $f \in C_C^{\infty}(\mathbb{R})$. Then $$ \hat{f}(\xi) = \int_{\mathbb{R}^n} f(x) e^{-2 \pi i \langle x, \xi \rangle } dx, $$ where $\alpha = (\alpha_1, \ldots, \alpha_n)$, $\xi = (\xi_1, \ldots, \xi_n)$ and $\partial^{\alpha} = \partial_{x_1}^{\alpha_1} \cdots \partial_{x_n}^{\alpha_n}$. We have $$ \widehat{\Delta f} (\xi) = \sum_{j=1}^n \widehat{ \frac{\partial^2 f}{\partial x_j^2} }(\xi) = \sum_{j=1}^n (2 \pi i)^2 \xi_j^2 \hat{f}(\xi) = - 4 \pi |\xi|^2 \hat{f}(\xi). $$ This means that $$ \widehat{ (- \Delta)f } (\xi) = 4 \pi |\xi|^2 \hat{f}(\xi). $$ We say that $4 \pi |\xi|^2$ is the Fourier multiplier of $- \Delta$. It is also true that $$ \widehat{ (-\Delta)^*f }(\xi) = (4 \pi |\xi|^2)^* \hat{f}(\xi) $$ and therefore $$ \widehat{ \phi(-\Delta)f }(\xi) = \phi(4 \pi |\xi|^2) \hat{f}(\xi). $$ But for which $\phi$? If $\phi(x) = x^{-1}$, we get $$ \widehat{ (- \Delta) f } (\xi) = \frac{1}{4 \pi |\xi|^2} \hat{f}(\xi), $$ which has a singularity so the function is not $C^1(\mathbb{C})$!
Now, let us first consider the case when function $f$ has compact support; $f \in C_C^2(\mathbb{C}) \subset \Omega$. If we denote $\zeta = \xi - z$, we get $$ F(z) = \frac{1}{2 \pi} \iint_{\mathbb{C}} \log | z - \xi | f(\xi) dA(\xi) = \frac{1}{2 \pi} \iint_{\mathbb{C}} \log |\zeta| f(\zeta + z) dA(\zeta). $$ This means that $$ \Delta F = \frac{1}{2 \pi} \iint_{\mathbb{C}} \log |\zeta| \Delta_z f(\zeta + z) dA(\zeta) = \frac{1}{2 \pi} \iint_{\mathbb{C}} \log | z - \xi| \Delta f(\xi) dA(\xi) = f(z). $$
Now consider the general case. Take $a \in \Omega$ and $r >0$ such that the closed ball around a with radius r is contained in $\Omega$; $\overline{B(a,r)} \subset \Omega$. Take the function $\chi \in C_C^{\infty}(\mathbb{C})$, such that
- $\chi \in [0,1],$
- $\text{supp } \chi \subset \Omega$,
- $ \chi \equiv 1$ on $\overline{B(a,r)}$.
Now we can write $f = f \cdot 1 = f \cdot \chi + f (1 - \chi)$. This gives us $$ F(z) = \frac{1}{2 \pi} \int \log | z - \xi| (\chi f)(\xi) dA(\xi) + \frac{1}{2 \pi} \int \log | z - \xi | \left( (1-\chi) f \right) dA(\xi) = F_1 + F_2. $$ Because we prove the theorem for functions with compact support, $\Delta F_1 (z) = f(z)$ for $z \in B(a,r)$. Since $(1-\chi)f \equiv 0$ on $B(a,r)$, it follows that $$ F_2(z) = \iint_{\Omega - B(a,r)} \log |z-\xi| (1- \chi)(\xi) f(\xi) dA(\xi). $$
This further implies that $$ (\Delta F_2)(z) = \iint_{\Omega - B(a,r)} \Delta_z (\log |z - \xi|)(1 - \chi)(\xi) f(\xi) dA(\xi) = 0, $$
since $\Delta_z(\log |z - \xi|) = 0$ where this integral is defined. QED.
Questions: The proof is so confusing that I don't even know where to begin.
- What the hell was all that Fourier transform stuff that was mentioned? I don't know how it proved anything from the statement or was in any way useful?
- How did we show that the function is $C^2$?