Here are some notations: \begin{align*} \mathbb{R}^n_+ &= \left\{(x_1,\dots,x_n)\in \mathbb{R}^n:x_n>0\right\} \\ T &= \left\{(x_1,\dots,x_n)\in \mathbb{R}^n:x_n=0\right\} \end{align*} and $$ |u|_{k,\alpha:\Omega}^{\prime}=\sum_{j=0}^{k}d^j\sup_{|\beta|=j}\left\|D^{\beta}u\right\|_{C^0(\Omega)} +d^{k+\alpha}\sup_{|\beta|=k}\left[D^\beta u\right]_{C^{0,\alpha}(\Omega)} $$ where $d=\mathrm{diam}(\Omega)$, $\left\|\cdot\right\|_{C^{0}(\Omega)}(=|\cdot|_{0:\Omega}$ in G-T's notation) is the uniform norm and $\left[\cdot\right]_{C^{0,\alpha}(\Omega)}$ is the Holder semi-norm of order $\alpha$.
Now let $x_0\in\overline{\mathbb{R}^n_+}, R>0,$ denote $B_1=B_R(x_0),B_2=B_{2R}(x_0)$ and $B_1^+=B_1\cap \mathbb{R}^n_+,B_2^+=B_2\cap \mathbb{R}^n_+,$ in theorem $4.11$ of G-T is given the result
Let $u\in C^2(B_2^+)\cap C^0 (\overline{B_2^+})$ and $f\in C^{\alpha} (\overline{B_2^+})$ satisfy $\Delta u=f$ in $B_2^+$ and $u=0$ on $T,$ then $u\in C^{2,\alpha}(\overline{B_1^+})$ and we have the estimate $$|u|_{2,\alpha:B_1^+}^{\prime}\leq C (|u|_{0:B_2^+}+R^2|f|^{\prime}_{0,\alpha:B_2^+})\tag{1}$$ where $C=C(n,\alpha).$
whose proof utilizes the function ($\Gamma$ is the fundamental solution to Laplace) $$w(x)=\int_{B_2^+}(\Gamma(x-y)-\Gamma(x^{\ast}-y))\,f(y)\,\mathrm d y,$$ with $x^{\ast}$ the reflection of $x$ about $T$ and the following estimate $$ |D^2w|_{0,\alpha:B_1^+}^{\prime}\leq C|f|_{0,\alpha:B_2^+}^{\prime}\tag{2}$$ In the remark after this theorem, he says when $u$ has compact support in $B_2^+\cup T,$ we can obtain from $(2)$ a simpler estimate as follows $$|D^2u|_{0,\alpha:B_2^+}^{\prime}\leq C|f|_{0,\alpha:B_2^+}^{\prime}\tag{3}$$ and the representation $$u(x)=w(x)=\int_{B_2^+}(\Gamma(x-y)-\Gamma(x^{\ast}-y))\,f(y)\,\mathrm dy\tag{4}$$ but I don't know how does $(3)$ happens, can someone help me?