Two questions on Schauder Interior Estimate (Theorem 6.2) from Gilbarg-Trudinger I'm using the $3rd$ edition of the book. I'm having trouble seeing the follow two estimates in the proof of Theorem 6.2 from pages 91 and 92.

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*For the first estimate I'm having trouble with is right afterthe sentence "On the other hand, if $|x_o-y_o|\geq d/2$,"
$$d_{x_o}^{2+\alpha}\frac{|D^2u(x_o)-D^2u(y_o)|}{|x_o-y_o|^\alpha}\leq \left(\frac{2}{\mu}\right)^\alpha[d_{x_o}^2|D^2u(x_o)|+d^2_{y_o}|D^2u(y_o)|].$$
Originally, I though this followed from the chain
$$(2^2 |x_o-y_o|)^\alpha \geq (2d)^\alpha \geq(d_{x_o})^\alpha,$$
but this can't be since $d_{x_o}/2\geq d$, not the other way around.


*The second question regards the first inequality of (6.18) on page 92,
$$|g|_{0,\alpha;B}^{(2)}\leq d^2|g|_{0;B}+d^{2,\alpha}[g]_{\alpha;B}.$$
 A: First question: By the triangle inequality, $$ d_{x_0}^{2+\alpha} \frac{\vert D^2u(x_0)-D^2u(y_0)\vert}{\vert x_0-y_0\vert^\alpha} \leqslant \bigg ( \frac{d_{x_0}}{\vert x_0-y_0\vert}\bigg )^\alpha \big ( d_{x_0}^2\vert D^2u(x_0)\vert + d_{x_0}^2 \vert D^2u(y_0)\vert \big ). $$ On one hand, they assume that $d_{x_0}=d_{x_0,y_0}=\min \{ d_{x_0},d_{y_0}\}$, so $d_{x_0} \leqslant d_{y_0}$. Hence, $$ d_{x_0}^2\vert D^2u(x_0)\vert + d_{x_0}^2 \vert D^2u(y_0)\vert \leqslant d_{x_0}^2\vert D^2u(x_0)\vert + d_{y_0}^2 \vert D^2u(y_0)\vert. $$ On the other hand, since $\vert x_0-y_0\vert \geqslant d/2$ and $d=\mu d_{x_0}$, we have that $$\bigg ( \frac{d_{x_0}}{\vert x_0-y_0\vert}\bigg )^\alpha \leqslant \bigg ( \frac{2d_{x_0}}d\bigg )^\alpha =  \bigg ( \frac{2}\mu\bigg )^\alpha .$$ These two observations give you that $$\frac{\vert D^2u(x_0)-D^2u(y_0)\vert}{\vert x_0-y_0\vert^\alpha} \leqslant \bigg ( \frac{2}\mu\bigg )^\alpha \big (d_{x_0}^2\vert D^2u(x_0)\vert + d_{y_0}^2 \vert D^2u(y_0)\vert \big ) $$ as required (I believe that the $\alpha$ on the denominator of the RHS of your question is a typo and should be $\mu$.)
Second question: By definition, $$\vert g \vert^{(2)}_{0,\alpha;B}= \vert g \vert^{(2)}_{0;B}+[ g ]^{(2)}_{0,\alpha;B}.$$ Then $$\vert g \vert^{(2)}_{0;B} =\sup_{x\in B}  d_{\partial B}^2(x)\vert g(x)\vert$$ where $d_{\partial B}(x) = \operatorname{dist}(x,\partial B)$. Since $B=B_d(x_0)$, we have that $d_{\partial B}(x) \leqslant d$, so $$ \vert g \vert^{(2)}_{0;B} \leqslant d^2\sup_{x\in B}  \vert g(x)\vert= d^2\vert g\vert_{0;B}.$$ Similarly, $$[ g ]^{(2)}_{0,\alpha;B} =\sup_{\substack{x,y\in B \\ x\neq y}} \bigg \{ d_{x,y}^{2+\alpha} \frac{\vert g(x)-g(y)\vert}{\vert x-y\vert^\alpha}\bigg \} \leqslant d^{2+\alpha} \sup_{\substack{x,y\in B \\ x\neq y}} \bigg \{ \frac{\vert g(x)-g(y)\vert}{\vert x-y\vert^\alpha}\bigg \} =d^{2+\alpha} [g]_{0,\alpha;B}$$
