I am trying to understand the proof of interior and global Holder estimate for the solutions of second order elliptic equations from Gilberg and Trudinger book. This is basically theorem 17.14 of the mentioned book. In that book authors proved that \begin{equation}\label{1} \omega(R)\leq \delta \omega(2R)+R|Dg|_{0,\Omega}+R^{2}|D^{2}g|_{0,\Omega}\tag{1} \end{equation} where $\omega(R)=osc_{B_{R}}w=\sup_{B_{R}}w-\inf_{B_{R}}w$ and $\omega=u_{\gamma\gamma}$(where $\gamma=(\gamma_{1},\gamma_{2},...,\gamma_{n})$ is the unit vector in $\mathbb{R}^{n}$ of the type $e_{i}=(0,0,..1,...0)$ or $\gamma=\frac{1}{\sqrt{2}}(e_{i}\pm e_{j})$ ) is the second order derivative of $u$ in the direction $\gamma.$ Moreprecisely, $u_{\gamma\gamma}(x)=\gamma^{T}D^{2}u(x)\gamma=\sum^{n}_{i,j=1}\frac{\partial^{2}u}{\partial x_{i}\partial x_{j}}\gamma_{i}\gamma_{j}.$ In paerticular, if $\gamma=e_{i}$ then $w=u_{\gamma\gamma}=u_{ii}$ and if $\gamma=\frac{1}{\sqrt{2}}(e_{i}+ e_{j})$ then $u_{\gamma\gamma}=\frac{1}{2}D_{ii}+D_{ij}+\frac{1}{2}D_{jj}$ and similarly, for $\gamma =\frac{1}{\sqrt{2}}(e_{i}- e_{j})$ then $u_{\gamma\gamma}=\frac{1}{2}D_{ii}-D_{ij}+\frac{1}{2}D_{jj}.$ Later on authors employed the Lemma 8.23 of the same book to get \begin{equation}\label{2} osc_{B_{R}} D^{2}u\leq C\Big(\frac{R}{R_{0}}\Big)^{\alpha}\{osc_{_{B_{R_{0}}}}D^{2}u++R_{0}|Dg|_{0,\Omega}+R_{0}^{2}|D^{2}g|_{0,\Omega}\}.\tag{2} \end{equation} Now I have the following questions:\
How to define $osc_{B_{R}} D^{2}u.$\
Even If we employ Lemma 8.23 of the Book of Gilberg and Trudinger which states that:~~Lemma 8.23 Let $\omega$ be a non-decreasing function on an interval $(0,R_{0}]$ satisfying for all $R\leq R_{0},$ the inequality $$\omega(\tau R)\leq \gamma \omega(R)+\sigma(R),$$ where $\sigma$ is a nondecreasing function and $0<\gamma,\delta<1.$ Then for any $\mu\in(0,1)$ and $R\leq R_{0}$ we have~~~$$\omega(R)\leq~C\Bigg(\Big(\frac{R}{R_{0}}\Big)^{\alpha}\omega(R_{0})+\sigma(R^{\mu}R_{0}^{1-\mu})\Bigg),$$ where $C=C(\gamma,\tau)$ and $\alpha=\alpha(\gamma,\tau,\mu)$ are positive constants.\
Now my question is that even if we apply Lemma 8.23 with $\omega(R)=osc_{B_{R}}w$ and $\sigma(R)=R|Dg|_{0,\Omega}+R^{2}|D^{2}g|_{0,\Omega}$ then Lemma assert that for any $R\leq R_{0}$ and $\mu\in(0,1)$ we have \begin{align} \omega(R)&\leq~C\Bigg(\Big(\frac{R}{R_{0}}\Big)^{\alpha}\omega(R_{0})+\sigma(R^{\mu}R_{0}^{1-\mu})\Bigg),\\ &= C\Bigg(\Big(\frac{R}{R_{0}}\Big)^{\alpha}\omega(R_{0})+\Big(\frac{R}{R_{0}}\Big)^{\mu} R_{0}|Dg|_{0,\Omega}+\Big(\frac{R}{R_{0}}\Big)^{2\mu} R^{2}_{0}|D^{2}g|_{0,\Omega}\Bigg)\\ &=C\Big(\frac{R}{R_{0}}\Big)^{\alpha}\Bigg(\omega(R_{0})+\Big(\frac{R}{R_{0}}\Big)^{\mu-\alpha} R_{0}|Dg|_{0,\Omega}+\Big(\frac{R}{R_{0}}\Big)^{2\mu-\alpha} R^{2}_{0}|D^{2}g|_{0,\Omega}\Bigg). \end{align} Which is not smae as \eqref{2} with $osc_{B_{R}} D^{2}u$ replaced by $osc_{B_{R}} w$ and $osc_{B_{R_{0}}} D^{2}u$ repaced by $osc_{B_{R_{0}}} w$\ \
Suppse that if any how we obtained: \begin{align} osc_{B_{R_{0}}} w&\leq C\Big(\frac{R}{R_{0}}\Big)^{\alpha}\Bigg(osc_{B_{R_{0}}}w+R_{0}|Dg|_{0,\Omega}+R^{2}_{0}|D^{2}g|_{0,\Omega}\Bigg), \end{align} where $w=u_{\gamma\gamma},$ that is if $\gamma=e_{i}$ then $w=u_{\gamma\gamma}=u_{ii}$ and if $\gamma=\frac{1}{\sqrt{2}}(e_{i}\pm e_{j})$ then $w=u_{gamma\gamma}=\frac{1}{2} u_{ii}\pm u_{ij}+\frac{1}{2}u_{jj}.$ Then how we conclude \eqref{2} from this expression (particulrly how to define $osc_{B_{R_{0}}} D^{2}u$).\
Again suppose that \eqref{2} is proved then how to obtain following interior Holder estimate from \eqref{2}
\begin{equation}\label{3} |u|^{*}_{2,\alpha;\Omega}\leq C(|u|^{*}_{2,\Omega}+|g|^{*}_{2,\Omega}), \tag{3} \end{equation} where $|u|_{2,\alpha;\Omega}^{*}$ is the interior Holder norms\
5 What type of inequality is needed so that one can obatin the boundary estimate as above. In particular, what would be version of the equation \eqref{2} near the boundary. Please also explain the patchup of the inetrior and boundary estimate to get estimate like \eqref{3} valid up to the boundary.\
I would thank in advance for answering the above questions.