If we have an oscillation estimate of $u\in W^{2,n} (B^{+}) \cap C^0(\bar B^{+})$, and $u=0$ on $T = B \cap \partial R^{n}_{+}$, that $$osc_{B^{+}} \frac{u}{x_{n}} \leq C.$$ Then how comes that $u$ is differential on $T$ ? Here $B = B_{1}(0)$.
This is a question I met when reading the remark of Theorem 9.31 in the book "Elliptic Partial Differential Equations of Second Order" by Gilbarg and Trudinger, version of revision of the 1983 second edition(P254).
The theorem does not have just $C$ on the right: it has $CR^\alpha$, where $R$ is the radius of the ball. This implies $$u(x',x_n) = f(y')\,x_n + O((|x_n|+|x'-y'|)^{1+\alpha}),\qquad (x',x_n)\to (y',0) \tag{1}$$ where $f(y')$ is the limit of $u(x',x_n)/x_n$ as $(x',x_n)\to (y',0)$. From (1) it also follows that $$D u(x',0) = f(x') $$