According to Theorem 1.1' in this [*paper*](https://camath.fudan.edu.cn/camb/ch/reader/create_pdf.aspx?file_no=27B603&flag=1) we have the following estimate on classical solutions $u \in C^2(\overline{B\_1^+})$ of $-\Delta u = f \text{ in } B\_1^+ = B\_1 \cap \{x \_n \ge 0 \}$ and $u = 0 \text{ on } \partial B\_1^+ \cap \{x\_n = 0\}$ 


$$|D^2u(x) - D^2u(y)| \le C\left(r\lVert u \rVert\_{L^\infty(B\_{1}^+)} + \int\_0^{r} \frac{\omega\_f(t)}{t}\,dt + r\int\_{r}^{1} \frac{\omega\_f(t)}{t^2}\,dt\right) \tag{1}$$ $\forall \, x,y \in B\_{1/2}^{+}$ with $r = |x-y|$ where, $\omega\_f$ denotes the modulus of continuity of $f$ which we assume is Dini continuous.


The proof is supposed to be similar to proof of Theorem 1.1 which is the interior estimate. Following the lines of this proof we can get all analogous 'boundary' harmonic function estimates in $B\_{1}^+$ v.i.a. applying the estimates in Theorem 1.1 to the odd extensions of the harmonic functions to the full ball $B\_1$. But in the last step (going along the lines of eqn $(1.13)$ in the paper) it seems we need the estimate $$|D^2u\_0(x) - D^2u\_0(y)| \le C \lVert u \rVert\_{L^{\infty}(B\_1^+)}|x-y|, \, \forall \, x,y \in B\_{1/2}^+ \tag{2}$$ where, $u\_0$ satisfies $-\Delta u\_0 = f(0)$ in $B\_{1}^+$ and $u\_0 = 0$ on $\partial B\_1^+ \cap \{x\_n = 0\}$ where, $C$ is supposed to be independent of $f$.



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> I can't seem to get the estimate $(2)$ for half-ball. 
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In case of interior estimate we can consider $v\_0 := u\_0 - \frac{f(0)}{2n}(1 - |x|^2)$ which is harmonic in $B\_1$ and write \begin{align\*} |D^2u\_0(x) - D^2u\_0(y)| = |D^2v\_0(x) - D^2v\_0(y)| &\le r\lVert D^3v\_0\rVert\_{L^{\infty}(B\_{1/2})} \\ &\le Cr\lVert v\_0\rVert\_{L^{\infty}(\partial B\_{1})} = Cr\lVert u\_0\rVert\_{L^{\infty}(\partial B\_{1})}\end{align\*} where, $r = |x-y|$ which proves the interior analogue of $(2)$ using only gradient estimate for harmonic function $v\_0$. 


But a similar approach doesn't seem to be working for the boundary case (for example, considering $u\_0 - \frac{f(0)}{2}x\_n^2$ which is harmonic and applying odd extension to this.)



> 
> Is there a different way of approaching the estimate $(2)$?
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Any help is appreciated. Thanks.



A boundary Schauder estimate
One approach is to observe that
$$\|u\_0\|\_{L^{\infty}(B\_1^+)} \geq \frac{1}{16n}|f(0)|.$$
It suffices by linearity to prove this when $f(0) = 1$. Consider the barrier
$$b(x) = \frac{1}{2n}\left(\left|x - \frac{1}{2}e\_n\right|^2 - \frac{1}{8}\right).$$
Since $b \geq \frac{1}{16n}$ on $\partial B\_1^+$, either $u\_0 \geq b$ somewhere on $\partial B\_1^+$ and we are done, or $u\_0 \leq b$ on $\partial B\_1^+$ in which case the comparison principle implies that 
$$u\left(\frac{e\_n}{2}\right) \leq -\frac{1}{16n}$$ 
and we are again done.


With this estimate in hand, your suggestion to use reflection and interior estimates for the harmonic function $v\_0 = u\_0 - \frac{f(0)}{2}x\_n^2$ works because
$$\|v\_0\|\_{L^{\infty}(B\_1^+)} \leq (1+8n)\|u\_0\|\_{L^{\infty}(B\_1^+)}.$$
More precisely,
$$\|D^3u\_0\|\_{L^{\infty}(B\_{1/2}^+)} = \|D^3v\_0\|\_{L^{\infty}(B\_{1/2}^+)} \leq C(n)\|v\_0\|\_{L^{\infty}(B\_1^+)} \leq C(n)(1+8n)\|u\_0\|\_{L^{\infty}(B\_1^+)}.$$


