The following is related to my previous question [1], which I think can be resolved if this question is resolved. Is it true for $u\in C^{1,\alpha}(B-p)$ where $B=B_p(r)$ a function satisfying the spacetime Laplacian $\Delta u+K|\nabla u|=0,$ that the following estimate holds, $$r^2|\nabla\nabla u|\leq C r^{1+\alpha}?$$ Here, $K$ is the trace of some prescribed symmetric $(0,2)$ tensor $k$. For simplicity, let us assume $K=0$, so that $u$ is harmonic. Classical Schauder estimates from Gilbarg-Trudinger tell us that the following bound holds, $$r^2|\nabla\nabla u|\leq C |u|_{C^0(B-p)}.$$ Does it follow that for harmonic $u$, we have $|u|_{0;B-p}\leq C r^{1+\alpha}$?
References: [1]: PDE inequalities to show a gradient is integrable