If on a bounded smooth, pseudoconvex domain in $\mathbb{C}^n$, $\mathrm{det} ( \mathrm{Hess}(u)) = f$ ($f>0$, $\mathrm{Hess}(u)>0$, $u=0$ on the boundary), if $f \in C^{k, \alpha}$, is $u \in C^{k+2, \alpha}$ ? (I mean, is there an apriori estimate on $u$ with the Hölder exponents of $f$ and $u$ being the same (equal to $\alpha$?)
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I assume that by $\text{Hess}(u)$ you mean the complex Hessian of $u$, so $u$ is plurisubharmonic. If $k\geq 1$ then you can differentiate the equation and apply Evans-Krylov to get uniform interior $C^{k+2,\alpha}$ estimates for $u$.
However if $k=0$ the situation is much more delicate. Assume $f$ is $C^\alpha$ and positive and $u$ is $C^2$, say, and suppose you want to prove uniform interior $C^{2,\beta}$ bounds for $u$ (here $\beta<\alpha$). Then the best result to date, due to Y. Wang, says that you have such estimates but depending also on a pointwise upper bound for $\Delta u$. There are earlier results of Chen-He, Blocki-Dinew and Dinew-Zhang-Zhang. Some of these results get the same exponent $\alpha$ instead of $\beta<\alpha$, under stronger assumptions.
It is an open problem to show that if $u$ is just $L^\infty$ and plurisubharmonic, and $u$ solves the Dirichlet problem for the complex Monge-Ampère equation in the sense of Bedford-Taylor, with RHS $f$ positive and $C^\alpha$, then $u$ is indeed $C^{2,\alpha}$. This would be the direct analogue of a well-known result of Caffarelli for the real Monge-Ampère equation, but it is still open. An old paper of Schulz claims this result, but it contains a mistake.
