TITLE: Does Hölder continuity imply smoothness for the CMC equation: $u:D^2\rightarrow\mathbb{R}^n$, $\Delta u = 2H\partial_xu\times\partial_yu$, $H$ constant?
QUESTION [6 upvotes]: Context: I am currently reading through the freely available lecture notes from Tristan Riviere (here) on the applicability of integration by compensation in the analysis of various geometrically motivated PDEs.
I have attempted to find something in the vast literature to the following effect: suppose $u:D^2\rightarrow\mathbb{R}^n$, $u\in W^{1,2}$, $\Delta u = f(u,\nabla u) = 2H\partial_xu\times\partial_yu$. Then
$$u\in C^{0,\alpha}(D') \rightarrow u\in C^\infty(D''),$$
where $D'' \subset\subset D' \subset\subset D$.
In other words, an interior estimate.  With the little regularity on hand, it appears to be very difficult.  I find this surprising because most of the time, proving the Hölder regularity of the solution is the 'most difficult' part.  I have the feeling that I am missing an obvious reference or a well-known folklore argument.
Showing the Hölder continuity of the solution relies on deriving a Morrey-type estimate with the help of the Wente lemma.  In the process of doing this, one also shows that
$$\sup_{\rho < 1/2, p\in B_{1/2}(0)} \rho^{-\alpha} \int_{B_\rho(p)} |\Delta u|$$
is bounded.  This implies that $f\in\mathcal{H}^1$.  I just include this extra detail in case this question fits into a general framework of optimal interior regularity for Poisson's equation on a disk when the right hand side is Hardy.  (This is the reason for the earlier form of my question.)
Can anybody help?

REPLY [4 votes]: I'm too lazy to type-up the proof myself, so I'll send you to a reference. 
Chang, S.-Y. A., Wang, L. and Yang, P. C. (1999), "Regularity of harmonic maps". CPAM has the proof in Section 3. Once you get $C^{1,\gamma}$ you immediately get RHS is in $C^\gamma$ and the rest follow by standard elliptic regularity. 
Note that the structure of the equation (RHS being of the form $d(u\cdot du)$) is only used for Wente's lemma. For the upgrade of regularity one uses a Caccioppoli type inequality. 
(BTW, the Chang-Wang-Yang result bypasses the Hardy space estimates. For that the result can be found in the original paper of Helein, though I'd guess the material is also in his book if you don't read French.)