If a compact Kahler manifold $(M,g)$ has constant scalar curvature, is the metric $g$ real analytic?

Hi to all!

Perhaps it is a silly question, if so i'll delete this post. Suppose we have a compact Kahler manifold $(M,g)$ of complex dimension $m$ with constant scalar curvature with respect to its metric $g$. My question is: does the condition of constant scalar curvature imply that the metric $g$ automatically real analytic? When i say that the metric is real analytic i mean that in a holomorphic coordinate chart with coordinate functions

$$(z^1,\ldots, z^m)\quad \textrm{ with } z^j=x^j+iy^j \textrm{ for }1\leq j\leq m$$

the coefficients $(g_{i\bar{j}})_{1\leq i,j\leq m}$ are analytic functions w.r.t. $x^k,y^k$.

It's not a silly question, but there's a standard answer, and it's a purely local result: If the Kähler metric is $C^2$ and has constant scalar curvature, then it is real-analytic with respect to the real-analytic structure that underlies the complex-analytic structure. The reason is that setting the scalar curvature equal to a constant is an elliptic equation for the potential of the metric that is an analytic function of its arguments in the local real-analytic coordinates that you define, and so the elliptic regularity results of Hopf and Morrey apply.
Thank you for the answer! My doubt is this: if have a $C^{2}$ metric i can set the equation for constant scalar curvature on a small coordinate chart and i get an elliptic (nonlinear) equation of fourth order on the potential of the metric. By elliptic regularity i have that the solution is $C^{\infty}$ but how can i conclude that the solution of this equation is analytic? – Italo Oct 28 '12 at 12:37