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In their paper A Direct Approach to the Determination of Gaussian and Scalar Curvature Functions, Kazdan and Warner claim something along the lines of: if $g$ is a metric in $W^{2,p}$ ($p>n$) whose scalar curvature is smooth, then $g$ is smooth. Now, this probably isn't true as stated because you throw away too much information, but here is the precise claim (I think):

Choose a smooth metric $g$, and consider the linearization $A=\mathrm{scal}'(g)$ of the scalar curvature at $g$. Let $A^*$ be its formal adjoint. Now consider the function $Q$ from a neighborhood of $0$ in $W^{4,p}$ to $L^p$ defined by $$Q(u)=\mathrm{scal}(g+A^*u).$$ It is not hard to check that $Q'(0)=AA^*$ and so by the usual Fredholm type theorems, this is elliptic at $0$ and locally surjective. Now they claim that if $f\in C^\infty$ and $Q(u)=f$, then $u\in C^\infty$ "by elliptic regularity." However, the linearization of $Q$ at $u$ is $$\text{scal}'(g+A^*u)\cdot A^*,$$ which is not obviously elliptic to me. We know that $A^*$ is injective$^1$, but know nothing about the linearization $\mathrm{scal}'(g+A^*u)$, especially because the metric isn't smooth (so this is not actually a pseudodifferential operator and that theory doesn't help).

So, is this elliptic at $u$? If not, is there a trick to get regularity anyhow?


$^1$ There are some hypotheses on $g$ that imply this, though I can't imagine they are important for the regularity argument here.

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