Bound on $Q$ implies bound on $R$?

Let $(M,g)$ be a smooth, closed Riemannian manifold with dimension $n>4$. Define the $Q$-curvature through the formula

$Q = \Delta R + \frac{n^3-4n^2+16n-16}{(n-1)(n-2)^2} R^2 - \frac{8(n-1)}{(n-2)^2}|Ric|^2,$

where $\Delta = -div\nabla$, $R$ is the scalar curvature, and $|Ric|$ is the norm of the Ricci tensor. This quantity arises in many problems in conformal differential geometry. The question is whether or not a uniform sup-norm bound on a sequence of $Q$ curvatures implies there exists a uniform sup-norm bound on the corresponding scalar curvatures, provided that all of the scalar curvatures change sign?

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Why do you believe such a sup norm bound exists? Is there an analogous situation where it does? And how does "all of the scalar curvatures change sign" help? –  Deane Yang Jul 15 '11 at 2:03
Deane: The mode situation is where there is a sequence of metrics with constant $Q$ curvature, where the constant is bounded uniformly. The changing signs hypothesis is there to prevent situations where the scalar curvature simply "translates its way to infinity". –  Viktor Bundle Jul 15 '11 at 13:38
Could you explain in more detail what you mean by the scalar curvature "translates its way to infinity"? –  Deane Yang Jul 15 '11 at 13:51
I haven't checked the constants in $n$-dimensional case but I found a formula for $Q$ in dimension 4. In dimension 4 the maximum principle gives an upper bound for $R$ in terms of $Q$ and the traceless part of Ricci. I'm sure that's known to everybody who works with $Q$-curvature. Does this work in higher dimensions? –  Deane Yang Jul 15 '11 at 14:22
Deane: In four dimensions the formula is special because it contains the conformal Laplacian of the scalar curvature, so, yes, higher dimensions are different. By "translate to infinity" I mean to rule out $R$ becoming unbounded like the sequence $\{R+n\}$. The idea is that if the Laplacian of $R$ is controlled it my prevent blow-up. –  Viktor Bundle Jul 15 '11 at 22:19