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For my Master's thesis, I'd like to prove the following (but I'm not sure it's true):

On a two-dimensional Riemannian manifold (oriented and closed), for any smooth function $f$, it holds that $$ \int_M \left( 2 |\nabla^2 f |^2 + \text{Scal} |\nabla f|^2 \right) \text{d} V \geq 0, $$ where Scal denotes the scalar curvature. It feel like I have to use some divergence/integration by parts theorem and use that $M$ is Einstein but I just can't make it. Thanks in advance!

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  • $\begingroup$ Why are you trying to prove this? Is there some motivation? $\endgroup$
    – Igor Rivin
    Commented Jul 12, 2013 at 20:33
  • $\begingroup$ Look up Weitzenböck formulae and Bochner identities: these introduce a curvature term that will reduce to $\mathrm{Scal}$ in yr $2$-dimensional setting. $\endgroup$
    – Fran Burstall
    Commented Jul 14, 2013 at 11:30

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You have the following idenity on a general closed Riemannian manifold: $$\int_M (\Delta_g f)^2 \, dv_g = \int_M \vert \nabla^2_g f\vert^2 + Ricc(\nabla f,\nabla f) \, dv_g$$ You can porve it by integrating by part the first term and commuting the covariant derivatives which make appear the curvature. In your special case $Ricc= \frac{Scal}{2} g$, then your are done.

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