# Integral inequality involving the biharmonic operator

Let M be a hyperbolic surface and let f be a real smooth function on M.

Considering a geometric inequality which I conjectured to be true for different reasons, I get after a lot of computations that the inequality is in fact equivalent to the following

$$\int_{M}(f^2 - \frac{1}{4}\(Delta(f))^2)dM≤ \frac{1}{V}(\int_{M}fdM)^2$$,

where dm is the volume form on M and $V=\int_{M}dM$.

Is this inequality correct or false ?

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The laplacian term integrates to zero, so what you've written becomes a much simpler conjecture. Perhaps this question as is, is not appropriate for the site. –  Otis Chodosh Oct 7 '12 at 14:55
I suspect the parentheses are in the wrong place. –  Deane Yang Oct 7 '12 at 15:00
I apologize: the notation was unclear. Thank you. –  Yves Oct 7 '12 at 16:14
It still can't be correct as written can be seen by scaling the metric $g\to \lambda g$ for $\lambda>>1$ and taking any non-zero function $f$ orthogonal to the constants. –  Rbega Oct 7 '12 at 16:25
Yes, the OP is right. The Euler-Lagrange equation for the functional has the biharmonic operator as the top order term. –  Deane Yang Oct 7 '12 at 22:51

According to this paper of Randol which I found in this answer of BR. Compact surfaces with metrics of constant curvature $-1$ may have first eigenvalue $\lambda_1$ as small as one likes. An eigenfunction $\phi_1$ of this eigenvalue gives a counter-example to your claim