Is there a formula for $ (I - \Delta)^k (fg) $ where $ \Delta $ is the Laplacian and $k \in Z $ similar to the Newton-Leibnitz formula for the derivatives of a product?
Action of $ (I - \Delta)^k$ on a product of functions
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analysis
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1$I$ and $-\Delta$ commute so you can expand $(I - \Delta)^k$ using the binomial theorem. There is also product formula for the Laplacian $\Delta(fg) = f\Delta g + 2 \nabla f \cdot \nabla g + g \Delta f$. I think you should be able to derive a formula using these two facts (though it might be messy). – 2012-10-23