# N-laplacian and Kelvin transform

Hi guys,

it is well known that Kelvin transform is not longer available for the p-laplacian (with p not equal to 2) in R^n. But I believe that for the borderline case (the n-laplacian) it is available. Someone knows something more.

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Is it a conjecture, you have read it somewhere and you cannot remember, is it a hope? The Kelvin transform itself does not depend on an operator, so I guess you actually refer to the result stating equivalence of (super/sub)harmonicity of some function and of its Kelvin transform. Do you expect exactly the same result? And - most importantly: why do you refer to the case of $p=n$ as "the borderline case"? The above result holds for the (2-)Laplacian regardless of dimension! –  Delio M. Mar 7 at 11:21