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Suppose I have a Riemann surface with a shrinking geodesic which is degenerating towards a surface with a cusp, and I consider a neighborhood of the shrinking geodesic, $C_[a,b] = [a,b]\times \mathbb{T}^1$ and a harmonic function $g$ on $C_[a,b]$, and I want a bound on $|g|$ which is independent of the radius of the "cusp". $g$ here is determined up to an additive constant. I thought I could do this using Harnack, since by adding a constant we can assume $\inf_{C[-\log{2},1]} {g} = 1$ and then Harnack bounds $|g|$ on a smaller $C$. It was pointed out to me that Harnack won't work in this case since I only have one degree of freedom so I can choose to make $g$ positive or make the inf = 1 but not necessarily both at once. I'm still quite sure this should be true, but any ideas about how to prove it greatly appreciated.

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The question does not make sense to me: Your surface has no cusps. Moreover, think of your annulus as a rung in the compiles plane. Harmonic functions are real parts of holomorphic functions, so the is no bound on the absolute value in any sense. Do you mean a bound in terms of boundary values? Then just use maximum principle. –  Misha Sep 17 '12 at 16:06
    
Sorry, true, the setting is a surface degenerating to one with a cusp. Harnack would give a bound on norm based on value at one point, since $g$ is continuous on a closed set. –  Jeff McGowan Sep 17 '12 at 16:45
    
@Jeff: Then you should rewrite your question more carefully to reflect what you mean (I am still unsure what the actual question is). Note that the notion of "radius" is meaningless for cusps, since they are all conformally-equivalent to the unit disk minus the origin. –  Misha Sep 17 '12 at 18:35
    
@misha, Yes to the conformal equivalence, but the thing is that "radius" is not meaningless here - think of a short geodesic shrinking to length zero, with "radius" being the distance from the geodesic to the "thick part" (in the sense of Margulis) of the surface. I will however take your advice and rewrite the question to make this clearer, thanks. –  Jeff McGowan Sep 17 '12 at 19:42

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