According to Wikipedia (under properties) a function which is entire on the whole Riemann sphere must be constant, but I can't quite see why. Is it because the limit as |z| goes to infinity of f(z) exists, and therefore f is bounded?

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- October 17th 2012, 12:27 PMthesmurfmasterAnalytic function on entire Riemann sphere is constant
According to Wikipedia (under properties) a function which is entire on the whole Riemann sphere must be constant, but I can't quite see why. Is it because the limit as |z| goes to infinity of f(z) exists, and therefore f is bounded?

- October 17th 2012, 01:18 PMjohnsomeoneRe: Analytic function on entire Riemann sphere is constant
Basically, yes.

If f entire on the Riemann sphere, then in particular it's continuous, so |f| is continuous from the compact sphere into the reals, and hence achieves its maximum. Thus |f| is bounded on the Riemann sphere, and so |f| is also bounded on just the complex plane. Then apply Louisville's theorem, and extend continuously to the point at infinity, to show that f must be a constant function.

You could also think of it in terms of Laurent series:

Take the Taylor series for f at z=0. Since f is entire, there are no singularities, hence the radius of convergence is infinity. But f analytic at the point at infinity means that f(1/z) is analytic at z=0 - and using that original power series for f you can see that that means that only the constant term can be non-zero.

In a Laurent series, the z-powers are to the point at infinity (i.e. "bad" points, singularities) as the (1/z)-powers are to the point z=0 (i.e. "bad" points, singularities). Thus for a Laurent series to have no bad points (meaning it represents an entire function on the whole Riemann sphere), it can't have any non-zero (1/z)-powers (since it's analytic at z=0) and it can't have any non-zero z-powers (since it's analytic at z=the point at infinity). Thus the only non-zero term it can have is the constant term.