I can't find on complex analysys texts the precise definition of continuity at $\infty$. In particular, my lecturer said: all entire non costant functions aren't continous at infinity, since they are unbounded (by Liouville). I thought the following: take the complex plane with the point $\infty$, i.e. the Riemann sphere. This is a compact set. If f is entire non constant and i suppose $f$ to be continous also at $\infty$, then $f$ should be continous on $\mathbb C \cup\infty$, then bounded (by Weierstrass). Absurd. Is it right?
continuity at infinity
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complex-analysis
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2A quick definition for continuity at infinity is: define $g(z)=f(1/z)$. Is there a definition of $g(0)$ that makes $g$ continuous? Analytic at infinity would mean the same. You'll need conditions on the domain of $f$ for this definition to work, possibly different based on whether you are talking analytic or just continuous – 2012-12-27
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0Maybe you didn't express yourself too well, but I understand your idea and I think that it is completely correct. – 2012-12-30
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Continuity at $\infty$ is usually understood to be continuity at the opposite of $0$ in Riemann sphere, that is correct (in general, study of meromorphic functions etc. makes sense in that context, in some ways more so than in the complex plane alone).
I don't understand what you mean by "absurd" -- you've simply shown that a complex-valued function holomorphic on $\bf C$ and continuous at $\infty$ is constant, and that is correct.
This may not be true if you allow the values to be $\infty$, too. Identity on the entire Riemann sphere is certainly holomorphic everywhere and nonconstant.
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0yes, i forgot to assume $f$ non constant, i've edited – 2012-12-27
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0what does it mean "opposite of zero" in Riemann sphere? – 2012-12-27
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0@FedericaMaggioni: That means $\infty$ in the Riemann sphere. It is opposite of zero if you look at it geometrically. :) Or algebraically, $\infty=1/0$. – 2012-12-27
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3So, to be explicit: $f$ is continuous at $\infty$ iff one of two things happens: 1) Either $\lim_{z\to\infty} f(z)$ exists and is a complex number $t$. (This is as saying that $\lim_{zeta\to 0}g(\zeta)=t$, where $g(\zeta)=f(1/z)$, so $g$ can be extended to a continuous function at $0$.) 2) Or $\lim_{z\to\infty} |f(z)|=+\infty$. (This is as saying that $\lim_{\zeta\to 0} h(\zeta)=0$, where $h(\zeta)=1/f(1/z)$.) – 2012-12-27
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0In particular, in order for $f$ to be continuous at $\infty$, we need that $f(z)$ be defined for all complex $z$ in a neighborhood of infinity, that is, there is an $R$ such that $f(z)$ is defined for all $z$ with $|z|\ge R$. (One can tweak this definition, to include other situations where this neighborhood condition fails. But this is the standard meaning.) – 2012-12-27
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1But if you allow the identity function to be continuous at infinity, you must also allow any function with a pole to be continuous at the pole. – 2012-12-27
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0Yeah, if continuity at infinity allows the value $\infty$, then the theorem is not true, since any polynomial would be continuous at $\infty$ – 2012-12-27
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0(Hmmm... I just saw all the typos in my first comment. Sorry about that.) – 2012-12-28