Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $f(z)$ is some analytic function which is bounded near $0$. Then $f(1/z)$ is bounded near $\infty$. What exactly does that last statement mean practically?

Does it mean $|f(1/z)|$ is bounded somehow?

share|improve this question

1 Answer 1

It means precisely the first statement: a function $g(z)$ bounded "at $z=\infty$" when $g(1/z)$ is bounded in a neighborhood of the origin $z=0$. Thus to say $f(1/z)$ is bounded at infinity is to say that $f(z)$ is bounded at zero. One also says functions are holomorphic at infinity, et cetera.

To understand this intuitively, just picture the point infinity located on the Riemann sphere; looking at a neighborhood of this point and examining a function $f$ there is equivalent to flipping the sphere upside down (done by taking the reciprocal) and examining $f(1/z)$ at the origin ($1/z$ at the origin corresponds to $\infty$ on the sphere).

share|improve this answer
    
Thanks! I'm reading something that says $f(z)=g(1/z)=z^nh(1/z)$ where $h(1/z)$ is bounded near $z=\infty$. It then says from this that for some $M$, $|f(z)|\leq M|z|^n$, which seems to follow from $|h(1/z)|\leq M$ outside a large enough circle. How can this conclusion be seen, at least intuitively? Does being bounded near $\infty$ always imply that a function is bounded by some constant if we get sufficiently far out? –  MJY Mar 22 '12 at 11:28
    
@MJY: Just as you say, $h(1/z)$ is bounded by (say) $M$ outside of a large enough circle, so $|f(z)|=|z|^n|h(1/z)|\le M|z|^n$ outside it as well. (Presumably $f$ is also holomorphic and thus bounded inside the circle as well; if you put the two bounds together...) –  anon Mar 22 '12 at 11:33
    
sorry, I see why $|f(z)|$ would be bounded, but I don't see why $h(1/z)$ being bounded at $z=\infty$ means it is bounded by $M$ outside a large enough circle. Unless that's just what the statement means by definition... –  MJY Mar 22 '12 at 11:34
    
@MJY: The intuition here is that any loop going around infinity on the Riemann sphere is also a loop in the complex plane; the side of the loop where $\infty$ is located, on the sphere, will be "outside" the loop when viewed in the plane. This is why any neighborhood of infinity is "outside a large enough circle" in the complex plane. –  anon Mar 22 '12 at 11:36

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.