Showing a function is bounded in the complex plane.

[mancini0]

Homework Statement

Hi everyone. I must show that if f is a continuous function over the complex plane, with
limit as z tends to infinity = 0, then f is in fact bounded.

The Attempt at a Solution

Since f is continuous and lim z --> infinity f(z) = 0, by definition of limit at infinity I know
for all epsilon > 0, there exists N > 0 such that for all |z| > N, |f(z)-0| < epsilon.

The problem gives the hint that I would be wise to consider epsilon = 1. With that I see that
|f(z)| < 1 for all |z| > N.

Then |f(z)| is bounded. I need to extend this to f(z). I know from absolute convergence that if the absolute value of a series converges, the series itself converges. But we are not dealing with series, nor have we yet learned about series in the complex plane.

 

[HallsofIvy]

What do you mean "extend to f(z)"? The definition of f(z) being bounded is precisely that |f(z)| is bounded. (We can't talk about m< f(z)< M as we might for real valued functions since the complex numbers are not an ordered field.)

 

[mancini0]

I see. Thank you. The follow up asks me if f is bounded for |z|<= N.
Since f is bounded and continuous, f is constant by Liouville's Theorem. Doesn't this mean f(z) is bounded for all z then?

 

[HallsofIvy]

By Liouville's theorem, a function analytic in the entire complex plane which is bounded is a constant. You said nothing before about f being entire.

But if you have already shown that f is bounded on the complex plane, then it is bounded on any subset so certainly on |z|<= N. That second part really does not make sense.