Opening mapping theorem in complex analysis

tttcomrader · #1 $Old$ May 23rd, 2009, 12:53 PM

There is a theorem in my book that says:

If a function f is analytic on a connected open set U, then f is either a constant function or an opening mapping.

I read the proof in wikipedia, but I really don't understand it. The book provided a proof that look easier, but I got really lost.

What is the idea of the proof? Why is this true?

Thank you.

ThePerfectHacker · #2 $Old$ May 23rd, 2009, 11:23 PM

Quote:

Originally Posted by tttcomrader $View Post$

There is a theorem in my book that says:

If a function f is analytic on a connected open set U, then f is either a constant function or an opening mapping.

I read the proof in wikipedia, but I really don't understand it. The book provided a proof that look easier, but I got really lost.

What is the idea of the proof? Why is this true?

Thank you.

Requirements] Here is a proof that I know and like. You need to know the minimum-modulos theorem. The maximum modulos theorem says (let us just say for disks for simplicity) that if $f$ is a continous complex function on a closed disk that is analytic on the interior of the disk then $f$ assumes it maximum on the boundary of the disk. The minimum-modulos theorem is similar, it says $f$ (under the same assumptions) assumes it minimum value on the boundary provided that $f$ is non-zero at any point inside the disk. You also need to know the identity-theorem, if $f$ (under the same assumptions) has a sequence of distinct points $\{z_n\}$ inside the disk that are zeros of $f$ (i.e. $f(z_n) = 0$ ) and $\{z_n\}$ converges to some point inside the disk then $f$ must be the zero function on the disk.

Basic Idea] Let $f$ be an analytic function on an open set $U$ . Remember we want to show that $f(U)$ is open, this means for any point in $f(U)$ we can find an open disk containing that point that lies wholly in $f(U)$ (that is just the definition of being 'open'). Let $\beta \in f(U)$ , be some point. Therefore, $\beta = f(\alpha)$ for some $\alpha \in U$ . We need to show there is some disk containing $\beta=f(\alpha)$ that lies in $f(U)$ . By using translations we can safely assume that $f(\alpha) = 0$ (why?). What remains now is that we have to show there is an open disk containing $0$ that lies wholly in the image of $f$ (i.e. $f(U)$ ).

Special Circle] By the identity-theorem there must be a circle around $\alpha$ (which is contained in $U$ ) so that $f$ does not vanish on this circle. To see why this must be true, assume to the contrary that for any circle you draw around $\alpha$ (that lies in $U$ ) there is always some point for which $f$ vanishes. We know for some $\tfrac{1}{r}>0$ we can draw a circle around $\alpha$ that is contained in $U$ . On this circle (as we are assuming) there is a point $z_0$ so that $f(z_0)$ . Now draw a smaller circle around that point of radius $\tfrac{1}{r+1}>0$ and again by assumption there is a point on this circle $z_1$ so that $f(z_1)=0$ . Now draw an even smaller circle around that point of radius $\tfrac{1}{r+2}>0$ and again there is a $z_2$ on the circle so that $f(z_2) = 0$ . In general we can create $z_n$ by drawing a circle of radius $\tfrac{1}{r+n}>0$ around $\alpha$ . Thus, we have a sequence of point $\{z_n\}$ . The limit of these points is $\alpha$ (because they shrink down to $\alpha$ in this limiting process). Now look at the identity theorem. We found a convergent sequence of distinct zeros that converge to a point ( $\alpha$ ) which is inside $U$ and so it forces $f$ to be the zero function on $U$ (since $U$ is an open connected set, I realized I stated the identity theorem for disks above but it works for open connected sets as well). But $f$ is not constant. Therefore, $f$ must have a circle, call it $C$ , on which it does not vanish.

Minimum] We know that $|f|$ attains a mimimum on $C$ (it is a compact set and continous functions on compact sets attain minimums and maximums). Let $2m = \min \{ |f(z)| : z \in C \}>0$ . We will show that the disk $D$ around $0$ of radius $m$ lies in the image of $f$ on the disk bounded by $C$ around $\alpha$ (and so definitely an open disk around $0$ is contained in $f(U)$ ). Let $\xi \in D$ and if $z\in C$ then $|f(z) - \xi| \geq ||f(z)|-|\xi|| \geq |f(z)| - |\xi| \geq 2m - m = m$ . However, $|f(\alpha) - \xi| = |-\xi| < m$ , so $|f(\alpha) - \xi| < |f(z) - \xi|$ . Therefore, on the closed disk around $\alpha$ with boundary $C$ we see that $|f(z) - \xi|$ does not attain a minimum on the boundary, and so by minimum-modulos theorem it means $|f(z^*) - \xi| = 0$ for some $z^*$ inside this disk, and so $\xi = f(z^*)$ . Thus, we have proven that any point in $D$ is the image of some point in that disk bounded by $C$ and centered at $\alpha$ . Thus, $\xi \in f(U)$ . Thus, $f(U)$ is open.