# Opening mapping theorem in complex analysis

• May 23rd 2009, 10:53 AM
Opening mapping theorem in complex analysis
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.
• May 23rd 2009, 09:23 PM
ThePerfectHacker
Quote:

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

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

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

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