# If $f$ is a non-constant analytic function on a compact domain $D$, then $Re(f)$ and $Im(f)$ assume their max and min on the boundary of $D$.

This is a homework problem I got, my attempted proof is:

Since $f$ is non constant and analytic, $f=u(x)+iv(y)$ where neither $u$ nor $v$ is constant(by Cauchy Riemann equations) and $u v$ are both analytic in $D$.

Therefore $u$ and $v$ both assume their max on the boundary of $D$ (by maximum modulus theorem).

Also, $u$ and $v$ have no minimums in the interior of $D$ unless they are $0$. I'm stuck here and don't know how to show that $u$ and $v$ are nonzero.

I looked at the answer at the back of the book. They used the Open Mapping Theorem(the image of an open set under a nonconstant analytic mapping is an open set):

According to the Open Mapping Theorem, the image under f of any open set D containing z0 in its interior is an open set containing $f (z_0)$ in its interior. Hence, both Re f and Im f assume larger and smaller values in D than the values $Re f (z_0)$ and $Im f (z_0)$.

I don't understant the proof given by the book, can someone explain a bit? Also, what do you think about my proof?

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Do you mean in your second paragraph that each of  u,v is harmonic in D? – AQP Mar 6 '12 at 0:59
er...I don't know what that means. In the second paragraph I  mean that by the Cauchy Riemann equations, if u is constant, then  u_x=u_y=0, then so is v_x and v_y(and vice versa), so f is constant. – Scharfschütze Mar 6 '12 at 1:03
If you use polar coordinates for a circle, and Cauchy's formula may help you prove the result . Then the value at the center of a ball is the average of the values on the perimeter. – AQP Mar 6 '12 at 1:08
I don't see how that helps...can you explain more? – Scharfschütze Mar 6 '12 at 1:16
I see, that is correct if that's what you meant. I understood something else. – AQP Mar 6 '12 at 1:16
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## 1 Answer

Here's the book's proof with a tad more detail: Suppose $z_0$ is not on the boundary. We will show neither the real nor the imaginary parts of $f$ are maximized at $z_0$. Since the real and complex parts are continuous, they obtain their maxima and minima somewhere on $D$ because $D$ is compact. Hence they must obtain their minima and maxima on the boundary.

Take a small neighborhood of $z_0$ inside your domain. The image of this open neighborhood is an open neighborhood of $f(z_0)$ by the open mapping theorem. Let $U$ be the neighborhood of $f(z_0)$ just described. For small values of $\varepsilon$, $U$ contains the points $f(z_0)+\varepsilon$ , $f(z_0)-\varepsilon$, $f(z_0)+i\varepsilon$, and $f(z_0)-i\varepsilon$, whose real and imaginary parts are more/less than those of $f(z_0)$.

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For this your wrote:Since the real and complex parts are continuous, they obtain their maxima and minima somewhere on D because D is compact. Hence they must obtain their minima and maxima on the boundary. I don't understand why the minima must also be on the boundary, because the minimum modulus theorem says it's possible if f is 0 at some point inside D. – Scharfschütze Mar 6 '12 at 1:38

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Max/min value of the function $\ne$ max/min modulus. – Did Mar 6 '12 at 6:16
The proof above was for the max/min values of the real and imaginary parts, but it is similar to the proof of the max/min modulus theorems: For small enough $\delta$, $(1+\delta)f(z_0)$ and $(1-\delta)f(z_0)$ are contained in $U$. Then the max/min modulus cannot be obtained at $z_0$ unless $f(z_0)$=0. So the max and min modulus are each either 0 or obtained on the boundary. The max modulus is only 0 if $f$ is 0 everywhere (in which case it is obtained on the boundary, among other places). – Brett Frankel Mar 6 '12 at 18:18
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