# Theorom from Serge Lang's Complex Analysis - pages 89-90

• Feb 20th 2010, 04:55 PM
Bernhard
Theorom from Serge Lang's Complex Analysis - pages 89-90
I am a math hobbyist/amateur studying complex analysis from Serge Lang's book Complex Analysis.

I need some help regarding Theorem 1.1 on Page 89

The theorem and is proof as given by Lang are as follows:

"Theorem 1.1 Let U be a connected open set, and let f be a holomorphic function on U. If f ' = 0 then f is constant.

Proof: Let $\alpha \ , \beta$ be two points in U and suppose first that $\gamma$ is a curve joining $\alpha$ to $\beta$ so that

$\gamma$(a) = $\alpha$ and $\gamma$(b) = $\beta$

The function

t --> f( $\gamma$(t))

is differentiable, and by the chain rule, its derivative is

f ' ( $\gamma$(t)) $\gamma$'(t) = 0 .... (1)

Hence this function is constant, and therefore

f( $\alpha$) = f( $\gamma$(a)) = f( $\gamma$(b)) = f( $\beta$)

Next suppose that $\gamma$ = { $\gamma_1, \gamma_2, ..... \gamma_n$} is a path joining $\alpha$ to $\beta$, and let $z_j$ be the end point of $\gamma_j$ , putting

$z_0 = \alpha , \ \ \ z_n \ = \ \beta$

By what we have just proved

f( $\alpha$) = f( $z_0$) = f( $z_1$) = f( $z_2$) = ..... = f( $\beta$)

thereby proving the theorem

QUESTIONS

(1) Lang does not describe the nature of the elements of U - but I am assuming that he is taking them as complex numbers? Is that right?

(2) At first sight - to someone familiar with elementary real analysis, the conclusion of the theorem that f ' = 0 implies f is constant does not seem surprising? Any comments?

(3) The statements in the proof:

"
The function

t --> f( $\gamma$(t))

is differentiable, and by the chain rule, its derivative is

f ' ( $\gamma$(t)) $\gamma$'(t) = 0 .... (1)

Hence this function is constant,"

seem to assume the proof since after assuming f ' = 0 Lang just states, "hence this function is constant".

• Feb 20th 2010, 05:57 PM
davismj
1) U is an open connected set. Open meaning it contains none of its boundary points, and connected meaning that it cannot be represented as two disjoint nonempty open subsets. In other words, if you partitioned the set (say in half), each point $\delta$ on the boundary of the subsets would be in either one set or the other.

U is the set of all complex numbers in this arbitrary set, that satisfies the above properties.

2/3) Theorem 1.1 seems to indicate that it is a pretty basic Theorem?
• Feb 20th 2010, 06:22 PM
Bernhard
Theorem in Complex Analysis - What is Proved?
Thanks.

Yes, it does seem a very basic theorem, but my Question 3 about what is actually "proved" is my real worry. We are supposed to show that f ' = 0 implies f is constant. Where in the proof is this actually demonstrated - it seems to be just stated as true in Lang's proof?

Examining the statements in the proof, we find the following:

"
The function

t --> f(http://www.mathhelpforum.com/math-he...65c1de79-1.gif(t))

is differentiable, and by the chain rule, its derivative is

f ' (http://www.mathhelpforum.com/math-he...65c1de79-1.gif(t))http://www.mathhelpforum.com/math-he...65c1de79-1.gif'(t) = 0 .... (1)

Hence this function is constant,"

seem to assume the proof since after assuming f ' = 0 Lang just states, "hence this function is constant".
• Feb 20th 2010, 07:00 PM
davismj
Quote:

Originally Posted by Bernhard
"

The function

t --> f(http://www.mathhelpforum.com/math-he...65c1de79-1.gif(t))

is differentiable, and by the chain rule, its derivative is

f ' (http://www.mathhelpforum.com/math-he...65c1de79-1.gif(t))http://www.mathhelpforum.com/math-he...65c1de79-1.gif'(t) = 0 .... (1)

Hence this function is constant,"

seem to assume the proof since after assuming f ' = 0 Lang just states, "hence this function is constant".

Okay, from my little knowledge, I'm assuming he's proving the complex case by presenting it in real components and using what we know to be true on reals.

Either that or he's proving this: U is a complex domain (as described), and f is a function on U. So, he proves that if f'(z) = 0, then f is constant on all of U.
• Feb 20th 2010, 08:01 PM
Bernhard
More on the Theorem on page 89-90 of Serge Lang "Complex Analysis"
I think it is true that he is "proving" what you say - namely

"Either that or he's proving this: U is a complex domain (as described), and f is a function on U. So, he proves that if f'(z) = 0, then f is constant on all of U."

My worry is that his proof seems to consist only of an assertion in the sentence -

"The function t --> f(http://www.mathhelpforum.com/math-he...65c1de79-1.gif(t)) is differentiable, and by the chain rule, its derivative is f ' (http://www.mathhelpforum.com/math-he...65c1de79-1.gif(t))http://www.mathhelpforum.com/math-he...65c1de79-1.gif'(t) = 0 .... (1)

Hence this function is constant," <--- ISN't THIS JUST AN ASSERTION?
• Feb 21st 2010, 08:39 AM
vince
Quote:

Originally Posted by Bernhard
I think it is true that he is "proving" what you say - namely

"Either that or he's proving this: U is a complex domain (as described), and f is a function on U. So, he proves that if f'(z) = 0, then f is constant on all of U."

My worry is that his proof seems to consist only of an assertion in the sentence -

"The function t --> f(http://www.mathhelpforum.com/math-he...65c1de79-1.gif(t)) is differentiable, and by the chain rule, its derivative is f ' (http://www.mathhelpforum.com/math-he...65c1de79-1.gif(t))http://www.mathhelpforum.com/math-he...65c1de79-1.gif'(t) = 0 .... (1)

Hence this function is constant," <--- ISN't THIS JUST AN ASSERTION?

The point of the proof is to prove the constancy of $f$ throughout the set, and not just locally -- that is, it doesnt "jump" anywhere in the set. It is for this reason he picks arbitrary points then considers the behaviour of $f$ first throughout the region connecting the points, then partitioning the region into arbitrarily small pieces. The key here is that $f$ is holomorphic in the entire set, so that one can evaluate $f^{'}$ anywhere in the set.
• Feb 21st 2010, 09:13 AM
Opalg
Quote:

Originally Posted by Bernhard
(1) Lang does not describe the nature of the elements of U - but I am assuming that he is taking them as complex numbers? Is that right?

Since the book is entitled "Complex analysis", and U is a set on which a holomorphic function is defined, I guess that Lang felt justified in taking it for granted that U is a subset of the complex numbers. (Smirk)

Quote:

Originally Posted by Bernhard
(2) At first sight - to someone familiar with elementary real analysis, the conclusion of the theorem that f' = 0 implies f is constant does not seem surprising? Any comments?

(3) The statements in the proof:

"The function t --> f( $\gamma$(t)) is differentiable, and by the chain rule, its derivative is f'( $\gamma$(t)) $\gamma$'(t) = 0 .... (1)

Hence this function is constant,"

seem to assume the proof since after assuming f' = 0 Lang just states, "hence this function is constant".

The point here is that the function $t\mapsto f(\gamma(t))$ is a function of the real variable t on the interval [a,b]. Lang proves (he does not "assume") that this function of t has zero derivative. As you say, it is then a familiar result from real analysis that this implies that the function is constant. That result is then used to deduce that the function f on the complex domain U is constant.

To put it briefly, the idea of the proof is that in this context a known result about a function of a real variable can be used to prove an analogous result for a function of a complex variable. I hope that helps to explain the structure of the proof.
• Feb 21st 2010, 09:28 AM
vince
Quote:

Originally Posted by Opalg
The point here is that the function $t\mapsto f(\gamma(t))$ is a function of the real variable t on the interval [a,b]. Lang proves (he does not "assume") that this function of t has zero derivative..

Opalq, well said. it's key that you noted the parametrization of $f$ to be a function of a real variable, which allows him to conclude that $f(\gamma(t))$ is constant, once he assumes its derivative is 0. he then propragates the argument throughout smaller pieces of the region connecting the points.
• Feb 21st 2010, 11:50 AM
Bernhard
Vince, Opalg

Thanks so much for your help - clear and thoughtful!

Bernhard