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Let $f$ be analytic on the closed unit disk centered at the origin and $|f(z)| < 1$ for $|z| = 1$. Show that $f$ has exactly one fixed point inside the open unit disk. That is, there exists a unique number $z_0$ with $|z_0| < 1$ such that $f(z_0) = z_0$. We must prove 1 there exists at least on solution, 2 there is at most one solution. 1) Having problems with at least one solution. 2)By definition a is a fixed point if f(a) = a. Assume that f has more then one fixed point a,b and that a

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Do you know the contraction mapping theorem? – froggie May 12 '12 at 19:58
What about Rouche's theorem? – froggie May 12 '12 at 20:01
ssume that f has more then one fixed point a,b and that a<b. By the Mean Value theorem, there is a number c that exists on the interval (a,b) s.t. f(c) = f(b) - f(a) / b - a = b-a/b-a = 1. However, |f(z)| < 1 thus there is at most one fixed point. – Joe Stevenson May 12 '12 at 20:30
@JoeStevenson: This version of MVT applies to real-valued functions, and does not generalize to complex-valued functions. We certainly can't say "$a<b$" and have it make any sense, as the complex numbers cannot be ordered in a way compatible with field operations. We can however talk about $(a,b)$ for any distinct points $a,b$, but in this case it simply denotes the line segment from $a$ to $b$ with the endpoints removed. – Cameron Buie May 14 '12 at 2:29
The version of MVT applicable to complex-valued functions in one complex variable is as follows: If $U$ is an open subset of $\mathbb{C}$, $f:U\to\mathbb{C}$ analytic, and $a,b\in\mathbb{C}$ are such that the segment $[a,b]\subset U$, then $f(b)-f(a)=\int_a^bf'(z)dz=(b-a)\int_0^1f'\bigl(a+t(b-a)\bigr)dt$. – Cameron Buie May 14 '12 at 2:36

Let $g(z)=f(z)-z$ and $h(z)=z$. Then on $\{|z|=1\}$, we have $$|g(z)+h(z)|=|f(z)|<|g(z)|+|h(z)|$$ This is because $|f(z)|<1$ and $|h(z)|=1$ on the boundary of the disk. Also, $f(z)-z$ and $z$ have no zeros on the boundary. Then Rouche's Theorem applies and $f(z)-z$ and $z$ have the same number of zeros inside the disk. We know $h(z)=z$ has one zero, so $f(z)-z$ has one zero. Hence there exists $z_0$ with $|z_0|<1$ and $f(z_0)=z_0$.

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Ya I saw that answer before. Can you explain how Rouche's theorem applies a little more – Joe Stevenson May 12 '12 at 20:56
The statement of Rouche's Theorem (as I've learned it) says that if $f$ and $g$ are analytic on a neighborhood of $\overline{B}(a; R)$ with no zeros on the circle $\{|z-a|=R\}$ and they satisfy $$|f+g|<|f|+|g|\text{ on }\{|z-a|=R\}$$ then they have the same number of zeros in $B(a; R)$. – Connor May 12 '12 at 21:13
I see what your saying now. Thanks. – Joe Stevenson May 12 '12 at 23:58

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