Holomorphic map with a fixed point has Jacobian with all eigen values in the unit disk

Question

Suppose $U\subset \mathbb{C}^n$ is a bounded domain and $f$ is a holomorphic map from $U$ to itself such that $f(a)=a$ for some $a\in U$. Prove that all eigen values $\lambda$ of $Df(a)$ lie in the unit disk.

I honestly have no idea how to start. Tried for $n=2$ for simplicity but dont know where to go from the complicated Jacobian. At least a hint is appreciated.

Every sequence has a locally uniformly convergent subsequence. Recall Montel's theorem. — Daniel Fischer, Mar 02 '17 at 17:02
Ok. So if I take such a family such that $a$ is a fixed point of each, then I get that the limit function also holomorphic with $a$ as a fixed point. Now how do I proceed to eigen values? I m sorry I dont see how it relates. — Extremal, Mar 02 '17 at 17:09
The specific family you should consider is ${ f, f\circ f, f\circ f\circ f, f\circ f\circ f \circ f,\dotsc}$. If $f^k$ denotes the $k$-fold iteration of $f$, what do you know about $Df^k(a)$? — Daniel Fischer, Mar 02 '17 at 18:14
So if $\lambda$ is an eigenvalue of $Df(a)$, then $\lambda^k$ is an eigenvalue of $Df^k(a)$. Next, since $f^k \colon U \to U$, what can you deduce about $Df^k(a)$? — Daniel Fischer, Mar 02 '17 at 18:44
If I can say that eigen values of $Df^k(a)$ are in the unit disk I'm done. But that is the exact thing im stuck at. — Extremal, Mar 02 '17 at 18:49

score 4 · Accepted Answer · answered Mar 02 '17 at 19:38

Since $U$ is bounded, the family $\{ f^k : k \in \mathbb{N}\}$ is a uniformly bounded family of holomorphic maps. By Cauchy's integral formula it follows that the families of partial derivatives

$$\left\{\frac{\partial(f^k)_i}{\partial z_j} : k \in \mathbb{N} \right\}$$

are locally uniformly bounded for all $i,j$. In particular, the components of $Df^k(a) = \bigl(Df(a)\bigr)^k$ are bounded independently of $k$. Therefore there is a constant $C$ such that $\lvert\mu\rvert \leqslant C$ whenever $\mu$ is an eigenvalue of some $Df^k(a)$. Since for an eigenvalue $\lambda$ of $Df(a)$, $\lambda^k$ is an eigenvalue of $Df^k(a)$, it follows that $\lvert\lambda\rvert^k \leqslant C$ for all $k$, and this implies $\lvert\lambda\rvert \leqslant 1$.

So all eigenvalues of $Df(a)$ lie in the closed unit disk. Of course eigenvalues of modulus $1$ are possible (e.g. for $f = \operatorname{id}_U$).

Why does it follow that "there is a constant $C$ such that $|\mu| \leq C$ whenever $\mu$ is an eigenvalue"? — Clyde, Jul 28 '17 at 00:10
@Nick A bound for the components of a matrix implies a bound for the operator norm. And the operator norm of a matrix is a bound for its eigenvalues. — Daniel Fischer, Jul 28 '17 at 12:18

Holomorphic map with a fixed point has Jacobian with all eigen values in the unit disk

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