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Let $D\subseteq\Bbb C$, $f:D\to\Bbb C$ holomorphic on $D$. Let $U$ be an open subset strictly contained in $D$: in this way $\partial U$ would be contained in $D$.

So I was asking myself if $\partial f(U)=f(\partial U)$ is always true. It seems obvious simply passing to the limit; am I right?

Joe
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1 Answers1

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No, $\partial f(U) = f(\partial U)$ need not be true. As an example, take $D = \Bbb C$, $$U = \{ z : |z| < 1 \text{ and } \operatorname{Re} z > 0 \} $$ and $f(z) = z^4$. Then $$ f(U) = \{ z : |z| < 1 \} \setminus \{ 0 \} $$ and $$ f(\partial U) = \{ z : |z| = 1 \} \cup [0, 1) $$ is a strict superset of $$\partial f(U) = \{ z : |z| = 1 \} \cup \{ 0 \} \, . $$

If $U$ is bounded then $\partial f(U) \subset f(\partial U)$ holds.

Proof: $\overline U$ is compact, therefore $f(\overline U)$ is compact as well. It follows that $$ \overline{f(U)} \subset f(\overline U) \, . $$ A (non-constant) holomorphic function is an open mapping, therefore $f(U)$ is an open set, so that $$ \partial f(U) = \overline{f(U)} \setminus f(U) \subset f(\overline U) \setminus f(U) \subset f(\overline U \setminus U) = f(\partial U) \, . $$

If, in addition, $f$ is injective (or more generally, a proper map from $U$ to $f(U)$) then equality holds.

Proof: For a proper map it holds that $$ \tag{*} f(U) \cap f(\partial U) = \emptyset $$ and therefore $$ f(\partial U) \subset f(\overline U) \subset \overline{f(U)} \\ \Longrightarrow f(\partial U) \subset \overline{f(U)} \setminus f(U) = \partial f(U) \, . $$

Proof of $(*)$: Assume that $w_0 \in f(U) \cap f(\partial U)$. Let $K$ be a compact disk such that $w_0 \in K \subset f(U)$. From $w \in f(\partial U)$ it follows that there is a sequence $(z_n)$, $z_n \in U$, such that $z_n \to z_0 \in \partial U$ and $f(z_0) = w_0$. Then $f(z_n) \to f(z_0) = w_0$ and therefore $f(z_n) \in K$ for $n \ge n_0$, so that $z_n \in f^{-1}(K)$ for $n \ge n_0$. If $f$ is a proper map then $ f^{-1}(K)$ is compact in $U$ which is a contradiction to $z_n \to z_0 \in \partial U$.

Martin R
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    ($f(\overline{U}) = \overline{f(U)}$ by the maximum modulus ?) – reuns Jun 14 '16 at 02:21
  • @user1952009: Yes, but (unless I am mistaken), the statement is true even for continuous proper maps, the holomorphy is not really needed. – Martin R Jun 14 '16 at 02:30
  • Actually, in your example $f(U)={z:0<|z|<1}$, since $0$ is not in $U$ or $f(U)$. Hence $\partial f(U)={z:|z|=1}\cup{0}$ as well. – Mario Carneiro Jun 14 '16 at 05:49
  • @MartinR: First of all thanks for the answer. Then: why does $f(U)\cap f(\partial U)=\emptyset$? – Joe Jun 14 '16 at 07:37
  • @MarioCarneiro: You are right, thanks! – Martin R Jun 14 '16 at 07:51
  • @Joe: I have expanded the argument. Actually there was an error previously, I hope that it is correct now. – Martin R Jun 14 '16 at 08:17
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    Continuity of f is definitely not enough for the inclusion, you need that f is open which is true for all noncombatant holomorphic maps. – Moishe Kohan Jun 14 '16 at 17:23
  • @studiosus: I hope that it is correct now ... – Martin R Jun 15 '16 at 03:06
  • @MartinR: does $\partial f(U)\subseteq f(\partial U)$ hold even when $U$ is unbounded? – Joe Jun 23 '16 at 10:55
  • @Joe: Not necessarily. $f(z) = e^z$ maps $U= { z : 0 < \operatorname{Im} z < \pi }$ to the upper half-plane, but $w = 0 \in \partial f(U)$, $w \notin f(\partial U)$. – It would work if you consider the extended complex plane $ \hat {\Bbb C}$because that is compact (with the spherical metric). – Martin R Jun 23 '16 at 11:18