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I am trying to understand the proof of the following result: Let $M$ be a convex domain in $\mathbb{C}^n$. Let $\mathbb{D}$ be the open unit disc in $\mathbb{C}$. If $\phi: \mathbb{D}\longrightarrow \bar{M}$ is holomorphic then either $\phi( \mathbb{D})\subset M$ or $\phi( \mathbb{D})\subset \partial M$.

The proof goes like this: Assume that $z\in \phi( \mathbb{D})\cap \partial M$. By convexity, we can choose a $\mathbb{C}$ linear functional $l$ such that Re $l(z)>$Re $l(w)$ for every $w\in M$. Hence $l(z)\in(l\circ \phi)( \mathbb{D})$ lies in the boundary of $l\circ \phi( \mathbb{D})$. By the open mapping theorem, $l\circ \phi$ is constant, So that $\phi( \mathbb{D})$ cannot contain points in $M$. Hence $\phi( \mathbb{D})\subset \partial M$.

I was able to understand the proof till the part where we get $l\circ \phi$ is constant. But why does it imply that $\phi( \mathbb{D})$ cannot contain points in $M$?

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If there exists $w\in \mathbb{D}$ such that $\phi(w)\in M,$ then $\mathrm{Re}\; l\circ \phi(w)<\mathrm{Re}\;l(z)$ by assumption. However, $z\in \phi(\mathbb{D})$ and thus, $l\circ \phi$ cannot be constant.

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