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Let us denote the maximum norm on $\mathbb{C}^n$ as $\lVert \cdot \rVert_{\infty}$ and the normal Euclidean norm on $\mathbb{C}^n$ (arising from the inner product) as $\lVert \cdot \rVert_2$. It is a well-known result in several complex variables that the unit ball $B = \{ z \in \mathbb{C}^n \ | \ \lVert z \rVert_2 < 1 \}$ and the unit polydisc $D = \{ z \in \mathbb{C}^n \ | \ \lVert z \rVert_{\infty} < 1\}$ are not biholomorphically equivalent. However, they are homeomorphic to one another which is shown via the homeomorphism $f : B \to D$ given by $$ f(z) = \begin{cases} \frac{\lVert z\rVert_2}{\lVert z \rVert_{\infty}}z & \text{ if } z \neq 0 \\ 0 & \text{ if } z = 0 \\ \end{cases} $$

Despite the famed result, I'm having trouble understanding why the homeomorphism $f$ isn't a (bi)holomorphic map. My explicit calculations don't seem to be working out how I expect they should (i.e., I expect to get a limit I can't calculate) and every time I think I've shown $f$ (or $f^{-1}$) is not holomorphic I question my working or find an error. I'm probably way overthinking it as it's getting late. Help would be greatly appreciated.

SubSpace626
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1 Answers1

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Let's take two variables for simplicity and consider the map from $\Omega=|z| < \frac{1}{2}$ to $\mathbb{C}$ given as $g=p \circ f \circ i$, where $i:\Omega \to B, i(z)=(z,\frac{1}{2}), p:D \to \mathbb{C}, p(z,w)=z$. Clearly $i,p$ are holomorphic, so if $f$ were holomorphic, $g$ would be too.

A simple computation shows that $g(z)=2z\sqrt{|z|^2+\frac{1}{4}}$ which is highly non-holomorphic on $\Omega$

Conrad
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