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The given problem says , Let $f$ be a complex holomorphic on the open unit disk $D$ such that $|f(z)|\longrightarrow1$ as $|z| \longrightarrow 1$, and $f$ is nonzero inside the open unit disk.Can such $f$ be extended uniquely to the boundary and also continuous on it?

I'm just able to extend $|f|$ uniquely continuous on the boundary.But not $f$!

Subhajit Saha
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2 Answers2

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If $f$ is holomorphic and non-zero in the unit disk with $|f(z)| \to 1$ for $|z| \to 1$, then $f$ is necessarily constant in the unit disk, see for example

It follows that such a function $f$ can always be extended continuously to the boundary.

Martin R
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Suppose that $f$ is not constant and suppose that there is a continuous function $g: \overline{D} \to \mathbb C$ such that $f=g$ on $D.$

By the maximum principle qwe get for $z \in D:$

$$ |f(z)| < \max \{ |g(z)| : z \in \partial D\}=1.$$

Since $f$ has no zeroes in $D$, the same arguments show that $\frac{1}{|f(z)|}<1$ for $z \in D.$ Hence $|f(z)>1$ for all $z \in D,$ a contradiction.

Conequence: the function $g$ exists only in the case of a constant $f$.

Fred
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  • I'm covinced with your ans, You assumed g as continuous on $cl D$ ,such that $f=g$ on $D$ ,but my question is how are you sure that such $g$ exists? – Subhajit Saha Apr 29 '19 at 14:31