# Very quick question: function extends to 1/z on the boundary of unit disc

How can one show there is no holomorphic function $f$ on the open unit disc $\mathbb{D}$ such that it extends continuously to $\frac{1}{z}$ on $\partial\mathbb{D}$? I mean $f$ takes value $\frac{1}{z}$ when $|z|=1$ and continuous here.

Thank you.

#### Solutions Collecting From Web of "Very quick question: function extends to 1/z on the boundary of unit disc"

Let us assume that $f:\mathbb{D}\rightarrow\mathbb{C}$ is holomorphic and extends continuously to $\partial\mathbb{D}$ with $f(z)=\frac{1}{z}$ for $z\in\partial\mathbb{D}$.

By the maximum principle, we have $|f(z)|\le 1$ for all $z\in\mathbb{D}$, since $|f(z)|=|\frac{1}{z}|=1$ for $z\in\partial\mathbb{D}$.

Consider the function $g(z)=zf(z)$. $g$ is also holomorphic, satisfies $|g(z)|\le 1$ and extends continously to $\partial\mathbb{D}$ with $g(z)=1$ for $z\in\partial\mathbb{D}$.

By the mean value property, we have

$$g(0)=\frac{1}{2\pi} \int_0^{2\pi} g(e^{i\phi})d\phi=\frac{1}{2\pi}\int_0^{2\pi} d\phi=1$$

By the maximum principle we get that $g$ must be constant. Since it is $1$ on the boundary, we have $g(z)=1$ for all $z\in\overline{\mathbb{D}}$.
Therefore

$$f(z)=\frac{g(z)}{z}=\frac{1}{z}$$

That is a contradiction, because $1/z$ has a pole at $0$. Hence there exists no such $f$.

Another approach: it’s straightforward to show that Cauchy’s integral theorem holds for functions that are continuous on $\bar{\mathbb{D}}$ and holomorphic on $\mathbb{D}$. Hence, if $1/z$ has a holomorphic extension, the integral of $1/z$ along the unit circle would vanish, which of course it doesn’t.