# How can I show this statement.

Show that there is no holomorphic fuction $f$ in the unit disc $D$ that extends continuously to boundary of $D$ such that $f(z)=\frac{1}{z} ~for~ z\in \partial( D)$.

I tried to apply maximum principle but I couln't find the way to prove it.

I just update the full statement and I think it probably assume it is not constant fuction.

Thank you.

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The constant function $f(z)=0$ is holomorphic on the unit disk and extends continuously to the boundary of the disk. Are you sure you haven't misunderstood or misquoted the exercise? – Henning Makholm Oct 10 '12 at 12:37
I think the word "non-constant" may be missing in the OP. – DonAntonio Oct 10 '12 at 13:24
The non-constant function $f(z)=z$ is holomorphic on the unit disk and extends continuously to the boundary of the disk.= – GEdgar Oct 10 '12 at 13:34
Could the upvoters explain how they understand the question? – Davide Giraudo Oct 10 '12 at 13:44

Suppose that $f(z)={1\over z}$ for $z\in\partial D$. Multiply by $z$, then $$zf(z)-1=0 {\rm\ for\ all\ } z\in \partial D$$ By the maximum principle, we must have $$zf(z)-1=0 {\rm\ for\ all\ } z\in D$$ as well. Now evaluate at $z=0$, and you have a contradiction.

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Can I generalize for every holomorphic function on unit disc? I think it is only for $f(z)=1/z$. Do I misundertands something? – Iikka Tillanen Oct 10 '12 at 21:25
I have assumed that $f(z)$ is holomorphic on the unit disk, of course, and that $f(z)=1/z$ on the boundary of the disk. A similar argument also works if $1/z$ is replaced by a function with a finite number of poles inside the unit disk. – Per Manne Oct 10 '12 at 21:39

If such a function existed, then for all $r < 1$ you'd have $\int_{|z| = r} f(z)\,dz = 0$ by Cauchy's Theorem. Taking limits as $r \rightarrow 1$ would give $\int_{|z| = 1} f(z)\,dz = 0$ as well, contradicting that $\int_{|z| = 1} {1 \over z}\,dz = 2\pi i$.

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$f(z)=\frac{1}{z}$ is just one example which cannot extends continuously to boundary of D ?? Can I define $f(z)=\frac{1}{z}~$ for every holomorphic function on the unit disc when it is on boundary of D? – Iikka Tillanen Oct 10 '12 at 22:36

By maximum principle we get that the function $g(z)=zf(z)-1$ can take the maximum value as $0$, but it can have the value less than $1$ just like in case if $z=0$, $g(0)=-1$, why it is a contradiction?

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