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Please help me solve this problem:

Let $g(z):\{z: |z|=1\}\to\mathbb{C}$ be a continuous function. Show that there is an analytic function $f(z)$ in the $\{z: |z|<1\}$ such that $f$ has continuous extension to $g$.

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This is not true. Take $g(z)=\dfrac1z$. Cauchy's integral theorem shows that $g$ can't be extended holomorphically.

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Beat me to it, and with a better example! – Potato Sep 28 '13 at 22:40
Is it impossible that there are another analytic function which its restriction to a unit circle is $\displaystyle\frac{1}{z}$. – Worawit Tepsan Sep 28 '13 at 22:43
@WorawitTepsan I don't think you understand. Take the function that is $1/z$ on the unit circle. This function cannot be extended. Your claim is not true. – Potato Sep 28 '13 at 22:44
@N.S. It's true for $|z|=r$ for any $r<1$ and if $f$ is continuous on $|z|\le 1$, uniform continuity allows us to push out to $|z|=1$. (It is true for any bounded domain with rectifiable boundary, but that's more difficult.) – mrf Sep 28 '13 at 23:00
@WorawitTepsan I think you are missing the point. There are many functions on the disc equal to $1/z$ on the circle but no analytic ones, since the integral along the circle is non-zero. – mrf Sep 28 '13 at 23:18

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