$f(z)$ is analytic in the unit circle - can you prove it's a constant function? $f(z)$ is analytic in the open unit disk and continuous on its edge.
Can you prove that if $f(z)=1$ on the upper half of the unit circle (for $z=e^{i\theta}, 0\le\theta\le\pi$) then $f(z)$ in constant in the unit disk?
 A: By the reflection principle the function
$$g(z):=\cases{f(z)\quad&$\bigl(|z|\leq 1\bigr)$\cr
&\cr
\overline{f\bigl(1/\bar z\bigr)} &$\bigl(|z|>1\bigr)$\cr}$$
is analytic in an open neighborhood $U$ of the point $i$ and constant on an arc through this point. It follows that $f$ is constant on $U\cap D$; therefore $f$ has to be constant on all of $D$.
A: Consider the function $g(z) = \overline{f(\bar z)}$. The function $g(z)$ is analytic in the unit disk and equals $1$ on the lower half of the unit circle. Now let $h(z) = (f(z) - 1) (g(z) -1)$. We have that $h(z) = 0$ both on the lower and upper half of the unit circle. By the maximum modulus principle, $h(z)$ is identically equal to $0$. Therefore, for every $z$ either $f(z) = 1$ or $g(z) =1$ (or both). Since functions $f(z)$ and $g(z)$ are continuous, one of them must be equal to $1$ in some neighborhood of $0$, and thus be equal to $1$ identically.
A: By the principle of permanence for analytic functions, any analytic function that is 0 on a set including one of its limit points must be 0 everywhere.
By subtracting the constant function equal to 1 everywhere from the function in question, we can apply the above principle to the difference and conclude that the original function is constant.
See e.g. https://mathworld.wolfram.com/PrincipleofPermanence.html.
