Looking for pathologic counterexample: Nonzero harmonic function which is zero on the unit circle except 1 From my Complex Variables class: Let $C_1$ be the unit circle, $B_1$ the open unit disc and $\Gamma = C_1 \backslash \{1\} $.
I am looking for a nonzero function $u \in C(B_1 \cup \Gamma)$ which is harmonic on $B_1$ and has the following property:

$u_{|\Gamma} = 0$

I have found out that $\lim_{z\rightarrow1}u(z) \neq 0$, because if this was the case, we could extend $u$ to $B_1 \cup C_1$ continuously and the Poisson kernel would tell us that $u = 0$ on the whole domain:

$u(z) = \frac{1}{2\pi}\int_{|\xi|=1}\underbrace{u(\xi)}_{=0}\cdot\frac{1-|z|^2}{|\xi-z|^2}|d\xi|=0$

Also, I know that $u$ can't be bounded (since we have to show this in the second part of the question).

The only harmonic nonzero function which satisfies $u_{|\Gamma}=0$ that I could come up with is $u(z) = \log |z|$ - but of course, this function is not defined for $z=0$.

Any other ideas?
 A: The Möbius transformation $T(z) = \dfrac{1+z}{1-z}$ maps the unit disk conformally onto the right
halfplane. $\Gamma = C_1 \backslash \{1\}$ is mapped onto the imaginary axis, and $\lim_{z \to 1} T(z) = \infty$.
Then $f(z) = T(z)^2$ maps the unit disk conformally onto 
$\mathbb C \setminus (-\infty, 0] $. $\Gamma$ is mapped to
the negative real axis, and
$\lim_{z \to 1} f(z) = \infty$.
Therefore
$$
u(z) = \text{Im } f(z) = \text{Im } \bigl( \frac{1+z}{1-z}\bigr)^2
$$
is harmonic in the unit disk and has the desired properties.
A: The boundary "function" cannot be a purely classical function, but maybe could be a constant multiple of a Dirac delta at $1$. This has Fourier expansion $\sum_{n\in \mathbb Z} 1\cdot e^{in\theta}$, which can be viewed as some sort of limit of $\sum_{n\ge 0}z^n + \sum_{n\ge 1}\overline{z}\,^n$, and these geometric series can be summed... (This sort of discussion can also give rise to the Poisson kernel...)
A: You already knew an example: The Poisson kernel based at $1,$ namely $z\to (1-|z|^2)/|1-z|^2.$
