Let $B= \{z \in \mathbb{C} : |z|< 1\}$ and $\overline{B}= \{z \in \mathbb{C} : |z|\leq 1\}$
Let $C(\overline{B},\mathbb{R})$ be the set of continuous functions from $\overline{B} \rightarrow \mathbb{R}$
and $g$ a continuous function define as following $g: \overline{B} \setminus B = \mathbb{S}^1 \rightarrow \mathbb{R}$.
I have to prove that the following set $X = \{ u \in C(\overline{B},\mathbb{R}) : u\restriction_{B} harmonic \space and \space u\restriction_{\overline{B} \setminus B} = g \}$ has only one element.
I thought this could be proved by defining two functions that satisfy the conditions of $X$ and then showing that is the same.
So let $u_1$ and $u_2$ be two functions in $X$ and $h: u1 - u2 $
Then $h\restriction_{\mathbb{S}^1} = u_1\restriction_{\mathbb{S}^1} - \space u_2\restriction_{\mathbb{S}^1} = g - g = 0$ so $u_1\restriction_{\mathbb{S}^1} = u_2\restriction_{\mathbb{S}^1} = g$.
Now I want to show that $h\restriction_{B} = u_1\restriction_{B} - u_2\restriction_{B} = 0$ using the fact that $u_1$ and $u_2$ are harmonic but I dont know how to do it.
Thank you in advance.
If $u_1,u_2: \overline{B}\to\mathbb{R}$ are both harmonic with $u_1|_{\partial B} = g = u_2|_{\partial B}$ then their difference $u_1 - u_2$ is also harmonic.
Their difference is also identically zero on the boundary. By the maximum principle, a harmonic function is extremized on the boundary of its domain. So $u_1 - u_2 = 0$.