Let $\mathbb{E}_x$ be the expectation associated with a probability measure such that $B_{t\geq0}$ is a Brownian motion started in x. I want to show that for $D\subset\mathbb{R}^2$ bounded, $y\in D, x\in D\setminus\{y\}$ and T the first exit time from D: $$\mathbb{E}_x[log|B(T)-y|] \text{ is harmonic in x}$$
I was trying to prove it with the following theorem, but failed at the local boundedness.
Let U be a domain, $B_t$ a Brownian motion started inside U and T the first hitting time of its boundary. Let f: $\partial U \rightarrow \mathbb{R}$ be measurable and such that the function $u:U \rightarrow \mathbb{R}$ with $$u(x)=\mathbb{E}_x[f(B_T)], \text{ for every }x\in U,$$ is locally bounded. Then u is a harmonic function.
Any help is appreciated.
As $D$ is open and $y \in D$, we have
$$r := d(y,\partial D) := \inf\{z \in \partial D; |z-y| \}>0.$$
On the other hand, the boundedness of $D$ implies that we can choose $R>0$ such that $|z| \leq R$ for any $z \in \partial D$. Consequently,
$$\log r \leq \log|B(T)(\omega)-y| \leq \log (R+|y|)$$
holds for any $\omega \in \Omega$. Taking expectation yields
$$\log(r) \leq \underbrace{\mathbb{E}^x \log(|B(T)-y|)}_{=: u(x)} \leq \log(R+|y|)$$
for any $x \in D$. This shows that $u$ is not only locally bounded, but bounded. Therefore, the harmonicity follows by applying the theorem mentioned in the question.