Proof: superharmonic function equal on $\partial D$ and at one point inside of D to its harmonic function, is harmonic on D (D compact)

Question

I am looking for a proof (literature or short idea) for the following statement, which I have found in several sources:

Let $M$ be a riemannian manifold, let $f:M\to\mathbb{R}$ be a superharmonic function, let $h$ be the unique harmonic function on the compact subset $D\subset M$ with $f(x) = h(x)$ for all $ x\in \partial D$ and $f(y) = h(y) $ for one $y\in \mathring{D}$. Then: $f|_{D} \equiv h$

Thank you all!

(please correct me if I have misstated the statement)

Answer 1

As both $f$ and $h$ are superharmonic, so is $f-h$. Now since $f(y) = h(y)$ and $f(x)\geq h(x)$ for all $x\in D$:

$$(f-h)(y) = 0 \leq (f-h)(x)\ \forall x \in \mathring{D}$$ So there exists a point $x_0=y\in D$ and a neighbourhood $U=D$ around $x_0$ s.t.

$$(f-h)(x_0) \leq (f-h)(x)\ \forall x \in U$$

Now with the minimum principle for superharmonic functions, it follows that $f-h$ is constant and therefore $f\equiv h$.