It is well known that all solutions of the Laplace equation $\nabla^2 u = 0$ satisfy the mean value theorem: the average value of $u$ over a sphere equals its value at the center of the sphere.
My question is: does the same theorem hold in case of vector laplacian? Let the vector field $\vec{v}$ satisfy $\nabla^2 \vec{v} = 0$. Is it true that the mean value of $\vec{v}$ over a sphere equals its value at the center of the sphere?
I would imagine the answer to be positive. For example, writing $\vec{v}$ in cartesian basis reduces the vector laplacian $\nabla^2 \vec{v} = 0$ to three ordinary laplacians: $\nabla^2 v_x = \nabla^2 v_y = \nabla^2 v_z = 0$ and makes the statement trivial. However, one can imagine using some other, curvilinear basis, in which the statement is far from trivial. Therefore, I would appreciate some insight or a coordinate-free proof of the theorem.