Divergence theorem and Green's identities Let $V$ be a simply-connected region in $\mathbb{R^3}$ and  $C^1$ functions $f,g:V\to \mathbb{R}$ . 
Prove that if $\nabla^2f=\nabla^2g=0$ then:
$f=g$ $\iff f(x)=g(x)$ for $x\in\partial V$
To prove $\Rightarrow$ is easy. If $f=g$ then for every $x$ in general $f(x)=g(x) $   let alone $x\in\partial V$.
I had some difficulties in proving $\Leftarrow$ . I tried to prove it using Green's first identity, but I got stuck.
 A: Probably you don't need Green's identity but similar idea as proof in Green's identity. The key technique is Divergence theorem. Consider identity:
$$\begin{align}
&\int_V\nabla\cdot(f\nabla f-f\nabla g)dV\\
=&\int_V(\nabla f\cdot\nabla f+f\Delta f-\nabla f\cdot\nabla g-f\Delta g)dV\\
=&\int_V\nabla f\cdot(\nabla f-\nabla g)dV\\
=&\oint_{\partial V}(f\nabla f-f\nabla g)\cdot dS\\
=&0
\end{align}$$
The third line uses $\Delta f=\Delta g=0$. And the fourth line applies divergence theorem. The fifth line uses $f=g$ on the boundary. We can conclude from the third line that
$$\nabla f\cdot(\nabla f-\nabla g)=0$$
Consider another identity
$$\begin{align}
&\int_V\nabla\cdot(f\nabla f-g\nabla g)dV\\
=&\int_V(\nabla f\cdot\nabla f+f\Delta f-\nabla g\cdot\nabla g-g\Delta g)dV\\
=&\int_V(\nabla f\cdot\nabla f-\nabla g\cdot\nabla g)dV\\
=&\oint_{\partial V}(f\nabla f-g\nabla g)\cdot dS\\
=&0
\end{align}$$
We can conclude from the third line that
$$|\nabla f|=|\nabla g|$$
Combine two conclusions, we have for any $x\in V$
$$\nabla(f-g)=\nabla f-\nabla g=0$$
Or there's some constant $C$ such that
$$f=g+C$$
Note that $f=g$ on the boundary, which requires $C=0$.

Actually, there's another easier way to look at this problem. Since
$$\Delta(f-g)=\Delta f-\Delta g=0$$
We know $h=f-g$ a harmonic function on $V$. Further we have $h=0$ is constant on the boundary. Based on mean value property of harmonic function, $h(x)=0$ for all $x\in V$.
A: Hint: Use Green's third identity with the Green's function which is identically zero on the boundary of $V$.
