Harmonic forms and functions on compact manifolds I hope my question is not stupid, I'm studying chapter 5.1 of Claire Voisin's book "Hodge theory and complex geometry".
Let $X$ be a compact manifold and $A^k(X)$ be the space of $C^\infty$ forms on $X$. Then we can define $\delta:=*d*:A^k(X)\rightarrow A^{k-1}(X)$, where $*$ is the hodge star operator. The laplacian $\Delta$ is equal to $\delta d - d \delta$ and $\omega\in A^k(X)$ is $harmonic$ (i.e. $\Delta\omega=0$) iff it's annihilated by both $d$ and $\delta$.
Now let $X$ be a compact subset of $\mathbb{R}^2$ and $f\in A^0(X)$ be a $C^\infty$ function on $X$. Does the previous definition coincide with the usual definition of $f$ harmonic if $\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}=0$?
My guess is no, because according to Voisin's definition it should be $df=0$ which to me is a stronger condition.. But then why are they called the same way?
Thank you!

Yes, except that for functions on a compact manifold with boundary, harmonic is no longer equivalent to being annihilated by $d$. Essentially, the boundary interferes with integrating by parts, as I'll explain.
On functions $f$ with domain (a subset of) $\mathbb{R}^2$, the classical Laplacian $\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}$ agrees with Voisin's $\Delta$ (possibly up to a minus sign). (Note that on functions, $\Delta = \delta d$ since $\delta$ and therefore $d \delta$ are zero on functions.) So indeed, $\Delta f = 0$ (using Voisin's $\Delta$) is equivalent to $\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0$.
But if the domain has a boundary, then it's no longer true that $\Delta f = 0$ implies that $df = 0$. Let's examine the argument: if $\Delta f = 0$, then 
\begin{align*}
0 =& \langle \Delta f, f \rangle \\
  =& \langle \delta d f, f \rangle \\
  =& \langle d f, d f \rangle + \text{(boundary term)}\\
\end{align*}
Here, $\langle \cdot, \cdot \rangle$ is the $L^2$-inner product on functions and/or differential forms. The last equality is by integration by parts. I haven't been very precise about what the "boundary term" is, but the point is this: if the domain is a compact manifold without boundary, or if $f$ is zero on the boundary, then the boundary term is zero, and we can conclude $0 = \langle d f, d f \rangle = ||df||^2$, which implies $df = 0$.
There is also an alternative argument showing $\Delta f = 0$ implies $df =0$ using the maximum principle, but this also relies crucially on the domain being a compact manifold without boundary, or on $f$ vanishing on the boundary. Note that compactness of the domain (or of the support of $f$) is essential to either argument. 
