Does the space of harmonic forms change continuously with the metric? Let $(M,g_0)$ be a closed $n$-dimensional Riemannian manifold. Let $1<k<n$ be fixed, and let $\Delta_{g_0}:\Omega^k(M) \to \Omega^k(M)$ be the $g_0$-Laplacian. Let $H^k_{g_0}=\text{ker} \Delta_{g_0}$. 

Suppose $g_{\epsilon}$ is close to $g_0$ in the $C^1$ sense. Is it true that $H^k_{g_0}$ is "close" to $H^k_{g_{\epsilon}}$ in some sense?

Here is one way of quantifying what does it mean for $H^k_{g_0}$ to be close $H^k_{g_{\epsilon}}$: (I guess there are other ways, this seemed natural to me).
Take the sup norm on $\Omega^k(M)$ (where the pointwise norm is the one induced by $g_0$), and let $S$ be the unit sphere of $\Omega^k(M)$ w.r.t this norm. Define $S_0=H^k_{g_0} \cap S, S_{\epsilon}=H^k_{g_{\epsilon}} \cap S.$ Set
$$ d(H^k_{g_{\epsilon}},H^k_{g_{0}}):=d_H(S_{\epsilon},S_0)$$
where $d_H$ is the Hausdorff distance of $H^k_{g_{\epsilon}},H^k_{g_{0}}$ inside $\Omega^k(M)$.

Is it true that if $g_{\epsilon}$ is close to $g_0$ then $H^k_{g_{\epsilon}}$ is close to $H^k_{g_{0}}$ in this sense?

I am not particularly fussed on this specific proximity measure, so results using 
other notions of distance are also welcomed.
 A: I think the answer is positive.
Let $D$ be the subspace of smooth closed $k$-forms on $M$. Equip $D$ with the supremum- $C^1$ norm:
$$
\| \omega \|_{C^1,sup}:=\max\{ \|\omega\|_{sup}, \|T\omega\|_{sup}  \},
$$
All the norms are w.r.t $g_0$.
Let $\delta_g$ be the codifferential of $d$ w.r.t the metric $g$. We consider $\delta_g$ as as family of bounded linear operators, where $g \to g_0$ in $C^1$.
$$
\delta_{g}:(D,\| \cdot \|_{C^1,sup}) \to (\Omega^{k-1}(M),\| \cdot \|_{sup})
$$
The image of $\delta_g$ is closed: Indeed, by the Hodge theorem
$$ \delta_g(\Omega^k(M)) \supseteq  \delta_g(D) \supseteq \delta_g\big(\text{Image}(d)\big) = \delta_g(\Omega^k(M))=(\text{Image}(d)\oplus H_g)^{\perp}$$ 
so $ \delta_g(D)= \delta_g(\Omega^k(M))  $ is closed in $\Omega^k(M)$ w.r.t the $L^2$ metric induced by $g$ hence (since $M$ is compact)  also w.r.t to the the $L^2$ metric induced by $g_0$. This implies it's closed w.r.t the uniform norm $\|\cdot\|_{sup}$.
Also, it is easy to see that
$$\| \delta_{g}-\delta_{g_0} \|_{op} \le C(g_0) \|g-g_0\|_{C^1,sup}, $$
so $\delta_{g} \to \delta_{g_0}$ when $g \to g_0$ in $C^1$.
Thus, we have a family of operators with closed images which converge to a limiting operator, where all the operators in the family have finite-dimensional kernels of the same dimension. This implies that "the kernels converge": we can choose bases for $H_g=\ker \delta_g$ which converge to a basis of $H_{g_0}=\ker \delta_{g_0}$.
The full argument for that claim is here.
