General "Hodge theorem" I know basically zero Hodge theory, so this question might be weird.
Let $$A \stackrel{S}{\longrightarrow} B \stackrel{T}{\longrightarrow} C$$
be a sequence of closed, densely defined maps of Hilbert spaces, with $T \circ S = 0$.  Note that the conditions imply that the adjoints of both maps are also densely defined.
Let $\Delta = SS^*+T^*T$ be the "Laplacian" of this complex.
Call an element $b$ of $B$ "harmonic" if $\Delta b = 0$.
Question 1 Is it true that every cohomology class $[b] = \{ b+ Sa : a \in \textrm{Dom}(S)\}$ has a harmonic representative?
I suspect the answer is "no", since otherwise proofs of the Hodge theorem would be presented in this generality.
Question 2 Can someone give me a natural example of where this fails?
Question 3 At this level of generality, are there any additional hypotheses which would yield a Hodge theorem? 
Some motivation: 
I frequently use the lemma that, in this situation, all closed forms are exact precisely when $\langle \Delta b,b \rangle \geq C\lVert b \rVert ^2$ for some constant $C$.
The proof of this is a slightly tricky application of Hahn-Banach.
If the "Hodge theorem" was true, the lemma would be a corollary of it, since the inequality implies 0 is the only harmonic form, and thus that the cohomology must be trivial.

Response to questions 1 and 2:
Take $A=B = L^2(\mathbb{Z})$ and $C=0$. Let $S((a_i)) = (a_i - a_{i-1})$ and $T$ be the zero map. Since $\sum_i (a_i- a_{i-1})^2 \leq \sum_i (2 a_i^2 + 2 a_{i-1}^2) = 4 \sum_i a_i^2$, the operator $S$ is bounded and hence closed. Note that any sequence $(b_i)$ in the image of $S$ obeys $\sum_{i=-\infty}^{\infty} b_i =0$, so $S$ is not surjective.
We have $S^{\ast}(b_i) = (b_i - b_{i+1})$, so $\Delta(b_i) = (-b_{i-1} + 2 b_i - b_{i+1})$. The kernel of $\Delta$ is sequences of the form $b_i = u i + v$. However, such a sequence is in $L^2(\mathbb{Z})$ if and only if $u=v=0$. So there are no nonzero harmonic functions in $B$.
I think you should be able to make a similar example with $L^2(\mathbb{R})$ and differentiation, but my lack of confidence in functional analysis terminology lead me to use the discrete version.
