Let $\mathcal{H}_i$, for $i=1,2$, be Hilbert spaces, and $T:{\frak Dom}(T) \subseteq \mathcal{H}_1 \to \mathcal{H}_2$ a densely-defined closed operator. If the kernel of $T$, and the kernel of its adjoint $T^*:{\frak Dom}(T^*) \subseteq \mathcal{H_2} \to \mathcal{H}_1$, are both finite dimensional, then does it follow that $T$ is Fredholm, which is to say, will $T$ have closed range?
I would guess not, but cannot produce an example.
No. Consider $P:\ell^2(\mathbb Z)\supset D(P)\rightarrow\ell^2(\mathbb Z)$ defined by $P(e_n) = e^n e_{n} $ with $$D(P)=\{ (\xi_n)\in\ell^2(\mathbb Z) : \sum_{n=-\infty}^\infty e^{2n} |\xi_n|^2 < \infty \}.$$ Then $P$ is positive, self-adjoint (closed and densely defined) injective, and $P$ does not have closed range.
Let $K$ be a finite-dimensional Hilbert space, and define $T$ on $K\oplus\ell^2(\mathbb Z)$ to be $P$ on $\ell^2(\mathbb Z)$ and $0$ on $K$ (so $D(T) = \{(\xi,\eta) : \xi\in K, \eta\in D(P) \}$). Then $T^*=T$ and $T$ has finite dimensional kernel, but the image of $T$ is not closed.
