0
$\begingroup$

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.

$\endgroup$
1

1 Answer 1

2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.