Dense Operators: Spectrum

Question

This thread is Q&A.

Given a Banach space $E$.

Consider closed operators: $$T:\mathcal{D}(T)\subseteq E\to E:\quad T=\overline{T}$$

Then for the domain: $$\sigma(T)\neq\mathbb{C}\implies\overline{\mathcal{D}(T)}=E$$

How can I prove this?

Answer 1

Given the Hilbert space $\ell^2(\mathbb{N}_0)$.

Consider the left shift: $$L_0:\mathcal{D}(L_0)\subseteq\ell^2(\mathbb{N}_0)\to\ell^2(\mathbb{N}_0):\quad L_0:=L$$

For nondense domain: $$\mathcal{D}(L_0):=\ell^2(\mathbb{N}):=\{\varphi\in\ell^2(\mathbb{N}_0):x_0=0\}$$

But it has an inverse: $$0\not\in\sigma(L_0):\quad RL_0=1_0\quad L_0R=1$$

Concluding counterexample.