Unbounded operator whose spectrum is the entire complex plane?

Question

The question is simple: how to find an unbounded operator $T:H\to H$ where $H$ is a Hilbert space such that $\text{Sp} T = \mathbb C$? This seems a very basic thing, but I have not found an example in the literature.

In some proofs, we need to consider this case separately. This example should be quite important.

Answer 1

Yes, such an example can be found on page 254 of Reed & Simon's Methods of Modern Mathematical Physics I.

Let $AC[0,1]$ be the family of all absolutely continuous functions on $[0,1]$ whose derivatives are in $L^2[0,1]$. Let $T:L^2[0,1] \to L^2[0,1]$ be the densely defined operator $i \frac{d}{dx}$ whose domain is the set $$ D(T) = \{ f \in L^2[0,1] : f \in AC[0,1] \}. $$ $T$ is then a closed operator whose spectrum is the entire complex plane $\Bbb C$. Indeed, observe that $$ (\lambda I - T)e^{-i\lambda x} = 0 $$ and the function $f(x) = e^{-i\lambda x}$ belongs to $D(T)$ for each $\lambda \in \Bbb C$.

Answer 2

Here is an elementary example.

Let $a_n$ be a sequence dense in the complex plane. For example $a_n$ is an enumeration of $\mathbb{Q}\oplus i\mathbb{Q}.$ Consider the operator $T$ on $\ell^2(\mathbb{N})$ given by $$T\{x_n\}=\{a_nx_n\}$$ with domain $$D(T)=\left\{\{x_n\}\,:\, \sum |a_nx_n|^2<\infty\right \}$$ Then $T$ is closed and the spectrum is equal to the entire complex plane.

Another example: let $$(Tf)(x,y)=(x+iy)f(x,y)$$ act on $L^2(\mathbb{R}^2)$ with domain $$D(T)=\left\{f\,:\, \iint (x^2+y^2)|f(x,y)|^2 \,dx\,dy<\infty\right \}$$

Answer 3

Consider $L^2(\mathbb C,dA(z))$ with $dA(z)$ the area measure and take $$ Tf(z)=zf(z). $$ Recall that the spectrum is the (essential) range of the function defining a multiplication operator.