Take $X=L^2(\mathbb{R})$ and for every $t \in \mathbb{R}$ and $z \in X$ define:
$$(\mathbb{T}_tz)(x) = z(t+x), \forall x \in \mathbb{R}$$
Such operator is unitary and it is called bilateral left-shift group. It then says that the generator is A such that:
$$D(A) = H^1(\mathbb{R}) \text{ (Sobolev space),}$$ $$A = \frac{d}{dx}$$
Now, I don't understand why $\sigma(A) = i\mathbb{R}$. It seems to me that they are not eigenvalues since the related eigenvectors will be in the form $z(x) = e^{ikx}, k \in \mathbb{R}$, which are not in the Sobolev space $H^1(\mathbb{R})$ (because not $L^2$ functions). So, if it is not point spectrum, what kind of spectrum it is?