Proposition 6

Let E be a complex Hilbert space and let $u$ be an endomorphism of E.

a) The set of eigenvalues of $u$ is contained in $\iota (u)$;

b) The spectrum of $u$ is contained in the closure of $\iota (u)$ in $\mathbf{C}$.

Let $\lambda$ be an eigenvalue of $u$ and let $x\in E$ be a nonzero vector such that $u(x) =\lambda x$. Replacing $x$ by $x/\|x\|$ if necessary, one may suppose that $\|x\|= 1$. Then $\langle x|u(x)\rangle =\lambda$, hence $\lambda \in \iota (u)$.

Let us prove b). Considering $u-\lambda 1_E$, we are reduced to proving that if 0 belongs to the spectrum of $u$, then 0 is adherent to $\iota (u)$.

Suppose first that there exists a real number $c >$ 0 such that $\|u(x)\|\geqslant c$ for every $x$ of norm 1 in E. Then the endomorphism $u$ is injective and closed (Lemma 8 of I, p. 107). Since it is not invertible by hypothesis, it is not surjective. Consequently, the orthogonal complement of the kernel of $u^*$ is not equal to E (EVT, V, p. 41, Prop. 4), which proves that the kernel of $u^*$ is not reduced to 0. Thus 0 belongs to $\iota (u^*)$, hence to $\iota (u)$.

If the preceding assumption is not valid, then for every integer $n\geqslant 1$, there exists a vector $x_n$ of norm 1 in E such that $\|u(x_n)\|\leqslant 1/n$. We then have $|\langle x_n|u(x_n)\rangle |\leqslant 1/n$, which implies that 0 belongs to the closure of $\iota (u)$.

Proposition 7 (Hausdorff-Toeplitz Theorem)

Let E be a complex Hilbert space and let $u\in \mathscr{L}(E)$. The numerical range $\iota (u)$ is a convex subset of $\mathbf{C}$.

We shall need two lemmas to prove this proposition.