Theorem 4

Let $u$ be a self-adjoint partial operator on E. The essential spectrum of $u$ is the intersection of the sets Sp($u+v$), where $v$ runs through the set of compact hermitian endomorphisms of E.

The partial operator $u+v$ is self-adjoint for every hermitian endomorphism $v$ of E, since $(u+v)^*=u^*+v^*($cf. IV, p. 236).

Let us prove that if $v$ is compact, then Sp$_e(u)\subset$ Sp$_e(u+v)$. Let $\lambda \in$ Sp$_e(u)$, and let $(x_n)_{n\in\mathbf{N}}$ be an orthonormal sequence in E such that $u(x_n)-\lambda x_n$ converges to 0 (Prop. 18). The sequence $(x_n)$ converges weakly to 0 in E (Lemma 13), and since the endomorphism $v$ is compact, the sequence $(v(x_n))_{n\in\mathbf{N}}$ converges to 0 in E (Cor. of Prop. 6 of III, p. 6). Consequently the sequence $(u+v)(x_n)-\lambda x_n$ converges to 0 in E, and Prop. 18 implies that $\lambda \in$ Sp$_e(u+v)$.

The essential spectrum of $u$ is therefore contained in the intersection of the sets Sp($u+v$) for hermitian $v\in \mathscr{L}^c(E)$.

Conversely, let $\lambda \in$ Sp$_s(u)$. Let $E_{\lambda}$ be the proper subspace of $u$ relative to $\lambda$, and $p_{\lambda}$ the orthoprojector with image $E_{\lambda}$; it is the spectral projector of $u$ relative to $\{\lambda \}$ (Cor. of Prop. 9 of IV, p. 278). By definition of the sensitive spectrum, the projector $p_{\lambda}$ is of finite rank, hence compact. Put $w=u+p_{\lambda}$; it is a self-adjoint partial operator. To conclude the proof, let us verify that $\lambda$ belongs to the resolvent set of $w$.

We have $E_{\lambda}\subset$ dom($u$), and the spaces $E_{\lambda}$ and dom($u$)$\cap E^{\circ}_{\lambda}$ are invariant under $u$ (Prop. 9 of IV, p. 278).

Let $x\in$ dom($u$). Write $x=p_{\lambda}(x)+y$, where $y\in$ dom($u$)$\cap E^{\circ}_{\lambda}$. From the above, we have

Let V be an open neighbourhood of $\lambda$ not meeting the spectrum of $u$, and let $c >0$ be such that the disk with center $\lambda$ and radius $c$ is contained in V. Let $\mu_y$ be the spectral measure of $y$ relative to $u$. Since $y$ is orthogonal to $E_{\lambda}$, the support of $\mu_y$ does not meet V (loc. cit.). We then have

It follows from (21) and (22) that $\|w(x)-\lambda x\|^2\geqslant$ inf($c^2,1$)$\|x\|^2$; since $w$ is self-adjoint and $\lambda \in \mathbf{R}$, the conclusion then follows from Lemma 5 of IV, p. 248.