Proposition 20

Let E be a complex Banach space and $u$ a closed partial operator on E. Let $\varepsilon \in \mathbf{R}^*_+$. For every $\lambda \in \mathbf{C}$, the following conditions are equivalent:

(i) One has $\lambda \in$ PSp$_{\varepsilon}(u)$;

(ii) Either $\lambda \in$ Sp($u$), or there exists $x\in$ dom($u$) such that $\|x\|= 1$ and $\|(\lambda 1_E-u)(x)\|< \varepsilon$;

(iii) There exists $v\in \mathscr{L}(E)$ such that $\|v\|< \varepsilon$ and $\lambda \in$ Sp($u+v$).

One may suppose that E is nonzero. Condition (ii) is a reformulation of the definition, and hence of condition (i).

Suppose that condition (i) is satisfied and let us prove (iii). If $\lambda$ belongs to the spectrum of $u$, one can take $v= 0$ in (iii).

Suppose therefore that $\lambda \notin$ Sp($u$). By definition of PSp$_{\varepsilon}(u)$, there exists $y\in E$ such that $\|y\|= 1$ and $\|R(u, \lambda )y\|> \varepsilon^{-1}$. Put $x= R(u, \lambda )y$. One has $x\not = 0$. By the Hahn–Banach theorem (EVT, II, p. 67, cor. 2), the linear form $\ell$ on $\mathbf{C}x$ such that $\ell (x) = 1$ admits a continuous extension $\ell_1\in E'$ such that $\|\ell_1\|=\|\ell \|$; hence $\|\ell_1\|=\|x\|^{-1}< \varepsilon$. For every $e\in E$, put $v(e) =\ell_1(e)y$. One has $v\in \mathscr{L}(E)$ and $v(x) =y$. It follows that $(u+v)x=u(x) +y=\lambda x$, therefore $\lambda \in$ Sp($u+v$). Since moreover $\|v\|=\|\ell_1\|< \varepsilon$, condition (iii) is satisfied.

Conversely, let $v\in \mathscr{L}(E)$ such that $\|v\|< \varepsilon$ and $\lambda \in$ Sp($u+v$). The partial operator $\lambda 1_E-(u+v)$ is therefore not injective with a continuous inverse; by Lemma 4 of IV, p. 243, applied to the injective partial operator $\lambda 1_E-u$ and to $-v$, one therefore has $\|v\|\geqslant \|R(u, \lambda )\|^{-1}$. It follows that condition (iii) implies (i).