One has $|\lambda_i|\leqslant \|u\|$ for every $i$, whence sup$_{i\in I}|\lambda_i|\leqslant \|u\|$. Since moreover

$\|u(x)\|^2=\sum_{i\in I}|\lambda_i|^2|\langle e_i|x\rangle |^2\leqslant ($sup$_{i\in I}|\lambda_i|^2)\|x\|^2$

for every $x\in E$ (EVT, V, p. 22, prop. 5), it follows that $\|u\|=$ sup$|\lambda_i|$.

b) For every bounded family $\lambda = (\lambda_i)_{i\in I}$, the family $(\lambda_ip_i)_{i\in I}$ is summable in the space $\mathscr{L}(E)$ endowed with the topology of simple convergence, and the sum of this family is the unique mapping $u\in \mathscr{D}_B(E)$ such that $\alpha (u) =\lambda$;

c) Let $u\in \mathscr{D}_B(E)$ and let $\lambda$ be the family of its eigenvalues relative to B. The spectrum of $u$ is the closure in K of the set of values of $\lambda$;

d) If K = $\mathbf{C}$, then the continuous functional calculus mapping of $\mathscr{C}$ (Sp($u$)) into $\mathscr{L}(E)$ associates with a continuous function $f\in \mathscr{C}$ (Sp($u$)) the endomorphism $\alpha (f\circ \lambda )$ in $\mathscr{D}_B(E)$.

By Lemma 1, the family of eigenvalues of an endomorphism diagonal in B is bounded and the mapping of $\mathscr{D}_B(E)$ into $\mathscr{B}(I)$ thus defined is a continuous isometric morphism of unital Banach algebras.

Let $i\in I$. For every $j\in I$, one has $\langle u^*(e_i)|e_j\rangle =\lambda_j\langle e_i|e_j\rangle =\langle \overline{\lambda_j}e_i|e_j\rangle$. It follows that $u^*(e_i) =\overline{\lambda}_ie_i$. The adjoint of $u$ is therefore the endomorphism diagonal in the basis B whose family of eigenvalues is $(\lambda_i)_{i\in I}$. Assertion a) follows.

Let $\lambda = (\lambda_i)_{i\in I}\in \mathscr{B}(I)$. For every $x\in E$, the family $(|\langle e_i|x\rangle |^2)_{i\in I}$ is summable, with sum $\|x\|^2$ (EVT, V, p. 22, prop. 5), whence, for every finite subset J of I:

$\|\sum_{i\in J}\lambda_ip_i(x)\|^2=\sum_{i\in J}|\lambda_i|^2|\langle e_i|x\rangle |^2\leqslant ($sup$_{i\in I}|\lambda_i|^2)(\sum_{j\in J}|\langle e_i|x\rangle |^2$

$\leqslant ($sup$_{i\in I}|\lambda_i|)^2\|x\|^2$.

Consequently, the family $(\lambda_ip_i)$ is summable in $\mathscr{L}(E)$ endowed with the topology of simple convergence. Its sum $u_{\lambda}$ satisfies $u_{\lambda}(e_i) =\lambda_ie_i$; it is therefore an endomorphism diagonal in the basis B, with eigenvalues $\lambda$. Assertion b) follows.

The last assertions result from a), from Example 3 of I, p. 17 and from Example 4 of I, p. 111.

a) The following conditions are equivalent:

(i) One has $\lambda \in \mathscr{C}_0(I; K)$;

(ii) The endomorphism $u$ is compact ;

(iii) The family $(\lambda_ip_i)_{i\in I}$ is summable in the Banach space

$\mathscr{L}(E)$. Its sum is then equal to $u$.

b) Suppose that $u$ is compact and denote by Λ the set of values of $\lambda$. The set of the $i\in I$ such that $\lambda_i\not = 0$ is countable. If E is of infinite dimension, one has Sp$_s(u) = \Lambda -\{0\}$ and Sp($u$) $= \Lambda \cup \{0\}$. If E is of finite dimension, then Sp$_s(u) =$ Sp($u$) $= \Lambda$.

By Lemma 1, one has $\|\sum_{j\in J}\lambda_jp_j\|=$ sup$_{j\in J}|\lambda_j|$ for every finite subset J of I. It follows, by the Cauchy criterion, that the family $(\lambda_ip_i)$ is summable in $\mathscr{L}(E)$ if and only if the family $\lambda$ tends to 0 at infinity, which implies that conditions (i) and (iii) are equivalent.

Condition (iii) implies that $u$ is compact (corollary of Proposition 2 of III, p. 4). Conversely, suppose that the endomorphism $u$ is compact. Since the family B is bounded in E, its image by $u$ in E is relatively compact in E (III, p. 2), and hence precompact. Let $\varepsilon >0$ and J be a finite subset of I such that the image of B by $u$ is contained in the union of the balls of radius $\varepsilon$ and centre $u(e_j)$ for $j\in J$. Let $i\in I$ - J. There exists $j\in J$ such that $\|u(e_i)-u(e_j)\|\leqslant \varepsilon$, hence

Consequently, one has $\lambda \in \mathscr{C}_0(I; K)$, that is, condition (i).

Finally, the spectrum of $u$ and its sensitive spectrum are calculated as functions of $\lambda$ by using prop. 1, c) and Proposition 5 of III, p. 90.

The set Sp$_s(u)$ is countable, and it does not contain 0 if E is of infinite dimension (III, p. 90, prop. 5, b)). For every element $\lambda \in$ Sp$_s(u)$, denote by $N_{\lambda}$ the nullspace of $u-\lambda 1_E$. It is of finite dimension (loc. cit.) and, since $u$ is normal, it coincides with the eigenspace of $u$ relative to $\lambda$ (EVT, V, p. 43, cor. of prop. 8). The spaces $N_{\lambda}$ are pairwise orthogonal (I, p. 132, n$^o5$).

Let F be the subspace of E Hilbert sum of the spaces $N_{\lambda}$ for $\lambda \in$ Sp$_s(u)-\{0\}$. It is a space of countable type, stable under $u$ since each subspace $N_{\lambda}$ is stable under $u$. Since $N_{\lambda}$ is also the eigenspace of $u^*$ corresponding to $\overline{\lambda}$ (EVT, V, loc. cit.), the endomorphism $u$ induces an endomorphism $\widetilde{u}$ of $F^{\circ}$ by passing to subspaces. The endomorphism $\widetilde{u}$ is compact (Proposition 3 of III, p. 5) and normal (Lemma 4 of I, p. 135). By construction, the point spectrum of $\widetilde{u}$ is contained in $\{0\}$ (for, any eigenvector for $\widetilde{u}$ would be one for $u$, hence would belong to one of the spaces $N_{\lambda}$ if the corresponding eigenvalue were non-zero). Thus the spectral radius of $\widetilde{u}$ is zero, whence $\widetilde{u}= 0$ since $\widetilde{u}$ is normal (Corollary 1 of I, p. 108). Hence $F^{\circ}\subset$ Ker($u$).

For every $\lambda \in$ Sp$_s(u)$, let $B_{\lambda}$ be an orthonormal basis of $N_{\lambda}$ and let $(e_j)_{j\in J}$ be the family union of the $B_{\lambda}$; it is an orthonormal basis of F. Let B be the union of $(e_j)_{j\in J}$ and of an orthonormal basis of $F^{\circ}$. It is an orthonormal basis of E and $u$ is diagonal in the basis B.

If $K =\mathbf{C}$, this follows at once from the theorem. Suppose $K =\mathbf{R}$. The space E is an $\mathbf{R}$-structure on $E_{(\mathbf{C})}($cf. A, II, p. 119). The endomorphism $u_{(\mathbf{C})}$ of $E_{(\mathbf{C})}$ is compact (Remark 4 of III, p. 2) and hermitian. Let $B = (e_j)_{j\in J}$ be an orthonormal basis of $E_{(\mathbf{C})}$ such that $u_{(\mathbf{C})}\in \mathscr{D}_B(E_{(\mathbf{C})})$ and $\lambda$ the family of eigenvalues of $u_{(\mathbf{C})}$ (Theorem 1). One has $\lambda \in \mathbf{R}^J($I, p. 106, Proposition 4) ; since the linear mapping $u_{(\mathbf{C})}$ is $\mathbf{R}$-rational, the eigenspace of $u_{(\mathbf{C})}$ corresponding to $\lambda_j$ is $\mathbf{R}$-rational for every $j\in J ($cf. A, V, p. 60, Proposition 6). Hence there exists a basis of it belonging to E, and a fortiori, there exists an orthonormal basis of it in E. The union of these bases is an orthonormal basis $B_{\mathbf{R}}$ of E such that $u\in \mathscr{D}_{B_{\mathbf{R}}}(E)$.

Let $v=u^*\circ u$. It is a compact endomorphism (III, p. 5, Prop. 3) and positive, hence hermitian, of E. By Corollary 1, there exists an orthonormal basis $(e_j)_{j\in J}$ of E such that $v$ is diagonal in this basis. The family $(\lambda_j)_{j\in J}$ of its eigenvalues is contained in $\mathbf{R}_+^J$. Put $\alpha_j=\surd\overline{\lambda_j}$ for all $j\in J$. Let I be the set of $j\in J$ such that $\alpha_j\not = 0$. It is a countable set since $v$ is compact. The family $(e_i)_{i\in I}$ is an orthonormal basis of the initial space of $v$, which is the initial space Ker($u$)$^{\circ}$ of $u$ (EVT, V, p. 43, Prop. 8). Put $f_i=\frac{1}{\alpha_i}u(e_i)$ for $i\in I$. Whatever $i$ and $j$ in I may be, one has

whence it follows that the family $(f_i)_{i\in I}$ is orthonormal in F. The corollary follows, since $u(e_i) =\alpha_if_i$ for all $i\in I$.

Let $B = (f_j)_{j\in J}$ be an orthonormal basis of E in which $u$ is diagonal (cor. 1 of IV, p. 150) and $(\lambda_j)_{j\in J}$ the family of eigenvalues of $u$ in the basis B. Let $J'$ be the set of $j\in J$ such that $\lambda_j$ belongs to the point spectrum of $u$.

For each $\lambda \in$ Sp$_s(u)$, there exists a bijection between the integers $n$ such that $0\leqslant n <M$ and $\lambda_n(u) =\lambda$ and the $j\in J'$ such that $\lambda_j=\lambda$, for these two sets have as cardinal the spectral multiplicity of $\lambda$. A choice of such bijections for each $\lambda$ defines a bijection $\iota$ of the set of integers such that $0\leqslant n <M$ onto the set $J'$. One defines the sequence $(e_n)_{0\leqslant n<M}$ in E by putting $e_n=f_{\iota(n)}$ for $0\leqslant n <M$. It is an orthonormal family in E.

In the case where M $<$ Card(I) $= +\infty$, the space F generated by $\{e_0, . . . , e_{M-1}\}$ is of finite dimension and its orthogonal complement $F^{\circ}$ is of infinite dimension; one chooses for $(e_n)_{n\geqslant M}$ an orthonormal family in $F^{\circ}$.

For every $x\in E$, one has

since $u(f_j) = 0$ when $j\in J$ - $J'$. If $n\in I$ satisfies $n\geqslant M$, one has $\lambda_n(u) = 0$, and one obtains the first formula of the proposition. The second results from it.

The endomorphism $f(u)$ is compact and positive and, for every $n\in I_E$, one has $\lambda_n(f(u)) =f(\lambda_n(u))$.

The endomorphism $f(u)$ is compact and positive by prop. 6, b) of III, p. 91 and prop. 15, a) of I, p. 117. The spectrum of $f(u)$ is the image under $f$ of the spectrum of $u$ (cor. 2 of I, p. 111). If $\lambda \in$ Sp$_s(f(u))$, the spectral multiplicity of $\lambda$ is the sum of the spectral multiplicities of the $\mu\in$ Sp($u$) such that $f(\mu) =\lambda$ (cor. 2 of III, p. 84). Since $f$ is increasing, the sequence $(f(\lambda_n(u)))_{n\in I_E}$ is decreasing. The assertion then results from the definition of the sequence $(\lambda_n(f(u)))_{n\in I_E}$.

a) For every subspace $F\in \mathscr{F}_{n+1}$ adapted to $u$, one has $\lambda_n(u) =r_F(u)$;

b) For every subspace $F\in \mathscr{F}_n$ adapted to $u$, one has $\lambda_n(u) = R_F(u)$.

Let $F\in \mathscr{F}_{n+1}$ be a subspace adapted to $u$, and $(f_i)_{0\leqslant i\leqslant n}$ an orthonormal basis of F such that $u(f_i) =\lambda_i(u)f_i$ for every $i$. For every $x$ in F, one has

with equality if $x=f_n$. This implies that $r_F(u) =\lambda_n(u)$, whence assertion a).

Let $F\in \mathscr{F}_n$ be a subspace adapted to $u$. The closed subspace $F^{\circ}$ is nonzero (for $n=$ dim(F) $<$ dim(E)) and stable under $u$; the endomorphism $\widetilde{u}$ of $F^{\circ}$ deduced from $u$ by passing to subspaces is compact and positive.

One has Sp($\widetilde{u}$)$\subset$ Sp($u$). Moreover, one has Sp($\widetilde{u}$)$\subset [0, \lambda_n(u)]$. In fact, it is enough to verify that $\lambda \leqslant \lambda_n(u)$ for every $\lambda \in$ Sp$_s(\widetilde{u})$. The number $\lambda$ is then an eigenvalue of $u$, hence there exists $j\in I$ such that $\lambda =\lambda_j(u)$. The eigenspace of $u$ relative to $\lambda_j(u)$ is therefore not contained in F, which implies that $\lambda_j(u)\leqslant \lambda_n(u)$.

Suppose that $\lambda_n(u)>0$. Then the eigenspace of $u$ relative to $\lambda_n(u)$ is not contained in F, and therefore $\lambda_n(u)$ belongs to the point spectrum of $\widetilde{u}$. It follows that sup(Sp($\widetilde{u}$)) $=\lambda_n(u)$. If $\lambda_n(u) = 0$, one has the same result since the spectrum of $\widetilde{u}$ is then reduced to $\{0\}$.

Moreover, by definition one has $R_F(u) = R_{\{0\}}(\widetilde{u})$, and finally $R_{\{0\}}(\widetilde{u}) =$ sup(Sp($\widetilde{u}$)) by Prop. 9 of I, p. 139, a). Assertion b) is proved.

$\lambda_n(u) =$ sup$_{F\in\mathscr{F}_{n+1}}r_F(u) =$ inf$_{F\in\mathscr{F}_n}R_F(u)$.

Let $(e_n)_{n\in I}$ be an orthonormal family having the property of Prop. 3 of IV, p. 152. For every integer such that $1\leqslant n <M + 1$, let $F_n\in \mathscr{F}_n$ be the subspace of dimension $n$ of E generated by $(e_0, . . . , e_{n-1})$; by construction, the space $F_n$ is adapted to $u$. By Prop. 5, one therefore has

Let $n\in I$. Let $F\in \mathscr{F}_{n+1}$. The restriction to F of the orthogonal projector onto $F_n$ is not injective, hence there exists $x\not = 0$ in F orthogonal to $F_n$. Since $x\in F^{\circ}_n$, one then has (Prop. 3 of IV, p. 152)

$m\geqslant n$

This proves that $r_F(u)\leqslant \lambda_n(u)$, whence in particular the inequality

(3) Fsup$_{\in\mathscr{F}_{n+1}}r_F(u)\leqslant \lambda_n(u)$.

Let $F\in \mathscr{F}_n$. The restriction to $F_{n+1}$ of the orthogonal projector onto F is not injective, hence there exists a vector $x\not = 0$ in $F_{n+1}$ orthogonal to F. Since $x\in F_{n+1}$, one has (loc. cit.)

and therefore $R_F(u)\geqslant \lambda_n(u)$. In particular, one obtains

(4) Finf$_{\in\mathscr{F}_n}R_F(u)\geqslant \lambda_n(u)$.

In view of formulas (2), (3) and (4), the proposition is proved.

a) One has $|\lambda_n(u)-\lambda_n(v)|\leqslant \|u-v\|$ for every $n\in I$;

b) If $u\leqslant v$, then $\lambda_n(u)\leqslant \lambda_n(v)$ for every $n\in I$.

Let $n\in I$ and let F be a vector subspace of dimension $n$ of E. For every $x\in F$, one has

hence the inequalities

Assertion a) follows from this by Proposition 6 of IV, p. 154.

If $u\leqslant v$, one has $\langle x|u(x)\rangle \leqslant \langle x|v(x)\rangle$ for every $x\in E$. For every $n\in I$ and every subspace F of dimension $n$, one therefore has $R_F(u)\leqslant R_F(v)$, whence $\lambda_n(u)\leqslant \lambda_n(v) ($loc. cit.).

a) One has $I_H\subset I_E$ and $\lambda_n(u_H)\leqslant \lambda_n(u)$ for all $n\in I_H$;

b) If H is of finite codimension $k\in \mathbf{N}$ in E, then one has $I_H+k\subset I_E$ and $\lambda_{n+k}(u)\leqslant \lambda_n(u_H)$ for all $n\in I_H$.

The endomorphism $u_H$ is compact (prop. 3 of III, p. 5). It is positive since $\langle x|u_H(x)\rangle =\langle i_H(x)|u(i_H(x))\rangle \geqslant 0$ for all $x\in H$.

Let $n\in I_H\subset I_E$. Let F be a subspace of dimension $n+ 1$ of H adapted to $u_H$. One has therefore $\lambda_n(u_H) =r_F(u_H)$ (prop. 5 of IV, p. 153, a)), and since moreover $r_F(u_H) =r_F(u)\leqslant \lambda_n(u)$ (prop. 6 of IV, p. 154), one obtains assertion a).

Suppose that H is of codimension $k\in \mathbf{N}$ in E and that $n\in I_H$. Let F be a subspace of H of dimension $n$ adapted to $u_H$. Its orthogonal complement in H is equal to $H\cap F^{\circ}$, and it is the orthogonal complement in E of the subspace $F + H^{\circ}$ of dimension $n+k$. Hence $n+k\in I_E$ and (prop. 5 of IV, p. 153, b))

$\lambda_n(u_H) =$ sup $\frac{\langle x|u_H(x)\rangle}{2}= R_{F+H^{\circ}}(u)$

$x\not =0$

whence $\lambda_n(u_H)\leqslant \lambda_{n+k}(u)$ (prop. 6 of IV, p. 154).

a) For $n\in I_E$, one has

$\alpha_n(u) =$ sup$_{F\in\mathscr{F}_{n+1}}$ inf$_{x\in F-\{0\}}\frac{\|u(x)\|}{\|x\|}=$ inf$_{F\in\mathscr{F}_n}$ sup$_{x\in F^{\circ-}\{0\}}\frac{\|u(x)\|}{\|x\|}$.

b) Let J be the set of $n\in I_E$ such that $\alpha_n(u)\not = 0$. There exist orthonormal families $(e_n)_{n\in J}$ in E and $(f_n)_{n\in J}$ in F such that for every $x\in E$, one has

Since $\langle x|u^*u(x)\rangle =\|u(x)\|^2$ for every $x\in E$, the definitions and prop. 6 of IV, p. 154, a), imply the equality of the first assertion.

Let $(e_n)_{n\in I_E}$ be an orthonormal family of E satisfying the conclusions of prop. 3 of IV, p. 152 applied to the positive compact endomorphism $u^*\circ u$ of E. Put $f_n=\alpha^{-1}_nu(e_n)$ for $n\in J$. By reasoning as in the proof of corollary 2 of IV, p. 150, one obtains assertion c).