The endomorphism $u$ is an isomorphism if and only if $u_i$ is an isomorphism for every $i\in I$. Applying this property to $u-\lambda 1_E$, we deduce that Sp($u$) is the union of the sets Sp($u_i$) for $i\in I$.

Let $f\in \mathscr{O}$(Sp($u$)). The endomorphism $f(u)$ of E belongs to the bicommutant of $u$ in $\mathscr{L}(E)$ (Theorem 5 of I, p. 74), and therefore commutes with the projectors $p_i$. It therefore stabilizes the spaces $E_i$. Consider the continuous unital morphism $\varpi$ of the product algebra $\prod_{i\in I}\mathscr{L}(E_i)$ into $\mathscr{L}(E)$ defined by $(v_i)_{i\in I}\mapsto \bigoplus_iv_i$. It maps the family $(u_i)$ to $u$. Proposition 7 of I, p. 75 then implies that Sp($u$)$\subset \bigcup_{i\in I}$ Sp($u_i$) and

which concludes the proof of the lemma.

Ker($u-\lambda 1_E$)$\subset$ Ker($f(u)-f(\lambda )1_E$).

Let $x\in E$ be nonzero. The set A of the $v\in \mathscr{L}(E)$ such that $x$ is an eigenvector of $v$ is a unital subalgebra of $\mathscr{L}(E)$.

The algebra A is full: if $v\in A$ is invertible in $\mathscr{L}(E)$ and if $x$ is an eigenvector of $v$ for the eigenvalue $\lambda$, then $\lambda \not= 0$ and $v^{-1}(x) =\lambda^{-1}x$, which proves that $v^{-1}\in A$.

The algebra A is closed in $\mathscr{L}(E)$. In fact, if $(v_n)_{n\in\mathbf{N}}$ is a sequence in A such that $v_n$ converges to $v\in \mathscr{L}(E)$, the sequence $(\lambda_n)_{n\in\mathbf{N}}$ such that $v_n(x) =\lambda_nx$ is bounded, and therefore admits a subsequence converging to a complex number $\mu$, so that $v(x) =\mu x$.

Suppose that $x$ is an eigenvector of $u$ and that $u(x) =\lambda x$. The algebra A contains $u$, and therefore contains the closed full unital subalgebra B generated by $u$, which is commutative. The mapping $\chi : B\rightarrow \mathbf{C}$ which associates with $v$ the eigenvalue of $v$ relative to $x$ is a character of B such that $\chi (u) =\lambda$. For every $f\in \mathscr{O}$(Sp($u$)), we have $f(u)\in B$ and $\chi (f(u)) =f(\chi (u)) =f(\lambda )$ by Proposition 7 of I, p. 75, whence the proposition.

One has Sp($u$) $= H_1\cup H_2($I, p. 128, Lemma 1). Since the sets $H_1$ and $H_2$ are compact, they are open and closed in Sp($u$). For every holomorphic function $f$ in a neighbourhood of Sp($u$), the endomorphism $f(u)$ leaves $E_1$ and $E_2$ invariant and coincides in $E_1$ with $f(u_1)$ and in $E_2$ with $f(u_2)$ (loc. cit.). Take in particular for $f$ the germ of holomorphic function $f_{H_1}$ which is equal to 1 in a neighbourhood of $H_1$ and 0 in a neighbourhood of $H_2$ (cf. No. 12 of I, p. 81); then $f_{H_1}(u_1)$ is the identity mapping of $E_1$ since $f_{H_1}$ = 1 in a neighbourhood of $H_1=$ Sp($u_1$), and $f_{H_1}(u_2)$ is zero since $f_{H_1}= 0$ in a neighbourhood of $H_2$. Hence $e_{H_1}(u) =f_{H_1}(u)$ is the projector with image $E_1$ and kernel $E_2$, and therefore one has $E_1= E_{H_1}$ and $E_2= E_{H_2}$.

Let E be a Banach space and $u$ an endomorphism of E. Let $(H_i)_{i\in I}$ be a finite family of open and closed subsets of Sp($u$), pairwise disjoint, and let H be their union. Relations (15) and (16) of I, p. 81 imply the following assertions :

a) The family of projectors $(e_{H_i}(u))_{i\in I}$ is orthogonal (that is to say, $e_{H_i}(u)e_{H_j}(u) = 0$ for all $(i, j)$ in $I^2$ such that $i\not=j$, cf. A, II, p. 18, Def. 7) and its sum is $e_H(u)$;

b) The vector space $E_H$ is the topological direct sum of the family $(E_{H_i})_{i\in I}$;

c) For every $j\in I$, one has $E_{Sp(u)-H_j}= E_{Sp(u)-H}\oplus \bigoplus_{i\not=j}E_{H_i}$;

d) The vector space $E_{Sp(u)-H}$ is the intersection of the family $(E_{Sp(u)-H_i})_{i\in I}$.

When H = Sp($u$), the decomposition as topological direct sum $\bigoplus_{i\in I}E_{H_i}$ of E is called the spectral decomposition of E associated with $u$ and with the finite partition $(H_i)_{i\in I}$ of Sp($u$).

Let $f$ be an element of $\mathscr{O}$(Sp($u$)). The spectrum of the endomorphism $f(u)$ of E is the image under $f$ of the spectrum of $u($I, p. 75, Prop. 8). Let L be a subset open and closed in Sp($f(u)$). The set H of elements $\lambda \in$ Sp($u$) such that $f(\lambda )$ belongs to L is open and closed in Sp($u$), and one has $e_L(f(u)) =e_H(u)$ since $f_L\circ f=f_H$ in $\mathscr{O}$(Sp($u$)) (loc. cit.).

a) We have Sp($u$) $=$ Sp($^tu$) ;

b) For every $f\in \mathscr{O}$(Sp($u$)), we have $f(^tu) =^tf(u)$ ;

c) We have $e_H(^tu) =^te_H(u)$ for every subset H of Sp($u$) which is open and closed.

For an endomorphism of E to be an automorphism, it is necessary and sufficient that its transpose be an automorphism of $E'$ (EVT, IV, p. 30, cor. 5), whence assertion a).

The mapping $v\mapsto^tv$ is a continuous unital homomorphism of the Banach algebra $\mathscr{L}(E)$ into the opposite Banach algebra of $\mathscr{L}(E')$ (EVT, IV, p. 7, prop. 8). Since the algebra $\mathscr{O}$(Sp($^tu$)) is commutative, the mapping $f\mapsto^tf(u)$ is a continuous unital homomorphism of the algebra $\mathscr{O}$(Sp($^tu$)) into the algebra $\mathscr{L}(E')$. This homomorphism maps the germ of the identity mapping of $\mathbf{C}$ to $^tu$, and therefore coincides with the homomorphism $f\mapsto f(^tu)$ (th. 5 of I, p. 74). This proves b).

Assertion c) results from b), applied to the function $f_H$ equal to 1 in a neighbourhood of H and to 0 in a neighbourhood of its complement in Sp($u$).

The proof follows that of lemma 1 of I, p. 128, using the remark of 6 of I, p. 110 and prop. 8 of I, p. 112.

Ker($u-\lambda 1_E$)$\subset$ Ker($f(u)-f(\lambda )1_E$).

The proof is analogous to that of proposition 1 of I, p. 128; let us take up its arguments again. The algebra A introduced in loc. cit. is here a unital involutive subalgebra of $\mathscr{L}(E)$ (EVT, V, p. 43, cor.). It therefore contains the unital involutive subalgebra B generated by $u$, which is commutative. The mapping $\chi : B\rightarrow \mathbf{C}$ which, to $v$, assigns the eigenvalue of $v$ relative to $x$ is a character of B such that $\chi (u) =\lambda$. For every $f\in \mathscr{C}$ (Sp($u$))), one has $f(u)\in B$ and $\chi (f(u)) =f(\chi (u)) =f(\lambda )$ by prop. 8 of I, p. 112, whence the assertion.

(i) The projector $p$ is an orthogonal projector, that is to say Ker($p$) $=$ Im($p$)$^{\circ}$ (EVT, V, p. 13);

(ii) The projector $p$ is hermitian;

(iii) The projector $p$ is normal;

(iv) One has Ker($p$)$\subset$ Ker($p^*$) ;

(v) One has Im($p$)$\subset$ Im($p^*$) ;

(vi) The projector $p$ is positive;

(vii) One has $\|p\|\leqslant 1$.

Let us first recall that Ker($p^*$) $=$ Im($p$)$^{\circ}$ and Ker($p$) $=$ Im($p^*$)$^{\circ}$ (EVT, V, p. 41, prop. 4). Moreover, the image of $p$ (resp. $p^*$) is closed, since it coincides with the kernel of $1-p$ (resp. $1-p^*$). Therefore one has

(1) Im($p$) $=$ Ker($p^*$)$^{\circ}$, Im($p^*$) $=$ Ker($p$)$^{\circ}$.

(i) $=\Rightarrow$ (ii)$:p^*$ is a projector with kernel Im($p$)$^{\circ}=$ Ker($p$) and whose image is Im($p^*$)$^{\circ}=$ Ker($p$) $=$ Im($p$)$^{\circ}$; hence $p^*=p$.

(ii) $=\Rightarrow$ (iii) since every hermitian endomorphism is normal.

(iii) $=\Rightarrow$ (iv) : since $p$ is normal, one has $\|p(x)\|^2=\|p^*(x)\|^2$ for all $x$ in E (EVT, V, p. 43, Prop. 7), whence the required inclusion.

(iv) $=\Rightarrow$ (v) follows from equalities (1) above.

(v) $=\Rightarrow$ (vi) : for all $x\in E$, one has $p(x)\in$ Im($p^*$), and consequently $\langle p(x)|x\rangle =\langle p^*(p(x))|x\rangle =\|p(x)\|^2\geqslant 0$.

(vi) $=\Rightarrow$ (vii) : let $x\in E$ and $y=x-p(x)\in$ Ker($p$). For all $t\in \mathbf{R}$, one has by hypothesis

which is possible only if $\langle y|p(x)\rangle = 0$. But then

and therefore $\|p\|\leqslant 1$.

(vii) $=\Rightarrow$ (i) : let $y\in$ Im($p$) ; denote by $z$ the orthogonal projection of $y$ onto Ker($p$)$^{\circ}$ and put $x=y-z\in$ Ker($p$). One has $p(z) =p(y) =y$, hence $\|y\|\leqslant \|z\|$ by hypothesis. But, since $x$ and $z$ are orthogonal, one has $\|y\|^2=\|x\|^2+\|z\|^2$, whence $\|x\|= 0$, that is to say $y=z$. Thus, Im($p$)$\subset$ Ker($p$)$^{\circ}$. Since moreover $\|p^*\|=\|p\|\leqslant 1$, one has analogously Im($p^*$)$\subset$ Ker($p^*$)$^{\circ}$, which yields the converse inclusion by (1).

a) For every open and closed subset H of the spectrum of $u$, the spectral projector $e_H(u)$ is an orthogonal projector whose kernel is the image of the spectral projector $e_{Sp(u)-H}(u)$ ;

b) If $H_1$ and $H_2$ are disjoint open and closed subsets of the spectrum of $u$, then the spectral subspaces $E_{H_1}$ and $E_{H_2}$ are orthogonal ;

c) If $\lambda \in \mathbf{C}$ is an isolated point of the spectrum of $u$, then $\lambda$ is an eigenvalue of $u$ and the image of the spectral projector $e_{\lambda}(u)$ is the eigenspace of $u$ relative to $\lambda$.

Let us prove a). Since the holomorphic functional calculus is compatible with the continuous functional calculus (I, p. 111, cor. 1), we have $e_H(u) =\varphi_H(u)$, where $\varphi_H\in \mathscr{C}$ (Sp($u$)) is the characteristic function of H. This implies $e_H(u)^*=\overline{\varphi}_H(u) =\varphi_H(u) =e_H(u)$, hence $e_H(u)$ is an orthogonal projector (lemma 3, (ii)). Its kernel is the image of the projector $1-e_H(u) =e_{Sp(u)-H}(u)$.

Let us prove b). The characteristic functions $\varphi_{H_1}$ and $\varphi_{H_2}$ of $H_1$ and $H_2$ in Sp($u$) are continuous and their product is zero, which implies $e_{H_1}(u)\circ e_{H_2}(u) =e_{H_2}(u)\circ e_{H_1}(u) = 0$. The inclusions $E_{H_2}(u)\subset E_{H_1}(u)^{\circ}$ and $E_{H_1}(u)\subset E_{H_2}(u)^{\circ}$ follow.

Finally, let us prove assertion c). The characteristic function $\varphi_{\lambda}$ of $\{\lambda \}$ is continuous and non-zero on Sp($u$) ; it satisfies $(z-\lambda )\varphi_{\lambda}(z) = 0$ for every $z\in$ Sp($u$). Hence we have $(u-\lambda 1_E)\varphi_{\lambda}(u) = 0$. The image of $\varphi_{\lambda}(u)$, which is non-zero, is therefore contained in the proper subspace of $u$ relative to $\lambda$. Since we have $e_{\lambda}(u) =\varphi_{\lambda}(u)$ and the image of $e_{\lambda}(u)$ contains the proper subspace of $u$ relative to $\lambda$, the assertion follows.

Since $u$ is normal, every proper vector of $u$ is also a proper vector of $u^*$ (EVT, V, p. 43, cor.). The assumption therefore implies that F is stable under $u$ and under $u^*$. By EVT, V, p. 41, prop. 4 (ii), we therefore have $u(F^{\circ})\subset F^{\circ}$ and $u^*(F^{\circ})\subset F^{\circ}$. It follows that the adjoint of $\widetilde{u}$ is the endomorphism of $F^{\circ}$ deduced from $u^*$. Since $u$ is normal, the endomorphism $\widetilde{u}$ is normal.

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.

Let F be the real affine subspace of $\mathscr{L}(E)_h$ formed by the elements of trace 1. The orthogonal projectors of rank 1 of E belong to F (Lemma 3, (ii)).

Let $u\in F$. We have $\|u-\frac{1}{2}1_E\|^2=$ Tr($u^2-u+\frac{1}{4}$) $=$ Tr($u^2$)$-\frac{1}{2}$. Consequently, $u\in S$ if and only if Tr($u^2$) $= 1$. Since 2 det($u$) $=$ Tr($u$)$^2-$ Tr($u^2$) $= 1-$ Tr($u^2$), this condition is equivalent to det($u$) $= 0$. By the Hamilton–Cayley theorem (A, III, p. 107, prop. 20), we therefore have $u\in S$ if and only if $u^2-u= 0$, which means that $u$ is a hermitian projector of rank 1 (loc. cit.), whence the result.

We reduce to the case where $u$ is linear and where B is the unit ball of E. We have $u(S)\subset u(B)$. Conversely, let $x\in B$ and let $y$ be a nonzero element of Ker($u$). The image of the continuous mapping $t\mapsto \|x+ty\|$ of $\mathbf{R}$ into $\mathbf{R}_+$ is an unbounded interval containing the real number $\|x\|\leqslant 1$. There therefore exists $t\in \mathbf{R}$ such that $\|x+ty\|= 1$. Then $x+ty\in S$ and $u(x+ty) =u(x)$, hence $u(x)\in u(S)$.

Let us prove prop. 7. Let $x$ and $y$ be elements of the unit sphere of E; let us prove that the segment with endpoints $\langle x|u(x)\rangle$ and $\langle y|u(y)\rangle$ is contained in the numerical image of $u$.

Let F be the subspace of E generated by $x$ and $y$. If dim(F) = 1, we have $\langle x|u(x)\rangle =\langle y|u(y)\rangle$, whence the assertion. Otherwise, dim(F) = 2; let $p$ then be the orthogonal projector of E with image F, and let us denote by $u_F$ the endomorphism of F given by $x\mapsto p(u(x))$. Since $p$ is hermitian (lemma 3 of I, p. 133), we have $\langle z|u_F(z)\rangle =\langle z|u(z)\rangle$ for every $z\in F$, so that $\iota (u_F)\subset \iota (u)$. We may therefore suppose that E = F.

For every element $z$ of E, let $v_z$ be the hermitian endomorphism of E defined by $t\mapsto \langle z|t\rangle z$; one has $\langle z|u(z)\rangle =$ Tr($u\circ v_z$). When $z$ runs through the unit sphere of E$,v_z$ ranges over the set of orthoprojectors of rank 1 of E, which is a sphere S in the real affine subspace V of $\mathscr{L}(E)$ formed by the hermitian endomorphisms of E of trace 1 (lemma 5). The numerical image of E is therefore the set of Tr($u\circ v$), for $v\in S$. The mapping $v\mapsto$ Tr($u\circ v$) of V into $\mathbf{C}$ is linear. Since dim$_{\mathbf{R}}(V) = 3>$ dim$_{\mathbf{R}}(\mathbf{C})$, it is not injective; it therefore follows from lemma 6 that $\iota (u)$ is convex.