The essential spectrum of $u$ consists of the complex numbers $\lambda$ such that $u-\lambda 1_E$ is not a Riesz endomorphism of E. In particular, if $u$ is compact, then every $\lambda \in$ Sp($u$)$-\{0\}$ is a sensitive point of the spectrum (cor. 2 of III, p. 77).

This follows from No.$^o3$ of I, p. 131 since, for every $\lambda \in$ Sp$_s(u)$, the subspaces $N_{\lambda}(u)$ and $I_{\lambda}(u)$ are respectively the image and the kernel of the spectral projector $e_{\lambda}(u)$.

a) There exists a decomposition of E into a topological direct sum $F\oplus$ G such that F is of finite dimension, that F and G are stable under every endomorphism commuting with $u$, and that the spectrum of the endomorphism of G induced by $u$ is contained in V;

b) Suppose V nonempty. There exists an element $v$ of the bicommutant of $u$ in $\mathscr{L}(E)$ whose spectrum is contained in V and such that $v-u$ is of finite rank.

Let H = Sp($u$)$\cap (\mathbf{C}-\mathring{V})$. The set H is contained in Sp$_s(u)$, hence is discrete; since it is closed in the compact space Sp($u$), it is finite. One can take for F and G the image and the kernel of the spectral projector $e_H(u)$ (Prop. 2).

Suppose that V is nonempty. Let $\mu\in V$ and let $v$ denote the endomorphism of E which agrees with the homothety of ratio $\mu$ in F and with $u$ in G. Its spectrum is contained in $\{\mu\} \cup$ (Sp($u$) - H), hence in V, and $v-u$ is of finite rank since F is finite-dimensional.

For every compact subset S of $\mathbf{C}$, let $\mathscr{O}(S)$ denote the algebra of germs of holomorphic functions in a neighbourhood of S with values in $\mathbf{C}$ (I, p. 49, §4, No. $1$).

The complex number $\mu$ belong to $f$(Sp($u$)), and hence to the spectrum of $f(u)$ (Prop. 8 of I, p. 75), if and only if H is nonempty. One then has $e_\mu(f(u)) =e_H(u)$ (No. $1$ of I, p. 127). The other assertions are deduced from this by Prop. 2.

a) The set Sp$_{\omega}(u)$ is compact. It is non-empty if E is infinite-dimensional;

b) Let $n\in \overline{\mathbf{Z}}$. The set Sp$_n(u)$ is the union of a family of bounded connected components of $\mathbf{C}-$ Sp$_{\omega}(u)$. It is open in $\mathbf{C}$, and its boundary in $\mathbf{C}$ is contained in Sp$_{\omega}(u)$.

The set $\mathbf{C}-$ Sp$_{\omega}(u)$ consists of the complex numbers $\lambda \in \mathbf{C}$ such that $u-\lambda 1_E$ is a strict morphism, whose kernel or cokernel has finite dimension. By props. 11 of III, p. 67 and 13 of III, p. 70, it is open. The set Sp$_{\omega}(u)$ is therefore closed. Since it is bounded, it is compact.

Let us prove b). Let $n\in \overline{\mathbf{Z}}$. The set Sp$_n(u)$ is contained in $\mathbf{C}-$Sp$_{\omega}(u)$. Let U be a connected component of $\mathbf{C}-$ Sp$_{\omega}(u)$ which meets Sp$_n(u)$. The mapping $\lambda \mapsto$ ind($u-\lambda 1_E$) from $\mathbf{C}-$ Sp$_{\omega}(u)$ into $\overline{\mathbf{Z}}$ being locally constant (cor. 1 of prop. 12 of III, p. 68 and cor. 1 of prop. 13 of III, p. 70), the index of $u-\lambda 1_E$ is equal to $n$ for every $\lambda \in U$. If $n\not = 0$, this implies that U is contained in Sp$_n(u)$. If $n= 0$, remark that the set U is a connected component of $\mathbf{C}-$ (Sp$_{\omega}(u)\cup$ Sp$_{-\infty}(u)\cup$ Sp$_{+\infty}(u)$). Since U meets Sp$_0(u)$ and hence Sp$_e(u)$, it follows then from Remark 2 that the set U is contained in Sp$_e(u)$, and consequently in Sp$_0(u)$. In all cases we conclude that Sp$_n(u)$ is the union of the connected components of $\mathbf{C}-$ Sp$_{\omega}(u)$ which meet Sp$_n(u)$. These are necessarily bounded since the set Sp($u$) is bounded. Consequently, Sp$_n(u)$ is open in $\mathbf{C}$ and its boundary is contained in Sp$_{\omega}(u)$. This proves b).

Suppose finally that the set Sp$_{\omega}(u)$ is empty. According to b), each of the sets Sp$_n(u)$, for $n\in \overline{\mathbf{Z}}$, is then empty. We therefore have Sp($u$) $=$ Sp$_s(u)$. The spectrum of $u$ is consequently discrete and compact, hence finite, and since all its points are of finite spectral multiplicity, the vector space E is finite-dimensional (III, p. 83, prop. 2). This completes the proof of a).

b) Any point adherent to Sp$_s(u)$ which does not belong to Sp$_s(u)$ belongs to Sp$_{\omega}(u)$.

Assertion a) is a direct consequence of assertion b) of th. 1. Let $\lambda$ be a point adherent to Sp$_s(u)$ which does not belong to Sp$_s(u)$. It belongs to the spectrum of $u$, since the latter is closed. It does not belong to any of the sets Sp$_n(u)$, for $n\in \overline{\mathbf{Z}}$, since these are open (loc. cit.) and disjoint from Sp$_s(u)$. We therefore have $\lambda \in$ Sp$_{\omega}(u)$, whence b).

a) The traces on $\mathbf{C}-\{0\}$ of the sets Sp($v\circ u$) and Sp($u\circ v$) $($resp. Sp$_s(v\circ u)$ and Sp$_s(u\circ v)$, resp. Sp$_n(v\circ u)$ and Sp$_n(u\circ v)$ for $n\in \overline{\mathbf{Z}}$, resp. Sp$_{\omega}(v\circ u)$ and Sp$_{\omega}(u\circ v))$ are equal;

b) Let $\lambda$ be an element of Sp$_s(v\circ u)$ distinct from 0. The spectral multiplicities of $\lambda$ for $v\circ u$ and for $u\circ v$ are equal.

Let $\mu$ be a nonzero complex number and let $n\in \overline{\mathbf{Z}}$. In order that $\mu1_E-v\circ u$ be an automorphism (resp. a Riesz endomorphism, resp. a strict morphism whose kernel or cokernel is of finite dimension and whose index is $n$), it is necessary and sufficient that $\mu1_F-u\circ v$ be one (III, p. 49, prop. 10). Assertion a) then follows from the definitions.

Let $\lambda$ be a point of Sp$_s(v\circ u)$ distinct from 0. One has

dim Ker(($\lambda 1_E-v\circ u$)$^n$) $=$ dim Ker(($\lambda 1_F-u\circ v$)$^n$)

for every $n\geqslant 0 ($loc. cit.), therefore the spectral multiplicities of $\lambda$ for $v\circ u$ and $u\circ v$ are equal.

a) One has Sp$_s(u) =$ Sp$_s(^tu)$, Sp$_n(u) =$ Sp$_{-n}(^tu)$ for every $n\in \overline{\mathbf{Z}}$ and Sp$_{\omega}(u) = Sp_{\omega}(^tu)$;

b) Every point of Sp$_s(u)$ has the same spectral multiplicity for $u$ and for $^tu$.

Assertion a) of prop. 3 of I, p. 131 proves that Sp($u$) $=$ Sp($^tu$), and assertion c) of loc. cit. implies that Sp$_s(u) =$ Sp$_s(^tu)$ and that assertion b) is valid.

For every $\lambda \in \mathbf{C}$, lemma 4 of III, p. 69 implies that $u-\lambda 1_E$ is a strict morphism if and only if $^tu-\lambda 1_{E'}$ is one, and that

dim Coker($^t(u-\lambda 1_E)$) $=$ dim Ker($u-\lambda 1_E$)

dim Ker($^t(u-\lambda 1_E)$) $=$ dim Coker($u-\lambda 1_E$)

in $\overline{\mathbf{Z}}$. Assertion a) follows from this, taking account of the definitions of the various parts of the spectrum (def. 3 of III, p. 85).

Let $\lambda \in \mathbf{C}$. In order that $u+h-\lambda 1_E$ be a strict morphism whose kernel (resp. cokernel) is of finite dimension, it is necessary and sufficient that $u-\lambda 1_E$ be so, by Theorem 1 of III, p. 72 (resp. Theorem 2 of III, p. 73).

The equalities

Sp$_{-\infty}(u+h) =$ Sp$_{-\infty}(u)$, Sp$_{+\infty}(u+h) =$ Sp$_{+\infty}(u)$,

Sp$_{\omega}(u+h) =$ Sp$_{\omega}(u)$

follow from this. Moreover, one has Sp$_n(u+h) =$ Sp$_n(u)$ by Theorem 3 of III, p. 73 for every $n\in \mathbf{Z}-\{0\}$.

One has Sp$_{\omega}(u+h) =$ Sp$_{\omega}(u)$ (Theorem 2), hence 0 belong to the unbounded connected component of $\mathbf{C}-$ Sp$_{\omega}(u+h)$. By the corollary to Theorem 1 of III, p. 87, either 0 does not belong to the spectrum of $u+h$, or else it is a sensitive point of this spectrum. In both cases, $u+h$ is a Riesz endomorphism of E.

Suppose first that E is of dimension 2. One may suppose that $E =\mathbf{R}^2$ endowed with the euclidean norm (TVS, I, p. 14, Theorem 2). Since X is denumerable, there exists a real number $r\in \mathbf{R}_+^*$ such that the circle C with center 0 and radius $r$ does not meet X. Let $x\in E$ - X; if $x \notin C$, there exists a point $y\in C$ such that the straight line $L_x$ joining $x$ and $y$ does not meet X, since X is denumerable. The set E - X is the union of C, which is connected, and of the connected sets $L_x\cup C$ for $x\in E$- $(X\cup C)$; these sets all contain C, and therefore E- X is connected (TG, I, p. 81, Proposition 2).

Consider the general case. Replacing X by $X-x$ for an element $x\in E$ - X, one is reduced to the case where $0\notin X$. Since E - X is the union of the sets F - $(X\cap F)$ when F runs through the set of 2-dimensional subspaces of E, and since these are connected by the preceding case and contain 0, the set E - X is connected (loc. cit.).

For every integer $i\geqslant 1$, the set $A_i$ of complex numbers $\lambda \in S$ such that $|\lambda |\geqslant \frac{1}{i}$ is compact and discrete in $\mathbf{C}$, hence finite. Let $a_i$ be its cardinal. Since S is infinite, the sequence $(a_i)$ tends to $+\infty$. Put $A_0=\emptyset$ and $a_0= 0$. For every $i\geqslant 1$, choose a bijection $n\mapsto \lambda_n$ of the interval $[a_{i-1}, a_i[$ of $\mathbf{N}$ onto $A_i-A_{i-1}$ such that the mapping $n\mapsto |\lambda_n|$ is decreasing on $[a_{i-1}, a_i[$. The sequence $(\lambda_n)_{n\in\mathbf{N}}$ satisfies the required properties.

Let E be a complete normable space of infinite dimension. The algebra $\mathscr{L}^c(E)$ is a non-unital subalgebra of $\mathscr{L}(E)$. Recall that for every compact endomorphism $u\in \mathscr{L}^c$(E), the spectrum Sp$'_{\mathscr{L}^c(E)}(u)$ is the spectrum of $u$ relative to the unital subalgebra $\mathscr{L}^c(E)\oplus \mathbf{C}1_E$ of $\mathscr{L}(E) ($I, p. 4, No.$^o4$).

a) If E is finite-dimensional, then Sp($u$) $=$ Sp$_s(u)$;

b) If E is infinite-dimensional, then Sp$_s(u) =$ Sp($u$)$-\{0\}$ and Sp$_{\omega}(u) =\{0\}$;

c) If Sp$_s(u)$ is infinite, it is the set of values of a sequence $(\lambda_n)_{n\in\mathbf{N}}$ of nonzero complex numbers, pairwise distinct, such that the sequence $(|\lambda_n|)_{n\in\mathbf{N}}$ is decreasing and converges to 0 ;

d) If E is infinite-dimensional, then Sp($u$) $=$ Sp$'_{\mathscr{L}^c(E)}(u)$.

Every element of Sp$_s(u)$ is an eigenvalue of finite spectral multiplicity (prop. 1 of III, p. 83).

Assertion a) is elementary (III, p. 85, example 1). Suppose now that E is infinite-dimensional.

Let $\lambda \in$ Sp($u$)$-\{0\}$. Then $u-\lambda 1_E$ is a Riesz endomorphism of E (cor. 2 of prop. 2 of III, p. 75), hence $\lambda$ belongs to Sp$_s(u)$. If E is infinite-dimensional, then Sp$_{\omega}(u)$ is not empty (III, p. 87, th. 1, a)). Necessarily one has Sp$_{\omega}(u) =\{0\}$, whence b).

The set Sp$_s(u)$ is discrete and bounded. Moreover, one has Sp$_s(u) =$ Sp($u$)$\cap (\mathbf{C}-\{0\})$ by b), hence Sp$_s(u)$ is closed in $\mathbf{C}-\{0\}$. Assertion c) therefore follows from Lemma 2.

The spectrum of $u$ is denumerable, and its complement in $\mathbf{C}$ is therefore connected (lemma 1). By the cor. to prop. 6 of I, p. 28, applied to the unital subalgebra $\mathscr{L}^c(E)\oplus \mathbf{C}1_E$ of $\mathscr{L}$ (E), one concludes that Sp($u$) $=$ Sp$'_{\mathscr{L}^c(E)}(u)$.

a) Let $f\in \mathscr{O}$(Sp($u$)) be such that $f(0) = 0$. The endomorphism $f(u)$ is compact;

b) Suppose that E is a complex Hilbert space and that $u$ is normal. Let $f$ be a continuous function on Sp($u$) such that $f(0) = 0$. The normal endomorphism $f(u)$ is compact.

Moreover, the converse assertions are valid if E is infinite-dimensional, and the condition $f(0) = 0$ is not necessary if E is finite-dimensional.

We may suppose that E is infinite-dimensional.

Let us prove a) and its converse. The endomorphism $u$ is an element of the Banach algebra $\mathscr{L}^c(E)$. Since E is infinite-dimensional, one has the equality Sp($u$) $=$ Sp$'_{\mathscr{L}^c(E)}(u)$ by prop. 5, d). The element $f(u)$ of the holomorphic functional calculus of the unital Banach algebra deduced from $\mathscr{L}^c(E)$ by adjunction of a unit element belongs to $\mathscr{L}^c(E)$ if and only if $f(0) = 0$ (I, p. 88). But moreover this element coincides with the element $f(u)$ of $\mathscr{L}(E)$ (prop. 7 of I, p. 75), hence $f(u)$ is compact if and only if $f(0) = 0$.

The proof of assertion b) and of its converse is exactly the same if one considers the continuous functional calculus of the involutive algebra $\mathscr{L}^c(E)$ (I, p. 110, def. 5).

Let $(j,|u|)$ be the polar decomposition of $u$ (def. 4 of I, p. 140). Since

$u=j|u|$ and $|u|=u^*u$ (prop. 10 of I, p. 139), the equivalence results from prop. 3 of III, p. 5 and assertion b) of the preceding proposition.

a) If $u$ is normal, then $u$ is a Riesz endomorphism if and only if $u$ is a Fredholm endomorphism of index 0.

b) If $u$ is hermitian, then $u$ is a Fredholm endomorphism if and only if $u$ is a Riesz endomorphism.

Every Riesz endomorphism is a Fredholm endomorphism of index 0 (proposition 5 of III, p. 46). Conversely, suppose that $u$ is a Fredholm endomorphism of index 0, and that $u$ is normal. Its nilspace then coincides with its kernel (EVT, V, p. 43, prop. 8), and is therefore of finite dimension. It then results from lemma 2 of III, p. 45 and from the definition that $u$ is a Riesz endomorphism.

Let us prove b). By a), it is enough to verify that the index of a hermitian Fredholm endomorphism $u$ is zero. But the orthogonal complement of the image of $u$ (which is closed) is then equal to the kernel of $u$ (EVT, V, p. 41, prop. 4), whence the assertion.

Both assertions result from the proposition and from the definition of the sets Sp$_n(u)$ for $n\in \overline{\mathbf{Z}}$ and of Sp$_{\omega}(u)$, which form a partition of the essential spectrum of $u$ (def. 2 of III, p. 83).

Let $h\in \mathscr{L}^c(E)$. Since Sp$_0(u)$ is empty (corollary above), Theorem 2 of III, p. 89 implies that Sp$_e(u+h) =$ Sp$_e(u)$. The intersection of the sets Sp($u+h$) therefore contains Sp$_e(u)$.

Let $\lambda \in$ Sp$_s(u)$. Let $E_{\lambda}$ denote the proper subspace of $u$ relative to $\lambda$, and $F_{\lambda}$ the image of the spectral projector associated with $u$ and $\mathbf{C}-\{\lambda \}$. The space E is the topological direct sum of $E_{\lambda}$ and $F_{\lambda}$. Let $h$ be the finite-rank endomorphism of E which is zero on $F_{\lambda}$ and coincides on $E_{\lambda}$ with the identity. The endomorphism $u+h$ is invertible, therefore $\lambda \notin$ Sp($u+h$). The theorem follows.

Thus, if $u$ is a normal endomorphism of E, and if $h\in \mathscr{L}^c$(E), the spectrum of $u+h$ can differ from the spectrum of $u$ only by isolated points having finite spectral multiplicity.

has a finite radius of convergence $r >0$. Let $D\subset \mathbf{C}$ be the open disc with center 0 and radius $r$. There does not exist a holomorphic function $\widetilde{f}$ defined on an open neighbourhood U of $r$ in $\mathbf{C}$ which coincides with $f$ on $U\cap D$.

Suppose there exists such a holomorphic function $\widetilde{f}$ defined on an open neighbourhood U of $r$. There exist real numbers $s < r$ and $\delta >0$ such that $s+\delta > r$ and such that the open disk $D'$ with center $s$ and radius $\delta$ is contained in U. The power series expansion of $\widetilde{f}$ at the point $s$ (VAR, R1, p. 26, 3.2.1) then converges for every $z$ in the disk with center 0 and radius $\delta$ (VAR, R1, p. 29, 3.3.4). This series is

$+\infty (n)+\infty (n)$

$\widetilde{f}_s(z) =\sum\widetilde{f}n$!$(s)z^n=\sum fn$!$(s)z^n$

$n=0n=0$ (VAR, R1, p. 27, 3.2.4). Since $s\in D$, we have

(VAR, R1, p. 28, 3.2.11). Let us take $z$ such that $0< z < \delta$. Since $a_k\geqslant 0$, we have

$\widetilde{f}_s(z) =\sum^{+\infty}(\sum^{+\infty}k(k-1)\cdots n($!$k-n+ 1)a_ks^{k-n})z^n$

$n=0k=n$ $+\infty k+\infty$

$=\sum(a_k\sum n$!($kk-$! $n$)! $s^{k-n}z^n)=\sum a_k(s+z)^k$

$k=0n=0k=0$

(TG, III, p. 40). When $s+z > r$, the convergence of this expression contradicts the assumption that the radius of convergence of the power series $f$ is equal to $r$.

Let E be a real Banach space. Let C be a pointed convex cone in E. The vector space generated by C is equal to $C-C$ (EVT, II, p. 12, cor. 1). In particular, the cone C is total in E if and only if $C-C$ is dense in E. Recall that C is said to be salient if $C\cap (-C)$ is reduced to 0 (EVT, II, p. 11).

The polar $C^{\circ}$ of C is the set of continuous linear forms $\ell \in E'$ such that $\ell (x)\geqslant 0$ for all $x\in C$ (EVT, II, p. 47, prop. 4). By the bipolar theorem (EVT, II, p. 48, th. 1), if C is a closed pointed convex cone, one has $C^{\circ \circ}= C$.

Indeed, if $u$ vanishes on C, then $u$ vanishes on $C-C$, hence on E, and therefore on $E_{(\mathbf{C})}$.

Let $E_{(\mathbf{C})}$ be the complexified space of E and $u_{(\mathbf{C})}$ the endomorphism of $E_{(\mathbf{C})}$ obtained by extension of the scalars from $u$. The spectral radius of $u_{(\mathbf{C})}$ is equal to $\varrho (u) ($I, p. 86); we denote it simply by $\varrho$. Since $\varrho >0$, the complex spectrum of $u$ is not reduced to 0. There therefore exists $\lambda_0\in$ Sp$_s(u_{(\mathbf{C})})$ such that $|\lambda_0|=\varrho$ (Prop. 5 of III, p. 90). Let $e_0$ be the spectral projector of $u_{(\mathbf{C})}$ associated with $\lambda_0$.

The resolvent $\lambda \mapsto R(u_{(\mathbf{C})}, \lambda ) = (\lambda -u_{(\mathbf{C})})^{-1}$ is holomorphic on the complement of the spectrum of $u_{(\mathbf{C})}$ (Theorem 1 of I, p. 24). The complex number $\lambda_0$ is a pole of the resolvent and its residue is the spectral projector $e_0($I, p. 131).

Let $y$ and $y'$ be elements of E such that $y+iy'\in E_{(\mathbf{C})}$ is an eigenvector of $u_{(\mathbf{C})}$ for the eigenvalue $\lambda_0$. Since $e_0(y+iy') =y+iy'\not = 0$, there exists an element $x_0\in C$ such that $e_0(x_0)\not = 0$ (Lemma 4). Since C is closed and salient, its polar $C^{\circ}$ is total (EVT, II, p. 48, Corollary 1), and there then exists a linear form $\ell_0\in C^{\circ}$ such that $\langle e_0(x_0), \ell_0\rangle \not = 0$.

Consider the function $f$ defined by $f(0) = 0$ and by

for $z\in \mathbf{C}$ such that $z^{-1}\notin$ Sp($u_{(\mathbf{C})}$). This function satisfies

for $|z|<1/\varrho$ (Theorem 1 of I, p. 24, d)) and is therefore holomorphic in the disc with centre 0 and radius $1/\varrho$.

There exists a holomorphic function $\widetilde{R}$ defined on an open neighbourhood U of $\lambda_0$ and with values in $\mathscr{L}(E)$ such that for $z$ belonging to U$-\{\lambda_0\}$, one has

(Prop. 17 of I, p. 83). For $z$ such that $z^{-1}\in U$ and $z\not = 1/\lambda_0$, one therefore has

The term of the series corresponding to $n= 0$ is $(z^{-1}-\lambda_0)^{-1}\langle e_0(x_0), \ell_0\rangle$. Since $\langle e_0(x_0), \ell_0\rangle \not = 0$, the uniqueness of the Laurent expansion (VAR, R1, p. 30, 3.3.9) implies that $f$ cannot be prolonged to a holomorphic function in a neighbourhood of $1/\lambda_0$. In particular, the radius of convergence of the power series expansion (1) of the function $f$ at the point 0 is equal to $1/\varrho$.

The coefficients of the power series (1) are $\langle u^n(x_0), \ell_0\rangle \geqslant 0$ since $u(C)\subset C$ and $\ell_0\in C^{\circ}$. By Lemma 3, the function $f$ cannot be extended to a holomorphic function in a neighbourhood of $1/\varrho$. The resolvent of $u_{(\mathbf{C})}$ therefore cannot be extended to a holomorphic function in a neighbourhood of $\varrho$, that is to say, $\varrho \in$ Sp($u$). This implies that $\varrho$ is an eigenvalue of $u_{(\mathbf{C})}$ (prop. 5 of III, p. 90). Since $\varrho$ is real, it is also an eigenvalue of $u$.

Let $d\geqslant 1$ be the order of the pole of the resolvent of $u_{(\mathbf{C})}$ at $\varrho$. Let $e$ be the endomorphism

$e=$ lim$_{z\rightarrow\varrho}(z-\varrho )^dR(u_{(\mathbf{C})}, z)$.

It is nonzero and commutes with $u_{(\mathbf{C})}$, and its image is contained in the eigenspace of $u_{(\mathbf{C})}$ corresponding to $\varrho ($cf. No.$^o3$ of I, p. 131). Let now $\ell$ be an element of $C^{\circ}$ and $x$ an element of C. By Theorem 1 of I, p. 24, d), one has

$\langle e(x), \ell \rangle =$ lim$_{zz>\varrho\rightarrow\varrho}(z-\varrho )^d\sum_{n\geqslant 0}\langle u^n(x), \ell \rangle z^{-n-1}\geqslant 0$.

The bipolar theorem (EVT, II, p 48, th. 1) implies that $e(x)\in C$ for all $x\in C$. Since C is total, there exists $x\in C$ such that $e(x)\not = 0$ (Lemma 4), and then $e(x)$ belongs to C and is an eigenvector of $u$ for the eigenvalue $\varrho$, as required.