Let F be a finite-dimensional vector subspace of E. Let S denote a topological supplement of the closure of $\{0\}$ in F, which is a direct subspace of E. The space $F_2= F\cap S$ is separated and finite-dimensional; there therefore exists an integer $n\geqslant 0$ such that $F_2$ is isomorphic to $K^n$ (TVS, I, p. 14, th. 2). There then exists a continuous linear mapping $p: S\rightarrow F_2$ extending the identity mapping of $F_2$ (TVS, II, p. 26, remark). The kernel of $p$ is a topological supplement of $F_2$ in S, and also of F in E.

Put $p=v\circ u$ and $q=u\circ v$. One has

therefore $p$ and $q$ are projectors. They are continuous. Let $E_1$ and $E_2$ be the image and the kernel of $p$, and let $F_1$ and $F_2$ be the image and the kernel of $q$. Since $u\circ v\circ u=u$, we have

Ker($u$)$\subset$ Ker($v\circ u$)$\subset$ Ker($u\circ v\circ u$) $=$ Ker($u$),

therefore Ker($u$) $=$ Ker($v\circ u$) $=$ Ker($p$) $= E_2$. Similarly, it follows from the inclusions

Im($u$)$\supset$ Im($u\circ v$)$\supset$ Im($u\circ v\circ u$) $=$ Im($u$)

that the image of $u$ is $F_1$. The spaces $E_2$ and $F_1$ are direct vector subspaces of E and F respectively.

The vector space E is the topological direct sum of $E_1$ and $E_2$, and $u$ defines by restriction a bijective linear mapping $u_1$ of $E_1$ onto $F_1$. For every $x\in E_1$, we have $v(u(x)) =p(x) =x$, and the mapping $u^{-1}_1$ of $F_1$ onto $E_1$ coincides with $v$ on $F_1$. Consequently, $u_1$ is an isomorphism of $E_1$ onto $F_1$ and $u$ is a strict morphism of E into F.

Let E and F be locally convex spaces and let $u\in \mathscr{L}(E; F)$. In order that $u$ be a direct injective morphism (resp. surjective), it is necessary and sufficient that there exist $v$ in $\mathscr{L}(F; E)$ such that $v\circ u= 1_E$ (resp. $u\circ v= 1_F$) (cf. TG, III, p. 47 and 48).

a) The set $\mathscr{I}(E; F)$ of the isomorphisms of E onto F ;

b) The set $\mathscr{M} \mathscr{D}(E; F)$ of injective direct morphisms of E into F, and more precisely, for every closed vector subspace $F_1$ of F, the set $\mathscr{M}_{F_1}(E; F)$ of elements of $\mathscr{M} \mathscr{D}(E; F)$ whose image is a topological complement of $F_1$;

c) The set $\mathscr{E} \mathscr{D}(E; F)$ of surjective direct morphisms of E onto F, and more precisely, for every closed vector subspace $E_1$ of E, the set $\mathscr{E}_{E_1}(E; F)$ of elements of $\mathscr{E} \mathscr{D}(E; F)$ whose kernel is a topological complement of $E_1$.

Moreover, the mapping $u\mapsto u^{-1}$ of $\mathscr{I}(E; F)$ onto $\mathscr{I}(F; E)$ is analytic.

By definition, $\mathscr{I}(E; E)$ is the set of invertible elements of the complete normed algebra $\mathscr{L}(E)$. By TG, IX, p. 40, Prop. 14, it is an open subset of $\mathscr{L}(E)$. The mapping $u\mapsto u^{-1}$ of $\mathscr{I}(E; E)$ into $\mathscr{I}(E; E)$ is analytic (LIE, III, § 1, n$^o1$, Example 2).

If the set $\mathscr{I}(E; F)$ is empty, it is open. Otherwise, let $u_0$ be an isomorphism of E onto F. The mapping $v\mapsto u_0\circ v$ is then an isomorphism of $\mathscr{L}(E)$ onto $\mathscr{L}(E; F)$ which transforms $\mathscr{I}(E; E)$ into $\mathscr{I}(E; F)$. Therefore the set $\mathscr{I}(E; F)$ is open in $\mathscr{L}(E; F)$. Moreover, if $u=u_0\circ v$ is an element of $\mathscr{I}(E; F)$, one has $u^{-1}=v^{-1}\circ u^{-1}_0$, and therefore the mapping $u\mapsto u^{-1}$ of $\mathscr{I}(E; F)$ onto $\mathscr{I}(F; E)$ is analytic.

Let $F_1$ be a closed vector subspace of F. For every $u\in \mathscr{L}(E; F)$, let $\overline{u}$ be the element of $\mathscr{L}(E\times F_1; F)$ defined by $\overline{u}(x, y) =u(x) +y$. The mapping $u\mapsto \overline{u}$ of $\mathscr{L}(E; F)$ into $\mathscr{L}(E\times F_1; F)$ is continuous, and $\mathscr{M}_{F_1}(E; F)$ is the set of elements $u$ of $\mathscr{L}(E; F)$ such that $\overline{u}$ belongs to $\mathscr{I}(E\times F_1; F)$. Since $\mathscr{I}(E\times F_1; F)$ is open in $\mathscr{L}(E\times F_1; F)$ by what precedes, the set $\mathscr{M}_{F_1}(E; F)$ is open in $\mathscr{L}(E; F)$. The same is true of $\mathscr{M} \mathscr{D}(E; F)$, which is the union of the $\mathscr{M}_{F_1}(E; F)$ when $F_1$ ranges over the set of closed vector subspaces of F.

Let $E_1$ be a closed vector subspace of E, and let $p$ be the canonical mapping of E onto the quotient Banach space $E/E_1$. In order that an element $u$ of $\mathscr{L}(E; F)$ belong to $\mathscr{E}_{E_1}(E; F)$, it is necessary and sufficient that the mapping $(u, p)$ of E into $F\times E/E_1$ belong to $\mathscr{I}(E; F\times E/E_1)$. As above, it follows that $\mathscr{E}_{E_1}(E; F)$ is open in $\mathscr{L}(E; F)$; the same is true of $\mathscr{E} \mathscr{D}(E; F)$, which is the union of the $\mathscr{E}_{E_1}(E; F)$.

The theorem follows from the following more precise statement:

(i) One has $\varphi (u_0) =v_0$;

(ii) For every $u$ in U, the mapping $\varphi (u)$ is a quasi-inverse of $u$, and in particular $u$ is a Fredholm mapping;

(iii) For every $u$ in U, one has ind($u$) $=$ ind($u_0$).

By prop. 2 of III, p. 42, (ii), there exist topological direct sum decompositions $E = E_1\oplus E_2$ and $F = F_1\oplus F_2$, and there exists $\alpha_0\in \mathscr{I}(E_1; F_1)$, such that $E_2$ and $F_2$ are finite-dimensional and that $u_0$ is represented by the matrix $\begin{pmatrix} \alpha_0 & 0 \\ 0 & 0 \end{pmatrix}$ relative to these decompositions.

Let U be the set of elements $u$ of $\mathscr{L}(E; F)$ such that, in the matrix representation $\begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix}$ of $u$ with respect to these decompositions, one has $\alpha \in \mathscr{I}(E_1; F_1)$. Since $\mathscr{I}(E_1; F_1)$ is open in $\mathscr{L}(E_1; F_1)$ (prop. 3 of III, p. 57), U is an open neighbourhood of $u_0$ in $\mathscr{L}(E; F)$. For $u=\begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix}$ in U, put

One has $\varphi (u_0) =v_0$ and the mapping $\varphi$ is analytic (loc. cit.). Modulo continuous linear mappings of finite rank, one has the congruences

Consequently, $\varphi (u)$ is a quasi-inverse of $u$. Every element $u$ of U defines by restriction an isomorphism of $E_1$ onto a topological supplement of $F_2$ in F. By prop. 3 of III, p. 44, one therefore has

ind($u$) $=$ codim$_F(u(E_1))-$ codim$_E(E_1)$

= dim(F$_2$)$-$ dim(E$_2$) $=$ ind($u_0$),

which concludes the proof.

Consider the continuous linear mappings $u:x\mapsto p(x)$ of Im($q$) into Im($p$) and $v:y\mapsto q(y)$ of Im($p$) into Im($q$). For every $x\in$ Im($q$), we have $x=q(x)$, whence

Hence $\|1_E-v\circ u\|\leqslant \|q-p\|^2<1$. By Corollary 1 of I, p. 22$,v\circ u$ is an automorphism of Im($q$). One proves analogously that $u\circ v$ is an automorphism of Im($p$). This implies that $u$ is an isomorphism of Im($q$) onto Im($p$), whence the first assertion of the lemma. The second follows by replacing $p$ and $q$ by $1_E-p$ and $1_E-q$ respectively.

The theorem follows from the following more precise statement:

(i) The endomorphism $\pi (u_0)$ is the projector with image N and kernel I;

(ii) For every $u\in U$, the linear mapping $\pi (u)$ is a projector of rank $d$ belonging to the bicommutant of $u$ and, in particular, commuting with $u$. Moreover, $u$ induces an automorphism of Ker($\pi (u)$);

(iii) Every element of U is a Riesz endomorphism whose nilspace has dimension $\leqslant d$.

If K = $\mathbf{C}$, let Sp($u_0$) denote the spectrum of $u_0$ relative to the algebra $\mathscr{L}(E)$. When K = $\mathbf{R}$, let Sp($u_0$) denote the complex spectrum Sp$_{\mathscr{L}(E_{(\mathbf{C})})}((u_0)_{(\mathbf{C})})$ of $u_0($I, p. 85, n$^o13$). By prop. 14 of III, p. 54, the point 0 is isolated in $\{0\} \cup$ Sp($u_0$). Let $r >0$ be a real number such that every element of Sp($u_0$)$-\{0\}$ has modulus $> r$. Let V denote the open set in $\mathbf{C}$ consisting of the complex numbers of absolute value $\not =r$; let $f$ be the holomorphic function on V defined by $f(z) = 0$ if $|z|> r$ and $f(z) = 1$ if $|z|< r$. If $K =\mathbf{C}$ (resp. $K =\mathbf{R}$), let $U'$ denote the set of elements $u$ of $\mathscr{L}(E)$ whose spectrum (resp. whose complex spectrum) is contained in V. The set $U'$ is an open neighbourhood of $u_0$ in $\mathscr{L}(E)$ and the mapping $u\mapsto f(u)$ of $U'$ into $\mathscr{L}(E)$ is holomorphic (I, p. 76, prop. 10 and I, p. 85, n$^o13$).

Let $u\in U'$. The endomorphism $f(u)$ is the spectral projector $e_0(u)$; it commutes with $u$, and $u$ induces by passing to the subspaces an automorphism of Ker($e_0(u)$) $($cf. III, p. 53, n$^o6$).

The projector $e_0(u_0)$ has image N and kernel I (III, p. 54, prop. 14); its rank is $d$. By lemma 1, the set U of elements $u\in U'$ such that $e_0(u)$ is of rank $d$ is an open neighbourhood of $u_0$. The set U and the mapping $\pi :u\mapsto e_0(u)$ of U into $\mathscr{L}(E)$ satisfy conditions (i) and (ii) of the proposition.

Let $u\in U$. Since $u$ induces an automorphism of Ker($e_0(u)$), which is a closed subspace of finite codimension of E, the endomorphism $u$ is a Riesz endomorphism of E (III, p. 48, prop. 8), the nilspace of $u$ is contained in the image of $\pi (u) =e_0(u)$, and is of dimension $\leqslant d$.

This is a consequence of formulas (2) and (5) and of the definition of the conorm of $u$.

$P =u(E)-u(B),Q =u(E)$ - $(\overline{u(B)}\cap u(E))$.

The conorm of $u$ is equal to the distance from 0 to P in F. If the normed space E is complete or if $u$ is a strict morphism, the conorm of $u$ is equal to the distance from 0 to Q in F.

Let N be the kernel of $u$; let $p: E\rightarrow E/N$ be the canonical surjection and let $v: E/N\rightarrow F$ be the mapping deduced from $u$ by passing to the quotient. The set $p(B)$ is the set of elements of $E/N$ of norm $<1$. We have

since the mapping $v$ is injective. In other words, P is made up of the elements of F of the form $v(y)$ with $y\in E/N$ and $\|y\|\geqslant 1$. Let $d_P$ denote the distance from 0 to P in F. We have

$d_P=$ inf$_{y\|\in yE\|\geqslant/N1}\|v(y)\|=$ inf$_{y\|\in yE\|=1/N}\|v(y)\|= ((u))$ by (6).

Suppose that $u$ is a strict morphism. Let $\varepsilon >0$. The set $\varepsilon u(B)$ is a neighbourhood of 0 in $u(E)$. The closure of $u(B)$ in $u(E)$ is equal to $\overline{u(B)}\cap u(E)$. It is contained in the set $u(B) +\varepsilon u(B)$, which is equal to $(1 +\varepsilon )u(B)$ since $u(B)$ is convex. Hence we have $(1 +\varepsilon )P\subset Q\subset P$, and the distance $d_Q$ from 0 to Q in F satisfies the inequalities $d_P\leqslant d_Q\leqslant (1 +\varepsilon )d_P$. Since this holds for every $\varepsilon >0$, we have $d_Q=d_P= ((u))$.

Suppose that $u$ is not a strict morphism, but that the normed space E is complete. Then $((u)) = 0$ (remark above). The closure of $u(B)$ in $u$(E), which is equal to $\overline{u(B)}\cap u$(E), is not a neighbourhood of 0 in $u(E)$ (EVT, I, p. 17, th. 1). There then exist points of Q arbitrarily near 0, whence $d_Q= 0 = ((u))$.

Let $u\in \mathscr{L}(E; F)$. We have to prove that for every real number $c >((u))$, the set of elements $v\in \mathscr{L}(E; F)$ such that $((v))< c$ is a neighbourhood of $u$. Let B denote the set of elements of E of norm $<1$. By Prop. 6, there exists $y\in E$ such that $u(y)\notin u(B)$ and $\|u(y)\|< c$. The distance $d$ from $u(y)$ to the closed set $u(B)$ is strictly positive. The set V of elements $v$ of $\mathscr{L}(E; F)$ satisfying the relations $\|v(y)\|< c$ and $\|u-v\|(1 +\|y\|)< d$ is a neighbourhood of $u$ in $\mathscr{L}(E; F)$. Let $v\in V$. For every $x\in B$, one has

Consequently $v(y)$ does not belong to $v$(B), and one has $((v))\leqslant \|v(y)\|< c$ by Prop. 6.

Let $j$ denote the canonical injection of the normed space F into its completion $\widehat{F}$. One has $((u)) = ((j\circ u))$. Since the linear mapping $^tj: (\widehat{F})'\rightarrow F'$ is bijective and isometric, one has likewise $((^tu)) = ((^t(j\circ u)))$. It is therefore enough to prove the proposition when the normed space F is complete, which we shall assume in the rest of the proof.

If $u$ is zero, one has $((u)) = ((^tu)) = +\infty$. If $u$ is not a strict morphism, then $^tu$ is not one either (EVT, IV, p. 29, Cor. 3), and one has $((u)) = ((^tu)) = 0$.

Assume henceforth that $u$ is a strict nonzero morphism. Let N denote the kernel of $u$ and I its image, and consider the canonical decomposition of $u:$

The kernel of $^tu$ is the orthogonal $I^{\circ}$ of I in $F'$ (EVT, IV, p. 27, Prop. 2), and $^ti$ defines by passing to the quotient an isometry $\iota$ of $F'/I^{\circ}$ onto $I'$ (EVT, IV, p. 9, Prop. 10). Moreover, $^tp$ defines an isometry $\pi$ of $(E/N)'$ onto the orthogonal $N^{\circ}$ of N in $E'$ (EVT, IV, p. 8, Prop. 9). The canonical decomposition of $^tu$ is therefore

where $v'=\pi \circ^tv\circ \iota$. One then has

(EVT, IV, p. 7, prop. 8), whence $((u)) = ((^tu))$ after formula (10).

It is enough to treat the case where the field K is equal to $\mathbf{R}$. Let $n$ be the dimension of G. Replacing F by a vector subspace of F of dimension $n+ 1$ containing G, we are reduced to the case where dim(F) $=n+ 1$. Let us reason by contradiction, supposing that the conclusion of the theorem is not satisfied. Let S be the unit sphere of F. For every $x\in S$, the distance from $x$ to G is then strictly less than $\|x\|= 1$, and one can choose an element $y(x)$ of G such that $\|x-y(x)\|<1$. Let $V_x$ denote the set of elements $z$ of S such that $\|z-y(x)\|<1$; this is an open neighbourhood of $x$ in S. There exists a continuous locally finite partition of unity $(\varphi_x)_{x\in S}$ on S subordinate to the covering $(V_x)_{x\in S}$ of S (TG, IX, p. 46, prop. 3 and p. 51, prop. 6). Let $f: S\rightarrow G$ be the continuous mapping given by

for every $z\in S$. Let $z\in S$. One has $\varphi_x(z)\geqslant 0$ for every $x\in S$, and there exists $x\in S$ such that $\varphi_x(z)>0$, since $\sum_{x\in S}\varphi_x(z) = 1$, hence

This property of $f$ contradicts Theorem 3.

When $u= 0$, one has $((u)) = +\infty$ and dim(E) $=n$, whence $d= +\infty$. Suppose henceforth that $u$ is non-zero, and therefore that $((u))<+\infty$. Let $v$ be an element of $\mathscr{L}(E; F)$ such that $\|u-v\|<((u))$ and let us prove that its kernel is of dimension $\leqslant n$. Let $x$ be an element of norm 1 of Ker($v$) and $y$ its image in $E/$ Ker($u$). One has (formula (5) of III, p. 61)

whence $\|y\|<1$. Now $\|y\|$ is the distance from $x$ to Ker($u$). Theorem 4 then implies that one has dim Ker($v$)$\leqslant$ dim Ker($u$) $=n$. This proves the inequality $((u))\leqslant d$. The converse inequality $d\leqslant ((u))$ results from the more precise lemma that follows.

Let $p$ denote the canonical mapping of E onto $E/$ Ker($u$). There exists $a\in E$ such that $\|p(a)\|= 1$ and $\|u(a)\|< c$ (formula (6) of III, p. 61). By the Hahn–Banach theorem (EVT, II, p. 24, cor. 2), there exists a continuous linear form $f$ on the normed space $E/$ Ker($u$) such that $f(p(a)) = 1$ and $\|f\|= 1$. The linear mapping $h:x\mapsto -(f\circ p)(x)u(a)$ of E into F is continuous, of rank 1, and one has $\|h\|\leqslant \|f\| \|p\| \|u(a)\|< c$. The kernel of the mapping $u+h$ contains that of $u$ and $a$; since $a\not \in$ Ker($u$), its dimension is therefore at least $n+ 1$. On the other hand, the linear mapping $u$ induces, by passing to subspaces, a linear mapping of Ker($u+h$) into Im($h$) with kernel Ker($u$)$\cap$ Ker($u+h$), so that dim(Ker($u+h$))$\leqslant$ dim(Ker($u$)) $+ 1$. The conclusion follows.