a) Every compact linear mapping from E into F is continuous ;

b) Every continuous linear mapping of finite rank from E into F is compact;

c) The set $\mathscr{L}^c(E; F)$ of compact linear mappings from E into F is a vector subspace of $\mathscr{L}(E; F)$.

Let us prove a). Let $v: E\rightarrow F$ be a compact linear mapping and U a neighbourhood of 0 in F. Choose a neighbourhood V of 0 in E such that $v(V)$ is relatively compact in F. Then $v(V)$ is bounded and there exists therefore a real number $\lambda >0$ such that the set $v(\lambda V)$ is contained in U; consequently, $v$ is continuous.

Let $u: E\rightarrow F$ be a continuous linear mapping of finite rank. Since the image $u(E)$ of $u$ is separated and of finite dimension, there exists a compact neighbourhood A of 0 in $u(E)$. The set $V =\overset{-1}{u}(A)$ is a neighbourhood of 0 in E and we have $u(V)\subset A$, hence the linear mapping $u$ is compact. This proves assertion b).

Let $u_1$ and $u_2$ be compact linear mappings of E into F. Let $V_1$ and $V_2$ be neighbourhoods of 0 in E such that the sets $u_1(V_1)$ and $u_2(V_2)$ are relatively compact in F. Put $V = V_1\cap V_2$ and $u=u_1+u_2$. Then $u(V)$ is contained in $u_1(V_1) +u_2(V_2)$, hence is relatively compact in F, and the linear mapping $u$ is compact. It is immediate that $\mathscr{L}^c(E; F)$ is stable under homothety. This proves that $\mathscr{L}^c(E; F)$ is a vector subspace of $\mathscr{L}(E; F)$, whence assertion c).

Let $u$ be an element of $\mathscr{L}(E; F)$ adhering to $\mathscr{L}^c(E; F)$ for the topology of bounded convergence. Denote by B the unit ball of E and by V a neighbourhood of 0 in F. Choose a neighbourhood W of 0 in F such that $W + W\subset V$. By hypothesis there exists an element $v$ of $\mathscr{L}^c(E; F)$ such that $(u-v)(B)\subset W$. The set $v(B)$ being relatively compact in F, there exists a finite subset M of F such that $v(B)\subset M + W$, whence $u(B)\subset M + V$. Since this holds for every V, the set $u(B)$ is precompact in F, hence its closure is also so (TG, II, prop. 1, p. 30). Since the latter is closed and bounded, it is complete by the hypothesis made on F, and consequently is compact (TG, II, cor., p. 30). It follows that $u(B)$ is relatively compact. This proves that $u$ belongs to $\mathscr{L}^c(E; F)$.

If F is locally convex, the hypothesis of the proposition means that F is quasi-complete (EVT, III, p. 8, déf. 6).

This follows from Propositions 1 and 2.

There exist Banach spaces E and F such that $\mathscr{L}^f(E; F)$ is not dense in $\mathscr{L}^c(E; F) ($cf. Remark 6 of III, p. 16 and Theorem 4 of III, p. 19, b)).

By hypothesis, there exists a neighbourhood V of 0 in E such that $u(V)$ is relatively compact in F. Put $U =\overset{-1}{v}(V)$. Then U is a neighbourhood of 0 in $E_1$ and its image by $w\circ u\circ v$ is contained in $w(u$(V)), hence is relatively compact in $F_1$. Consequently, the linear mapping $w\circ u\circ v$ is compact.

Let E be a separated topological vector space; according to Props. 1 and 3, $\mathscr{L}^c(E)$ is a two-sided ideal of the algebra $\mathscr{L}(E)$. When E is a Banach space, $\mathscr{L}(E)$ is a Banach algebra and $\mathscr{L}^c(E)$ is a closed two-sided ideal of $\mathscr{L}(E)$ (cor. to Prop. 2). It is a proper ideal if E is of infinite dimension (cf. Remark 3, p. 2).

In fact, the space $\mathscr{L}^c(E)$ is a closed, hence self-adjoint, two-sided ideal of $\mathscr{L}(E)$ (Lemma 15 of I, p. 122).

Identify F with a topological vector subspace of $\widehat{F}$ and denote by $\widehat{u}:\widehat{E}\rightarrow \widehat{F}$ the unique continuous linear mapping such that $\widehat{u}\circ j$ coincides with $u$ on E. Since $u$ is compact, there exist a neighbourhood V of 0 in E and a compact subset A of F such that $u(V)\subset A$. We have $\widehat{u}(j(V))\subset A$, whence $\widehat{u}(j(V))\subset A$. Now $j(V)$ is a neighbourhood of 0 in $\widehat{E}$ (TG, III, p. 24, Prop. 7). It follows that the image of $\widehat{u}$ is contained in F and the continuous linear mapping $v:\widehat{E}\rightarrow F$ deduced from $\widehat{u}$ by passing to the subspace is compact. Since $j(E)$ is dense in $\widehat{E}$, the mapping $v$ is the unique continuous linear mapping of $\widehat{E}$ into F such that $u=v\circ j$.

Let V be a neighbourhood of 0 in E such that $u(V)$ is relatively compact in F. Let $p$ be a continuous seminorm on E such that V contains the unit ball of $p$ (EVT, II, p. 26, cor.). Denote by $E_p$ the seminormed space obtained by endowing E with the seminorm $p$. The mapping $u$ is a compact linear mapping of $E_p$ into F. The separated completion G of $E_p$ is a Banach space. Denote by $v$ the canonical linear mapping of $E_p$ into G. According to Prop. 4, there exists a compact linear mapping $w: G\rightarrow F$ such that $u=w\circ v$. On the other hand, $v$ is a continuous mapping of E into G because the topology of E is finer than that of $E_p$.

The mapping $u$ is a continuous linear mapping of E into F, and also of $E_{\sigma}$ into $F_{\sigma}$. Its restriction to B is therefore a continuous mapping of the set B, endowed with the topology induced by $\sigma (E,E')$, into the space $F_{\sigma}$. Now the set $u(B)$ is contained in a compact subset C of F (remark 1 of III, p. 2) and the space $F_{\sigma}$ is separated, so that the topologies induced on C by those of F and of $F_{\sigma}$ coincide. The proposition follows.

Indeed, the set consisting of the point $x$ and the points of the sequence $(x_n)_{n\in\mathbf{N}}$ is a bounded subset of E (EVT, III, p. 3, cor. of prop. 2).

Indeed the set B, endowed with the topology induced by $\sigma (E',E)$, is compact (EVT, III, p. 17, cor. 2).

Indeed, the topology induced on B by $\sigma (E',E)$ is metrisable (EVT, III, p. 19, cor. 2). The assumption of the corollary therefore implies that the restriction of $u$ to B is a continuous mapping of B into F, when B is endowed with the topology $\sigma (E',E)$, and the corollary follows from the preceding proposition.

The space $E'_{\mathfrak{S}}$ is separated according to TG, X, p. 8, prop. 7; on the unit ball of $E'$, the $\mathfrak{S}$-topology coincides with the weak topology $\sigma (E',E)$ (EVT, III, p. 17, prop. 5), whence the corollary.

(i) The linear mapping $u$ is compact ;

(ii) The set $u(B)$ is a compact subset of F ;

(iii) The restriction of $u$ to B is a continuous mapping of the set B, endowed with the topology induced by $\sigma (E,E')$, into F ;

(iv) For every sequence $(x_n)_{n\in\mathbf{N}}$ of points of B which converges to 0 for the topology $\sigma (E,E')$, the sequence $(u(x_n))_{n\in\mathbf{N}}$ converges to 0 in F ;

(v) From every infinite sequence of points of $u(B)$, one can extract a sequence which converges in F ;

(vi) Every infinite sequence of points of $u(B)$ has an adherence value in F.

Since E is a reflexive space, B is a compact subset of $E_{\sigma}$ (TVS, IV, p. 17, prop. 6), hence (iii) implies (ii). Condition (ii) implies (i) by definition, and condition (i) implies (iii) by prop. 6. It is also elementary that (iii) implies (iv) and that (v) implies (vi).

We shall now prove that (iv) implies (v). Let $(x_n)$ be an infinite sequence of points of B. Since B is a compact subset of $E_{\sigma}$, there exists, by the theorem of $\breve{S}$mulian (TVS, IV, p. 36, th. 2), a sequence $(y_n)$, extracted from the sequence $(x_n)$, which converges in $E_{\sigma}$ to a limit $y$. The sequence $(y_n-y)$ is bounded in $E_{\sigma}$, hence in E (TVS, IV, p. 1, prop. 1); it is therefore contained in a homothetic set of B. If condition (iv) is satisfied, the sequence $(u(y_n-y))$ converges to 0 in F, and $(u(y_n))$, which is a sequence extracted from $(u(x_n))$, converges to $u(y)$. This proves that (iv) implies (v).

To conclude, let us prove that (vi) implies (ii). Let $j: F\rightarrow \widehat{F}$ be the canonical injection of F into its completion. Under hypothesis (vi), $u(B)$ is a precompact subset of F (TVS, IV, p. 32, prop. 1), hence $j(u(B))$ is a relatively compact subset of $\widehat{F}$ (TG, II, n$^{\circ}2$, p. 29) and $j\circ u$ is a compact linear mapping. By the equivalence already proved of conditions (i) and (ii), $j(u(B))$ is a compact subset of $\widehat{F}$. Consequently, $u(B)$ is a compact subset of F.

a) If the mapping $u$ is compact, then its transpose $^tu$ is a compact linear mapping of $F'_c$ into $E'_c$ and a continuous linear mapping of $F'_c$ into $E'_b$;

b) Suppose that the space E is a semi-normed space and that the space F is quasi-complete. If the transpose $^tu$ is a continuous mapping of $F'_c$ into $E'_b$, then the linear mapping $u$ is compact.

Suppose first that the linear mapping $u$ is compact. Then there exists a neighbourhood V of 0 in E whose image under $u$ is contained in a compact subset C of F. By definition (TVS, II, p. 47, def. 2 and p. 68, def. 1), the polar $C^{\circ}$ of C is a neighbourhood of 0 in $F'_c$ and one has $^tu(C^{\circ})\subset V^{\circ}$. Now $V^{\circ}$ is an equicontinuous subset of $E'$, closed for the weak topology, hence compact in $E'_c$ (TVS, III, p. 17, cor. 2 and prop. 5). This proves that $^tu$ is a compact linear mapping of $F'_c$ into $E'_c$.

Retain the preceding hypotheses and let U be a neighbourhood of 0 in $E'_b$. It contains the polar $A^{\circ}$ of a bounded subset A of E. Since the linear mapping $u$ is compact, the set $u(A)$ is relatively compact in F (remark 1 of III, p. 2), and $(^tu)^{-1}(A^{\circ}) =u(A)^{\circ}$ is a neighbourhood of 0 in $F'_c$. This proves that $^tu$ is a continuous linear mapping of $F'_c$ into $E'_b$.

Let us now place ourselves under the assumptions of b) and denote by B the unit ball of E. The polar $B^{\circ}$ of B is the unit ball of $E'_b$ and one has $(^tu)^{-1}(B^{\circ}) =u(B)^{\circ}$. If $^tu$ is a continuous mapping of $F'_c$ into $E'_b$, the set $u(B)^{\circ}$ is therefore a neighbourhood of 0 in $F'_c$; it therefore contains the polar $C^{\circ}$ of a compact subset C of F. By the bipolar theorem (EVT, II, p. 49, cor. 3), the set $u(B)$ is contained in the closed convex envelope of $C\cup \{0\}$; this is compact because the space F is quasi-complete (EVT, III, p. 8). This proves that the linear mapping $u$ is compact.

(i) The linear mapping $u$ of E into F is compact;

(ii) The linear mapping $^tu$ of $F'$ into $E'$ is compact;

(iii) The linear mapping $^tu$ of $F'_c$ into $E'$ is continuous.

The equivalence of (i) and (iii) results from prop. 9, a) and b); the implication (iii)$\Rightarrow$(ii) follows from the fact that the identity mapping of $F'$ into $F'_c$ is compact (cor. of prop. 7 of III, p. 7).

To conclude, let us prove that condition (ii) implies (i). Suppose that the linear mapping $^tu: F'\rightarrow E'$ is compact. It results from the implication (i)$\Rightarrow$(ii), applied to $^tu$, that $^{tt}u$ is a compact linear mapping of $E''$ into $F''$. Let us denote by $v$ the canonical linear mapping of E into $E''$; it is continuous. Since F is identified with a closed vector subspace of $F''$ and since $u$ coincides with $^t(^tu)\circ v$ on E, it results from remark 5 of III, p. 3 that the linear mapping $u$ is compact.

Let us denote by $B'$ the unit ball of $F'$, endowed with the topology induced by the topology $\sigma (F',F)$. Since F is a Banach space of countable type, $B'$ is a compact metrisable space (EVT, III, p. 19, cor. 2). Under the assumption made, the restriction of $^tu$ to $B'$ is a continuous mapping of $B'$ into $E'$ and $^tu(B')$ is a compact subset of $E'$. By cor. 1, the linear mapping $u$ is compact.

We shall deduce from it the following result.

One may suppose that X is a convex subset of a locally convex space E. Let N be the closure of $\{0\}$ in E and $\pi : E\rightarrow E/N$ the canonical surjection. Since X is a separated space, the restriction of $\pi$ to X is injective; since X is compact, it therefore defines a homeomorphism of X onto a nonempty compact convex subset of the separated locally convex space $E/N$. This allows us to suppose that the space E is separated.

Let $f: X\rightarrow X$ be a continuous mapping. The mapping $h: X\rightarrow E$ defined by $h(x) =f(x)-x$ is continuous; since X is compact, the image $h(X)$ is closed. It is therefore enough to prove that 0 adheres to $h$(X), that is to say that for every convex open neighbourhood U of 0 in E, there exists a point $b$ of X such that $f(b)-b\in U$.

Let U be a convex neighbourhood of 0 in E. For every $a\in X$, let $V_a$ denote the set of elements $x$ of X such that $f(x)-f(a)\in U$. It is open in X and contains $a$. Since X is compact and nonempty, there exists a nonempty finite subset A of X such that the sets $V_a$, for $a\in A$, cover X. Let $(\varphi_a)_{a\in A}$ be a continuous partition of unity subordinate to the covering $(V_a)_{a\in A}$ (TG, IX, p. 43, prop. 1 and p. 47, th. 3). Let F denote the vector subspace of E generated by $f(A)$. For every $x\in F\cap X$, put

This defines a continuous mapping $g: F\cap X\rightarrow E$. Its image is contained in the convex envelope of $f$(A), hence in $F\cap X$. Since the set $F\cap X$ is a nonempty compact convex subset of a finite-dimensional vector space, the mapping $g$ has a fixed point $b\in F\cap X$ by Theorem 1. We have

For every $a\in$ A such that $\varphi_a(b)\not = 0$, one has $b\in V_a$ so that $f(b)-f(a)\in U$. Since U is convex, it follows that $f(b)-b\in U$, which completes the proof.

(i) There exists no closed vector subspace of E, distinct from $\{0\}$ and E, which is stable under A;

(ii) The set A contains a nonzero compact endomorphism. Then the commutant of A is reduced to the homotheties.

Replacing A by the unital subalgebra of $\mathscr{L}(E)$ generated by A, one is reduced to the case where A is a unital subalgebra of $\mathscr{L}(E)$. Let then $h$ be a nonzero compact endomorphism of E belonging to A.

There exists an element $x_0$ of E such that $h(x_0)\not = 0$. Let V be a closed neighbourhood of $h(x_0)$ not containing 0. Since the endomorphism $h$ is compact, one can choose a convex open neighbourhood U of $x_0$ such that $h(U)$ is a relatively compact subset of V. The closure C of $h(U)$ is therefore a compact convex subset of E; since it is contained in V, it does not contain 0.

Let $y$ be a point of C. Let $Ay$ denote the set of images of $y$ under the elements of A. It is a vector subspace of E stable under A; it is nonzero since it contains $y$. By assumption (i) of the theorem, the set $Ay$ is dense in E. There therefore exists an element $b$ of A such that $b(y)\in U$. Since the set C is compact, there exists a

$-1$

finite family $(b_1, . . . , b_n)$ of elements of A such that the sets $b_i(U)$ cover C. There exists a continuous partition of unity $(\varphi_1, . . . , \varphi_n)$ on C subordinate to this covering (TG, IX, p. 47, th. 3). Define a mapping $f: C\rightarrow E$ by putting

for every $x\in C$; it is a continuous mapping. Since U is convex and $b_i(x)$ belongs to U when $\varphi_i(x)$ is not zero, we have $f(C)\subset U$ and $h(f(C))\subset C$. By Leray-Schauder's theorem (Theorem 2 of III, p. 11), there exists an element $x\in C$ such that $h(f(x)) =x$. Put then

We have $h(a(x)) =h(f(x)) =x$ and $x$ is nonzero since 0 does not belong to C, hence the kernel of $ha-1_E$ is nonzero.

Let us complete the proof of Theorem 3. Let $a$ be an element of A such that the kernel T of $ha-1_E$ is nonzero (Lemma 1). Let $i$ denote the canonical injection of T into E. The linear mapping $i$ is equal to $h\circ a\circ i$, and is therefore compact. Hence the dimension of T is finite (Remark 3 of III, p. 2). Let $u$ be an element of $\mathscr{L}(E)$ which commutes with A. Since $u$ commutes with $h\circ a$, we have $u(T)\subset T$. Since T is of nonzero finite dimension over the algebraically closed field $\mathbf{C}$, the endomorphism of T induced by $u$ has an eigenvalue $\lambda$.

Put F = Ker($u-\lambda 1_E$). It is a closed nonzero vector subspace of E; it is stable under A, since $u$ commutes with the elements of A. By assumption (i), we have F = E, whence $u=\lambda 1_E$.

By assumption (ii) of Theorem 3, the space E is not zero. It therefore contains a vector subspace F of dimension 1. This subspace is closed because E is assumed separated. By Theorem 3, every element of A is a homothety, and therefore stabilizes F. It then follows from assumption (i) of Theorem 3 that F is equal to E.

In fact, Corollary 1 shows that the set $A =\{u, h\}$ cannot satisfy assumption (i) of Theorem 3.

The case $u= 0$ is immediate. The case $u\not = 0$ follows from Cor. 2 by taking $h=u$.

The set A of orthogonal projections of finite rank in E is equicontinuous. Let $n\geqslant 1$ be an integer and $(x_1, . . . , x_n)$ a family of elements of E. Let V denote the vector subspace generated by the $x_i$ and $p_V$ the orthogonal projection with image V. We have $p_V(x_i) =x_i$ for $1\leqslant i\leqslant n$. It follows that $1_E$ is adherent to A for the topology of simple convergence, and hence that E is an approximation space (remark 3).

Other examples of approximation spaces will be given in No.$^o7$ of III, p. 20.

The definition of the topology of precompact convergence shows that the condition is sufficient. Conversely, suppose that E is an approximation space. Let P be a topological complement in E of the closure of $\{0\}($cf. remark 2). Then the set A of the $u\in \mathscr{L}^f(E)$ such that $u(E)\subset P$ is dense in $\mathscr{L}^f(E)$. Since P is separated, every element $u$ of A is of the form $x\mapsto \sum^n_{i=1}f_i(x)e_i$, where $n\in \mathbf{N},e_1, . . . , e_n$ are elements of E and $f_1, . . . , f_n$ continuous linear forms on E (EVT, I, p. 14, th. 2). This proves that the condition is necessary.

a) The set $\mathscr{L}^f(E; F)$ is dense in $\mathscr{L}_{pc}(E; F)$;

b) The set $\mathscr{L}^f(F; E)$ is dense in $\mathscr{L}_{pc}(F; E)$;

c) Let $\mathfrak{S}$ be a set of bounded subsets of F and $u\in \mathscr{L}(F; E)$. Suppose that the image under $u$ of every subset of F belonging to $\mathfrak{S}$ is precompact. Then $u$ is adherent to $\mathscr{L}^f(F; E)$ for the $\mathfrak{S}$-topology (EVT, III, p. 13).

Let $v$ be an element of $\mathscr{L}(E; F)$. The mapping $\varphi :w\mapsto v\circ w$ of $\mathscr{L}_{pc}(E)$ into $\mathscr{L}_{pc}(E; F)$ is continuous (TG, X, p. 5, prop. 3). We have $\varphi (1_E) =v$ and $\varphi (\mathscr{L}^f(E))\subset \mathscr{L}^f(E; F)$, therefore $v$ is adherent to $\mathscr{L}^f(E; F)$ in $\mathscr{L}_{pc}(E; F)$. This proves a).

Analogously, under the assumptions of c), the mapping $\psi :w\mapsto w\circ u$ of $\mathscr{L}_{pc}(E)$ into $\mathscr{L}_{\mathfrak{S}}(F; E)$ is continuous (loc. cit.). We have $\psi (1_E) =u$ and $\psi (\mathscr{L}^f(E))\subset \mathscr{L}^f(F; E)$, therefore $u$ is adherent to $\mathscr{L}^f(F; E)$ in $\mathscr{L}_{\mathfrak{S}}(F; E)$.

The image of a precompact subset of F by a continuous linear mapping of F into E is precompact, therefore b) follows from c).

This follows from prop. 11, c), for the image of a bounded subset of F by a compact linear mapping of F into E is relatively compact in E (remark 1 of III, p. 2), therefore precompact.

Let A be a precompact subset of E and U a neighbourhood of 0 in E. By hypothesis there exist $i\in I$ and a continuous semi-norm $p$ on $F_i$ such that U contains $(p\circ v_i)^{-1}([0,1])$. Put $F = F_i,v=v_i$ and $B =v$(A), and suppose that F is an approximation space. The set B is precompact in F. There therefore exist (lemma 2) an integer $n\geqslant 1$, elements $y_1, . . . , y_n$ of F and continuous linear forms $f_1, . . . , f_n$ on F, such that one has, for every $y\in B$,

$j=1$

Since B is bounded (EVT, III, p. 3, prop. 2), there exists a real number $M>0$ such that $|f_j(y)|\leqslant M$ for every $j$ such that $1\leqslant j\leqslant n$ and every $y\in B$. Moreover, since $v(E)$ is dense in F, there exists, for each integer $j$ such that $1\leqslant j\leqslant n$, an element $x_j$ of E such that $p(y_j-v(x_j))\leqslant (2nM)^{-1}$. The linear mapping

belongs to $\mathscr{L}^f(E)$. For every $x\in A$, one has

whence $p(v(x-u(x)))\leqslant \frac{1}{2}+\frac{nM}{2nM}= 1$, and therefore $x-u(x)\in U$. The proposition follows.

Let indeed $(E_i)_{i\in I}$ be a family of approximation spaces. For every finite subset J of I, let us put $E_J=\prod_{i\in J}E_i$ and denote by $v_J$ the canonical mapping of $E =\prod_{i\in I}E_i$ into $E_J$. The locally convex space $E_J$ is an approximation space (Remark 1). The corollary then follows from Prop. 12 applied to the locally convex space E, to the family $(E_J)$ of locally convex spaces, and to the continuous linear mappings $v_J: E\rightarrow E_J$.

Let $\mathscr{P}$ be the set of continuous prehilbertian seminorms on E. For each $p\in \mathscr{P}$, the hypothesis implies that the seminormed space $E_p$ obtained by endowing E with the seminorm $p$ is an approximation space (Corollary 2 and Proposition 10), and the identity mapping of E into $E_p$ is continuous. The corollary therefore follows from Proposition 12.

Since E is metrisable, there exists a fundamental system $(V_m)_{m\in\mathbf{N}}$ of balanced neighbourhoods of 0 in E such that $V_0= E$ and $2^{m+1}V_{m+1}\subset V_m$ for every $m\geqslant 0$. Since $(x_n)$ converges to 0, there exists a strictly increasing sequence $(N_m)_{m\in\mathbf{N}}$ of integers such that $N_0= 0$ and such that, for every $m\geqslant 0$, one has $x_n\in V_m$ for $n\geqslant N_m$. For every integer $n\geqslant 0$, there exists a unique integer $m\geqslant 0$ such that $N_m\leqslant n <N_{m+1}$, and one then puts $\lambda_n= 2^{-m}$ and $y_n= 2^mx_n$. The sequence $(\lambda_n)$ thus defined converges to 0. Moreover, since $y_n\in V_m$ for $n\geqslant N_{m+1}$, the sequence $(y_n)$ converges to 0 in E. Finally, one has $x_n=\lambda_ny_n$ for every $n$.

Recall (EVT, II, p. 28) that if E is a vector space and L a convex balanced subset of E, one denotes by $E_L$ the vector subspace of E generated by L, endowed with the seminorm whose unit ball is L. When the set L contains no line, this seminorm is a norm.

The set A is contained in the closed convex balanced hull of the set of points of a sequence $(x_n)_{n\in\mathbf{N}}$ of elements of E converging to 0 (EVT, IV, p. 24, Cor. 1). Let $(y_n)_{n\in\mathbf{N}}$ and $(\lambda_n)_{n\in\mathbf{N}}$ be sequences satisfying the conclusions of Lemma 3. The closed convex balanced hull L of the points of the sequence $(y_n)$ contains A and is a compact subset of E (EVT, III, p. 8). Hence $E_L$ is a Banach space (EVT, III, p. 8, Cor.). In this Banach space, the norm of $x_n$ is majorized by $\lambda_n$, therefore the sequence $(x_n)$ tends to 0. The closed convex balanced hull $\widetilde{A}$ of the sequence $(x_n)$ in $E_L$ is a compact subset of $E_L$ (EVT, III, p. 8). Since L is a bounded subset of E, the canonical injection of $E_L$ into E is continuous. The topologies induced on $\widetilde{A}$ by those of $E_L$ and of E coincide, and $\widetilde{A}$ is a compact, hence closed, subset of E. Since the set $\widetilde{A}$ is convex, balanced, and contains the sequence $(x_n)$, it contains A. This completes the proof.