(b) Eis locally compact;
(c) 0∈E has a totally bounded neighborhood inE.
Proof. Since each n-dimensional topological linear space is linearly homeo-morphic to Rn (Corollary 2.5.2), we have (a) ⇒ (b). Since every compact subset ofE is totally bounded, the implication (b)⇒(c) follows.
(c)⇒(a): LetU be a totally bounded neighborhood of0∈E. By Propo-sition 2.4.1, we have a circled neighborhood V of 0 such that V +V ⊂ U. Then,V is also totally bounded. First, we show the following:
Claim. For each closed linear subspaceF E, there is somex∈U such that (x+V)∩F =∅.
Contrary to the claim, suppose that (x+V)∩F = ∅ for everyx ∈ U. Since V =−V, it follows that U ⊂F +V, so we have V +V ⊂F +V. If (n−1)V ⊂F+V then
nV ⊂(n−1)V +V ⊂F+V +V ⊂F+F+V =F+V.
By induction, we have nV ⊂ F +V for every n ∈ N, which implies that V ⊂
n∈N(F+n−1V).
Takez∈E\F. SinceF is closed inE, we have a circled neighborhoodW of 0∈E such thatW ⊂V and (z+W)∩F =∅. The total boundedness of V implies the topological boundedness, hence V ⊂mW for some m∈N. On the other hand,k−1z∈V for some k∈N. Sincek−1z∈V ⊂F+ (km)−1V, it follows that z ∈ F +m−1V ⊂ F +W. This contradicts the fact that (z+W)∩F=∅.
Now, assume thatEis infinite-dimensional. Letv1∈U\{0}andF1=Rv1. Then,F1 is closed inE (Proposition 2.5.1) andF1=E. Applying the Claim above, we havev2∈U such that (v2+V)∩F1=∅. Note thatv2∈v1+V. Let F2=Rv1+Rv2. Since F2 is closed inE (Proposition 2.5.1) andF2=E, we can again apply the Claim to findv3∈U such that (v3+V)∩F2=∅. Then, note thatv3∈vi+V fori= 1,2. By induction, we havevn ∈U,n∈N, such that vn ∈vi+V for i < n. Then,{vn |n∈N} is not totally bounded. This is a contradiction. Consequently,E is finite-dimensional. ⊓⊔
By Theorem 2.5.9, every infinite-dimensional topological linear space is not locally compact.
2.6 Metrizability and Normability 111 2.6.1 Theorem A topological linear space E is metrizable if and only if 0∈E has a countable neighborhood basis.
In a more general setting, we shall prove a stronger result. A metricdon a groupGis said to beleft(resp.right)invariantifd(x, y) =d(zx, zy) (resp.
d(x, y) = d(xz, yz)) for each x, y, z ∈ G; equivalently, d(x, y) = d(x−1y,1) (resp. d(x, y) = d(xy−1,1)) for each x, y ∈ G. When both of two metricsd andd′on a groupGare left (or right) invariant, they are uniformly equivalent to each other if and only if they induce the same topology. It is said thatdis invariantif it is left and right invariant. Every invariant metricdon a group Ginduces the topology onGthat makesGa topological group. In fact,
d(x, y) =d(x−1xy−1, x−1yy−1) =d(y−1, x−1) =d(x−1, y−1) and d(xy, x′y′)d(xy, x′y) +d(x′y, x′y′) =d(x, x′) +d(y, y′).
It is easy to verify that a left (or right) invariant metric d on a groupG is invariant ifd(x, y) =d(x−1, y−1) for eachx, y∈G. Theorem 2.6.1 comes from the following:
2.6.2 Theorem For a topological groupG, the following are equivalent:
(a) Gis metrizable;
(b) The unit1∈Ghas a countable neighborhood basis;
(c) Ghas an admissible bounded left invariant (right invariant) metric.
Proof. Since the implications (a)⇒(b) and (c)⇒(a) are obvious, it suffices to show the implication (b)⇒(c).
(b)⇒(c):14We shall construct a left invariant metricρ∈Metr(G). Then, a right invariant metricρ′∈Metr(G) can be defined byρ′(x, y) =ρ(x−1, y−1).
By condition (b), we can find an open neighborhood basis {Vn |n ∈ N} at 1∈Gsuch that
Vn−1=Vn and Vn+1Vn+1Vn+1⊂Vn for eachn∈N.15 LetV0=G, and define
p(x) = inf{2−i|x∈Vi} ∈I for eachx∈G.
SinceVn=Vn−1for eachn∈N, it follows thatp(x) =p(x−1) for everyx∈G.
Note that
n∈ωVn ={1}.16 Hence, for everyx∈G,
14The idea of the proof is the same as that of Theorem 1.4.1 (b)⇒(a).
15Note that {Vnx | n ∈ } is an open neighborhood basis at x ∈ G. For each x, y∈Gandn∈, Vn+1x∩Vn+1y=∅implies Vn+1y⊂Vnx. Indeed,ux=vy for someu, v∈Vn+1, henceVn+1y=Vn+1v−1ux⊂Vnx. Thus, the metrizability ofGcan be obtained by the Frink Metrization Theorem 1.4.1. On the other hand, Vn={Vnx|x∈G} ∈cov(G) and stVn+1≺ Vn. Indeed, st(Vn+1x,Vn+1)⊂Vnx.
Thus, the metrizability ofGcan also be obtained by Corollary 1.4.4.
16It is assumed thatGis Hausdorff.
p(x) = 0 ⇔ x=1.
By induction onn, we shall prove the following:
(∗) p(x−01xn)2
n
i=1
p(x−i−11xi) for eachx0, x1, . . . , xn∈G.17 The casen = 1 is obvious. Assume (∗) form < n. Ifn
i=1p(x−i−11xi) = 0 or n
i=1p(x−i−11xi) 12, it is trivial. When 2−k−1 n
i=1p(x−i−11xi) <2−k for some k∈N, choose 1mnso that
m−1
i=1
p(x−1i−1xi)<2−k−1 and
n
i=m+1
p(x−1i−1xi)<2−k−1.
Note that p(x−1m−1xm) < 2−k. By the inductive assumption, p(x−10 xm−1) <
2−k and p(x−m1xn) < 2−k. Then, x−01xm−1, x−m1−1xm, x−m1xn ∈ Vk+1. Since Vk+1Vk+1Vk+1⊂Vk, it follows thatx−01xn ∈Vk, hence
p(x−01xn)2−k 2
n
i=1
p(x−i−11xi).
Now, we can define a metricρonGas follows:
ρ(x, y) = inf ni=1p(x−i−11xi)n∈N, xi ∈G, x0=x, xn=y . By the definition, ρis left invariant. Note thatρ(x, y)p(x−1y)1. Then, x−1y ∈ Vn implies ρ(x, y) p(x−1y) 2−n <2−n+1, which meansxVn ⊂ Bρ(x,2−n+1). On the other hand, ifρ(x, y)<2−n thenp(x−1y)2ρ(x, y)<
2−n+1 by (∗), which implies x−1y ∈Vn. Thus, Bρ(x,2−n)⊂xVn. Therefore, ρis admissible. ⊓⊔
In the above proof, a right invariant metricρ ∈ Metr(G) can be directly defined as follows:
ρ(x, y) = inf
n
i=1p(xi−1x−1i )
n∈, xi∈G, x0=x, xn=y
. Every metrizable topological linear spaceE has an admissible (bounded) metricρthat is not only invariant but also satisfies the following:
(♯)|t|1⇒ρ(tx,0)ρ(x,0).
To verify this, let us recall how to define the metric ρ in the above proof.
Taking a neighborhood basis {Vn |n ∈N} at 0∈E so that Vn =−Vn and Vn+1+Vn+1+Vn+1⊂Vn for eachn∈N, we define the admissible invariant metricρas follows:
17For eachx, y∈G, letδ(x, y) =p(x−1y). Then, this inequality is simply the one given in the Sketch of the direct proof for Corollary 1.4.4.
2.6 Metrizability and Normability 113 ρ(x, y) = inf ni=1p(xi−xi−1)n∈N, xi ∈E, x0=x, xn=y
, where p(x) = inf{2−i | x ∈ Vi}. Since E is a topological linear space, the condition that Vn = −Vn can be replaced by a stronger condition that Vn
is circled, i.e., tVn ⊂Vn for t ∈[−1,1]. Then,p(tx)p(x) for each x ∈E and t ∈ [−1,1], which implies that ρ(tx,0) ρ(x,0) for each x ∈ E and t∈[−1,1].
Let d be an invariant metric on a linear space E. Addition on a linear space E is clearly continuous with respect to d. On the other hand, scalar multiplication on E is continuous with respect to dif and only if dsatisfies the following three conditions:
(i) d(xn,0)→0 ⇒ ∀t∈R, d(txn,0)→0;
(ii) tn→0 ⇒ ∀x∈E, d(tnx,0)→0;
(iii) d(xn,0)→0, tn→0 ⇒ d(tnxn,0)→0.
Indeed, the “only if” part is trivial. To show the “if” part, observe tnxn−tx= (tn−t)(xn−x) +t(xn−x) + (tn−t)x.
Sincedis invariant, it follows that
d(tnxn, tx) =d((tn−t)(xn−x) +t(xn−x) + (tn−t)x,0)
d((tn−t)(xn−x),0) +d(t(xn−x),0) +d((tn−t)x,0), whered(tnxn, tx)→0 if tn→tand d(xn, x)→0. Thus, the above three conditions imply the continuity of scalar multiplication onE with respect tod.
It should be remarked that condition (♯) implies condition (iii).
An invariant metricdon E satisfying these conditions is called a linear metric. A linear space with a linear metric is called ametric linear space.
Then, every metric linear space is a metrizable topological linear space. Con-versely, we have the following fact:
Fact Every admissible invariant metric for a metrizable topological linear space is a linear metric.
For subsets of a metric linear space, the total boundedness coincides with that in the metric sense. On the other hand, the topological boundedness does not coincide with the metric boundedness. In fact, every metrizable topological linear spaceEhas an admissible bounded invariant metric. For instance, given an admissible invariant metricdforE, the following are admissible bounded invariant metrics:
min
1, d(x, y)
, d(x, y) 1 +d(x, y).
For a linear metricρonEwith the condition (♯), the functionalE∋x→ ρ(x,0)∈Ris called anF-norm. In other words, a functional·:E→Ron a linear spaceE is called anF-normif it satisfies the following conditions:
(F1) x0 for everyx∈E;
(F2) x= 0 ⇒ x=0;
(F3) |t|1 ⇒ txx for everyx∈E;
(F4) x+yx+yfor everyx, y∈E;
(F5) xn →0 ⇒ txn →0 for everyt∈R;
(F6) tn →0 ⇒ tnx →0 for everyx∈E.
Conditions (F3), (F5), and (F6) correspond to conditions (♯), (i), and (ii), respectively. The converse of (F2) is true because0= 0 by (F6). Then, x = 0 if and only if x =0. Condition (F3) implies that −x =x for every x ∈ E. Furthermore, conditions (F3) and (F4) imply condition (F5). Indeed, using (F4) inductively, we havenx nxfor everyn∈Æ. Eacht ∈ [0,∞) can be written as t = [t] +s for some s ∈ [0,1), where [t] is the greatest integer t. Since sx xby (F3), it follows that tx ([t] + 1)x. Because −x=x,tx ([|t|] + 1)xfor every t∈Ê. This implies condition (F5). Thus, condition (F5) is unnecessary.
A linear space E given an F-norm · is called an F-normed linear space. Every norm is an F-norm, hence every normed linear space is an F-normed space. An F-norm · induces the linear metric d(x, y) =x−y. Then, everyF-normed linear space is a metric linear space. AnF-norm on a topological linear spaceEis said to beadmissibleif it induces the topology forE. As we saw above, ifEis metrizable, thenEhas an admissible invariant metric ρ satisfying (♯), which induces the F-norm. Therefore, we have the following:
2.6.3 Theorem A topological linear space has an admissibleF-norm if and only if it is metrizable. ⊓⊔
For each metrizable topological linear space, there exists anF-norm with the following stronger condition than (F3):
(F3∗) x=0, |t|<1⇒ tx<x,
which implies thatsx<txfor eachx=0and 0< s < t. The following proposition guarantees the existence of anF-norm with the condition (F3∗):
2.6.4 Proposition Every (completely) metrizable topological linear spaceE has an admissible invariant (complete) metric dsuch thatd(tx,0)< d(x,0) ifx=0and|t|<1, which induces an admissible F-norm satisfying (F3∗). If an admissible invariant metricρforE is given,dcan be chosen so thatdρ (hence, if ρ is complete, then so is d). Moreover, if ρ is bounded, d can be chosen to be bounded.
Proof. Given an admissible (bounded) invariant metric ρ for E, we define d1(x, y) = sup0<s1ρ(sx, sy). Then,d1is an invariant metric onEwithd1ρ (ifρis bounded then so isd1). For eachε >0, since the scalar multiplication E×R∋(x, s)→sx∈Eis continuous at (0, s) andIis compact, we can find
2.6 Metrizability and Normability 115 δ >0 such thatρ(x,0)< δimpliesρ(sx,0)< εfor everys∈I, henceρ(x, y)<
δimpliesd1(x, y) = sup0<s1ρ(sx, sy)ε. Thus,d1is uniformly equivalent to ρ. In particular,d1is admissible. Forr >0, we define an admissible invariant metricdr forE bydr(x, y) =d1(rx, ry) (= sup0<srρ(sx, sy)). Observe that dr(tx,0)dr(x,0) for eachx∈E andt∈I.
Now, let Q∩(0,1] = {rn | n ∈ N}, where r1 = 1. We define d(x, y) =
n∈N2−n+1drn(x, y). Then, dis an invariant metric onE and ρ(x, y)d1(x, y)d(x, y)2d1(x, y),
hence d is admissible (if ρ is bounded then so is d). It also follows that d(tx,0) d(x,0) for each x ∈ E and t ∈ I. It remains to show that d(tx,0)=d(x,0) for eachx∈E\ {0}and 0< t <1. SinceQ∩(0,1) is dense in (0,1), it suffices to show thatd(tx,0)=d(x,0) for eachx∈E\ {0} and t∈Q∩(0,1). Assume that there exists somex∈E\{0}andt∈Q∩(0,1) such that d(tx,0) =d(x,0). Note thatdr(tx,0) =dr(x,0) for eachr∈Q∩(0,1).
Then, it follows that
dt(x,0) =dt(tx,0) =dt2(x,0) =dt2(tx,0)
=dt3(x,0) =dt3(tx,0) =· · · ,
so dt(x,0) =dtn+1(x,0) =dt(tnx,0) for every n∈N. Since limn→∞tn = 0, it follows that dt(x,0) = limn→∞dt(tnx,0) = 0, hence x = 0, which is a contradiction. ⊓⊔
The topological linear space RN = s (the space of sequences) has the following admissibleF-norms:
sup
i∈N
min
1/i,|x(i)| ,
i∈N
min
2−i,|x(i)| ,
i∈N
2−i|x(i)| 1 +|x(i)|, . . . . The first two do not satisfy condition (F3∗), but the third does.
We now consider the completion of metric linear spaces (cf. 1.3.10).
2.6.5 Proposition LetG be a topological group such that the topology is induced by an invariant metric d. The completion G = (G, d)˜ of (G, d) is a group such thatGis its subgroup andd˜is invariant. Similarly, the completion of a metric (F-normed or normed) linear spaceE is a metric (F-normed or normed) linear space containingE as a linear subspace.
Proof. We define the algebraic operations onG as follows: For eachx, y∈G, choose sequences (xi)i∈Nand (yi)i∈NinGso as to converge toxandy, respec-tively. Since d is invariant, (xiyi)i∈N and (x−i1) are Cauchy sequences in G.
Then, definexyandx−1 as the limits of (xiyi)i∈Nand (x−i 1)i∈N, respectively.
It is easily verified that these are well-defined. Since ˜d(x, y) = limi→∞d(xi, yi),
it is also easy to see that ˜d is invariant, which implies the continuity of the algebraic operations (x, y)→xyandx→x−1.
For the completionE of a metric linear spaceE, we can define not only addition but also scalar multiplication in the same way. To see the continuity of scalar multiplication, letx∈Eandt∈R. Choose a sequence (xi)i∈NinE so as to converge tox. For eachε >0, we can chooseδ0>0 (depending ont) so that
z∈E, d(z,0)< δ0, |t−t′|< δ0⇒d(t′z,0)< ε/4.
Then, we have n0 ∈ N such that d(xn, xn0) < δ0 for everyn n0. Choose δ1>0 so that δ1< δ0and
|s|< δ1⇒d(sxn0,0)< ε/4.
Now, let x′ ∈ E and t′ ∈ R such that ˜d(x, x′)< δ0 and |t−t′| < δ1. Take a sequence (x′i)i∈N in E so as to converge to x′ and choose n1 ∈ N so that n1 n0 and d(xn, x′n) < δ0 for every n n1. Then, for every n n1, it follows that
d(txn, t′x′n)d(txn, txn0) +d(txn0, t′xn0) +d(t′xn0, t′xn) +d(t′xn, t′x′n)
=d(t(xn−xn0),0) +d((t−t′)xn0,0)
+d(t′(xn0−xn),0) +d(t′(xn−x′n),0)
< ε/4 +ε/4 +ε/4 +ε/4 =ε.
WhenE is anF-normed (or normed) linear space, it is easy to see that the F-norm (or norm) forE naturally extends toE. ⊓⊔
Concerning the completeness of admissible invariant metrics, we have the following:
2.6.6 Theorem LetGbe a completely metrizable topological group. Every admissible invariant metric for G is complete. In particular, a metric linear space is complete if it is absolutelyGδ (i.e., completely metrizable).
Proof. Letdbe an admissible invariant metric forGandGbe the completion of (G, d). Note that G is a topological group by Proposition 2.6.5. It suffices to show that G =G. Since Gis completely metrizable, Gis a dense Gδ-set in G (Theorem 1.5.2), hence we can writeG\G=
n∈NFn, where each Fn
is a nowhere dense closed set inG. Assume G\G=∅ and takex0∈G\G.
Since x0x∈G\G for everyx∈G, it follows that G⊂
n∈Nx−10 Fn, where eachx−01Fn is also a nowhere dense closed set in G. Then, we have
G=
n∈N
Fn∪
n∈N
x−01Fn,
which is the countable union of nowhere dense closed sets. This contradicts the complete metrizability ofG (the Baire Category Theorem 1.5.1). ⊓⊔
2.6 Metrizability and Normability 117 2.6.7 Corollary LetGbe a metrizable topological group. Every completely metrizable Abelian subgroup H of G is closed in G. Hence, in a metrizable topological linear space, every completely metrizable linear subspace is closed.
Proof. By Theorem 2.6.2,Ghas an admissible left invariant metricd. Because H is an Abelian subgroup of G, the restriction of d on H is an admissible invariant metric forH, which is complete by Theorem 2.6.6. Hence, it follows that H is closed inG. ⊓⊔
It is said that anF-norm (or anF-normed space) iscompleteif the metric induced by theF-norm is complete. It should be noted that every metrizable topological linear space has an admissible F-norm (Proposition 2.6.4) and that every admissibleF-norm for a completely metrizable topological linear space is complete (Theorem 2.6.6). A completely metrizable topological linear space (or a completeF-normed linear space) is called anF-space. AFr´echet space is a locally convex F-space, that is, a completely metrizable locally convex topological linear space. Every Banach space is a Fr´echet space, but the converse does not hold. In fact, s =RN is a Fr´echet space but it is not normable (Proposition 0.2.1).
Concerning the quotient of anF-normed (or normed) linear space, we have the following:
2.6.8 Proposition Let E = (E, · ) be an F-normed (or normed) linear space andF a closed linear subspace ofE. Then, the quotient spaceE/F has the admissible F-norm (or norm) |||ξ|||= infx∈ξx, where if · is complete then so is|||·|||. Hence, ifEis (completely) metrizable or (completely) normable then so is E/F.
Proof. It is easy to see that||| · |||is anF-norm (or a norm). It should be noted that the closedness ofF is necessary for condition (F2). Letq:E→E/F be the natural linear surjection, i.e.,q(x) =x+F. Then, for eachε >0,
q(x)x< ε
=
ξ∈E/F |||ξ|||< ε ,
which means that q : E → (E/F,||| · |||) is open and continuous, so it is a quotient map. Then,||| · |||induces the quotient topology, i.e.,||| · |||is admissible for the quotient topology. It also follows that ifE is locally convex then so is E/F.
We should remark the following fact:
Fact |||ξ−ξ′|||= inf
x−x′x′ ∈ξ′
for eachx∈ξ.
Indeed, the left side is not greater than the right side by definition. For each x, y∈ξandy′∈ξ′,
y−y′=x−(y′+x−y)inf
¨
x−x′
¬
¬x′∈ξ′
©
becausey′+x−y∈ξ′. Thus, the left side is not less than the right side.
We shall show that if·is complete then so is|||·|||. To see the completeness of||| · |||, it suffices to prove that each Cauchy sequence (ξi)i∈NinE/F contains a convergent subsequence. Then, by replacing (ξi)i∈N with its subsequence, we may assume that|||ξi−ξi+1|||<2−i for eachi∈N. Using the Fact above, we can inductively choose xi ∈ ξi so thatxi−xi+1 <2−i. Then, (xi)i∈N
is a Cauchy sequence in E, which converges to some x∈ E. It follows that (ξi)i∈N converges to somex+F. ⊓⊔
In the above,E/F is called thequotientF-normed(ornormed)linear spacewith theF-norm (or norm)|||·|||, which is called thequotientF-norm (ornorm). Note thatE/F is locally convex if so isE. IfEis a Banach space, a Fr´echet space, or anF-space, then so isE/F for any closed linear subspace F ofE.
Recall thatA ⊂E is topologically bounded if, for each neighborhoodU of0∈E, there exists somer∈Rsuch thatA⊂rU.
2.6.9 Theorem A topological linear spaceEis normable if and only if there is a topologically bounded convex neighborhood of 0∈E.
Proof. The “only if” part is trivial. To see the “if” part, let V be a topo-logically bounded convex neighborhood of 0∈ E. Then, W = V ∩(−V) is a topologically bounded circled convex neighborhood of 0 ∈ E. Hence, the Minkowski functionalpW is a norm onE by Proposition 2.3.4. By Corollary 2.4.12,
x∈EpW(x)< ε
=εp−W1([0,1)) =εintW for eachε >0.
For each neighborhoodU of0∈E, we can chooser >0 such thatW ⊂rU.
Then,
x∈EpW(x)< r−1
=r−1intW ⊂r−1W ⊂U, hencepW induces the topology forE. ⊓⊔
For the local convexity, we have the following:
2.6.10 Theorem A (metrizable) topological linear space E is locally con-vex if and only if E is linearly homeomorphic to a linear subspace of the (countable) product
λ∈ΛEλ of normed linear spacesEλ.
Proof. As is easily observed, the product of locally convex topological linear spaces is locally convex, and so is any linear subspace of a locally convex topological linear space. Moreover, the countable product of metrizable spaces is metrizable. Then, the “if” part follows.
We show the “only if” part. By the local convexity,Ehas a neighborhood basis {Vλ | λ ∈ Λ} of 0 ∈ E consisting of circled closed convex sets (cf.
Proposition 2.4.2), where cardΛ=ℵ0 ifEis metrizable (Theorem 2.6.1). For eachλ∈Λ, letFλ be a maximal linear subspace ofE contained in Vλ. (The
2.6 Metrizability and Normability 119 existence ofFλis guaranteed by Zorn’s Lemma.) Then,Fλis closed inE. Let qλ : E → E/Fλ be the natural linear surjection, where we do not give the quotient topology toE/Fλbut we want to define a norm onE/Fλ.
Observe thatqλ(Vλ) is a circled convex set inE/Fλ and0∈coreqλ(Vλ).
Moreover,R+ξ⊂qλ(Vλ) for eachξ∈(E/Fλ)\ {0}. Indeed, takex∈E\Fλ
so thatqλ(x) =ξ. By the maximality ofFλ, Rx+Fλ ⊂Vλ, i.e.,tx+y∈Vλ
for somet∈Randy∈Fλ, where we can taket >0 becauseVλis circled. For eachz∈Fλ,
tx+y=12(2tx+z) +12(2y−z).
Since 2y−z∈Fλ ⊂Vλ, it follows that 2tx+z∈Vλ. Then, 2tξ =qλ(2tx)∈ qλ(Vλ).
By Proposition 2.3.4, the Minkowski functionalpλ=pqλ(Vλ):E/Fλ→R forqλ(Vλ) is a norm. Thus, we have a normed linear spaceEλ= (E/Fλ, pλ).
Observe that
0∈intVλ= coreVλ⊂q−1λ (coreqλ(Vλ))
=qλ−1 p−q1
λ(Vλ)([0,1))
= (pλqλ)−1([0,1)).
By Proposition 2.4.11, the sublinear functional pλqλ :E →R is continuous, which implies thatqλ:E→Eλ is continuous.
Leth:E→
λ∈ΛEλ be the linear map18 defined byh(x) = (qλ(x))λ∈Λ. Ifx=0∈Ethenx∈Vλ(sox∈Fλ) for someλ∈Λ, which impliesqλ(x)= 0, henceh(x)=0. Thus,his a continuous linear injection. To see that his an embedding, it suffices to show that
h(Vλ)⊃h(E)∩pr−λ1(p−λ1([0,12))) for eachλ∈Λ.
Ifpλ(prλh(x))< 12 then
qλ(2x) = prλh(2x)∈p−λ1([0,1))⊂qλ(Vλ), hence 2x−y∈Fλ for somey∈Vλ. Then, it follows that
x=12(2x−y) +12y∈Vλ, so h(x)∈h(Vλ). This completes the proof. ⊓⊔
Combining Theorem 2.6.10 with Proposition 2.6.5 and Corollary 2.6.7, we have the following:
2.6.11 Corollary A topological linear spaceEis a Fr´echet space if and only if E is linearly homeomorphic to a closed linear subspace of the countable product
i∈NEi of Banach spacesEi. ⊓⊔
18That is, a continuous linear function.