1.3. Inductive topologies and LF-spaces that U

that U_n+1 \E_n =U_n. Then, by induction, we get that for any k 2N there exists a convex neighbourhood U_n+k of the origin in (E_n+k,⌧_n+k) such that U_n+k \E_{n+k 1} = U_{n+k 1}. Hence, for any k 2 N, we get U_n+k \ En = U_n. If we consider now U := S₁

k=1U_n+k, then U \E_n = U_n and U is a neighbourhood of the origin in (E,⌧_ind). Indeed, for any m 2 N we have U\Em =S₁

k=1U_n+k\Em =S₁

k=m nU_n+k\Em, which is a countable union of neighbourhoods of the origin in ⌧_m as for k m nwe getn+k mand so⌧_n+k E_m=⌧_m. We can then conclude that⌧_n✓⌧_ind E_n,8n2N. Corollary 1.3.6. Any LF-space is a locally convex Hausdor↵. t.v.s..

Proof. Let (E,⌧_ind) be the LF-space with defining sequence{(E_n,⌧_n) :n2N}

and denote by F(o) [resp. Fn(o)] the filter of neighbourhoods of the origin in (E,⌧_ind) [resp. in (E_n,⌧_n)]. Then:

\

V2F(o)

V = \

V2F(o)

V \ [

n2N

E_n

!

= [

n2N

\

V2F(o)

(V \E_n) = [

n2N

\

Un2Fn(o)

U_n={o},

which implies that (E,⌧_ind) is Hausdor↵by Corollary 2.2.4 in TVS-I.

As a particular case of Proposition1.3.2 we easily get that:

Proposition 1.3.7.

Let (E,⌧_ind) be an LF-space with defining sequence {(E_n,⌧_n) : n 2 N} and (F,⌧) an arbitrary locally convex t.v.s..

1. A linear mappingu fromE intoF is continuous if and only if, for each n2N, the restriction u E_n of u to E_n is continuous.

2. A linear form on E is continuous if and only if its restrictions to each E_n are continuous.

Note that Propositions 1.3.5 and 1.3.7 and Corollary 1.3.6 hold for any countable strict inductive limit of an increasing sequence of locally convex Hausdor↵t.v.s. (even when they are not Fr´echet).

The next theorem is instead typical of LF-spaces as it heavily relies on the completeness of the t.v.s. of the defining sequence. Before introducing it, let us recall the concept of accumulation point of a filter on a topological space together with some basic useful properties.

Definition 1.3.8. Let F be a filter on a topological space X. A point x2X is called an accumulation point of F if x belongs to the closure of every set which belongs to F, i.e. x2M ,8M 2F.

Proposition 1.3.9. If a filter F of a topological space X converges to a pointx, then x is an accumulation point of F.

Proof. If x were not an accumulation point of F, then there would be a set M 2 F such that x /2M. Hence, X\M is open in X and contains x, so it is a neighbourhood of x. Then X\M 2 F as F ! x by assumption. But F is a filter and so M \ X\M 2 F and so M \ X\M 6=;, which is a contradiction.

Proposition 1.3.10. If a Cauchy filterF of a t.v.s. X has an accumulation pointx, then F converges to x.

Proof. Let us denote by F(o) the filter of neighbourhoods of the origin in X and consider U 2 F(o). Since X is a t.v.s., there exists V 2F(o) such that V+V ✓U. Then there existsM 2Fsuch thatM M ✓V asF is a Cauchy filter in X. Beingx an accumulation point of F guarantees thatx 2M and so that (x+V)\M 6= ;. Then M ((x+V)\M) ✓ M M ✓ V and so M ✓ V + ((x+V)\M) ✓V +V +x ✓ U +x. Since F is a filter and M 2 F, the latter implies that U +x 2F. This proves that F(x) ✓F, i.e.

F !x.

Theorem 1.3.11. Any LF-space is complete.

Proof.

Let (E,⌧_ind) be an LF-space with defining sequence{(E_n,⌧_n) :n2N}. LetF be a Cauchy filter on (E,⌧_ind). Denote by FE(o) the filter of neighbourhoods of the origin in (E,⌧_ind) and consider

G:={A✓E: A◆M+V for someM 2F, V 2FE(o)}. 1)G is a filter on E.

Indeed, it is clear from its definition that G does not contain the empty set and that any subset ofE containing a set inG has to belong to G. Moreover, for anyA₁, A₂ 2Gthere existM₁, M₂2F,V₁, V₂ 2FE(o) s.t. M₁+V₁ ✓A₁ and M₂+V₂ ✓A₂; and therefore

A₁\A₂ ◆(M₁+V₁)\(M₂+V₂)◆(M₁\M₂) + (V₁\V₂).

The latter proves thatA₁\A₂ 2G, since F andFE(o) are both filters and so M₁\M₂ 2F and V₁\V₂ 2FE(o).

18

2) G✓F.

In fact, for any A2G there exist M 2F and V 2FE(o) s.t.

A◆M+V M+{0}=M which implies that A2F sinceF is a filter.

3) G is a Cauchy filter onE.

Let U 2 FE(o). Then there always exists V 2 FE(o) balanced such that V +V V ✓U. AsF is a Cauchy filter on (E,⌧_ind), there existsM 2F such that M M ✓V. Then

(M +V) (M+V)✓(M M) + (V V)✓V +V V ✓U which proves that G is a Cauchy filter since M +V 2G.

It is possible to show (and we do it later on) that

9p2N: 8A2G, A\E_p6=;. (1.12) This property together with the fact that G is a filter ensures that the family

Gp:={A\Ep: A2G}

is a filter on E_p. Moreover, since G is a Cauchy filter on (E,⌧_ind) and since by Proposition 1.3.5 we have ⌧_ind Ep = ⌧p, Gp is a Cauchy filter on (E_p,⌧_p). Hence, the completeness of E_p guarantees that there exists x 2 E_p s.t. Gp ! x which implies in turn that x is an accumulation point for Gp

by Proposition 1.3.9. In particular, this gives that for any A 2 G we have x 2 A\E_p^⌧p = A\E_p^⌧ind ✓ A^⌧ind, i.e. x is an accumulation point for the Cauchy filter G. Then, by Proposition 1.3.10, we get that G ! x and so FE(x)✓G✓F. Hence, we proved thatF !x2E.

Proof. of (1.12)

Suppose that (1.12) is false, i.e. 8n2N, 9A_n 2G s.t. A_n\E_n =;. By the definition of G, this implies that

8n2N,9M_n2F, V_n2FE(o),s.t. (M_n+V_n)\E_n=;. (1.13) Since E is a locally convex t.v.s., we may assume that each Vn is balanced, convex, and such that V_n+1 ✓V_n. For eachn2N, define

W_n:=conv V_n[

n[1

k=1

(V_k\E_k)

! .

Moreover, if for some n 2 N there exists h 2 (W_n+M_n)\E_n then h 2 E_n and h 2 (W_n+M_n). Therefore, we can write h = x+w with x 2 M_n and w 2 Wn ✓ conv(Vn[(V1\En 1)). As Vn and V1\En 1 are both convex, we get that h = x+ty+ (1 t)z with x 2 M_n, y 2 V_n, z 2V₁\E_{n 1} and t2[0,1]. Thenx+ty=h (1 t)z2E_n, but we also havex+ty2M_n+V_n (since Vn is balanced). Hence, x+ty 2 (Mn+Vn)\En which contradicts (1.13), proving that

(W_n+M_n)\E_n=;,8n2N. Now let us define

W :=conv [1 k=1

(V_k\E_k)

! .

As W is convex and as W \E_k contains V_k \E_k for all k 2 N, W is a neighbourhood of the origin in (E,⌧_ind). Moreover, as (Vn)_n2N is decreasing, we have that for alln2N

W =conv

n[1

k=1

(V_k\E_k)[ [1 k=n

(V_k\E_k)

!

✓conv

n[1

k=1

(V_k\E_k)[Vn

!

=Wn. SinceFis a Cauchy filter on (E,⌧_ind), there existsB2Fsuch thatB B ✓W and so B B ✓W_n,8n2N. We also have thatB\M_n6=;,8n2N, as both B and M_n belong to F. Hence, for all n2Nwe get

B (B\M_n)✓B B ✓W_n, which implies

B ✓Wn+ (B\Mn)✓Wn+Mn

and so

B\En✓(Wn+Mn)\En (1.13)

= ;.

Therefore, we have got that B \En = ; for all n 2 N and so that B = ;, which is impossible asB 2F. Hence, (1.12) must hold true.

Example I: The space of polynomials

Letn2Nand x:= (x₁, . . . , x_n). Denote by R[x] the space of polynomials in the n variables x₁, . . . , x_n with real coefficients. A canonical algebraic basis forR[x] is given by all the monomials

x^↵:=x^↵₁¹· · ·x^↵_nⁿ, 8↵= (↵1, . . . ,↵n)2Nⁿ0. 20

For any d 2 N0, let Rd[x] be the linear subpace of R[x] spanned by all monomials x^↵ with|↵|:=Pn

i=1↵_id, i.e.

Rd[x] :={f 2R[x]|degf d}.

Since there are exactly ^n+d_d monomials x^↵ with |↵|d, we have that dim(Rd[x]) = (d+n)!

d!n! ,

and so that Rd[x] is a finite dimensional vector space. Hence, by Tychono↵

Theorem (see Corollary 3.1.4 in TVS-I) there is a unique topology ⌧_e^d that makesR^d[x] into a Hausdor↵t.v.s. which is also complete and so Fr´echet (as it topologically isomorphic toR^dim(Rd[x])equipped with the euclidean topology).

As R[x] := S₁

d=0R^d[x], we can then endow it with the inductive topol- ogy ⌧_ind w.r.t. the family of F-spaces (Rd[x],⌧_e^d) :d2N⁰ ; thus (R[x],⌧_ind) is a LF-space and the following properties hold (proof in Exercise Sheet 3):

a) ⌧_ind is the finest locally convex topology on R[x],

b) every linear map f from (R[x],⌧_ind) into any t.v.s. is continuous.

Example II: The space of test functions

Let ⌦✓R^d be open in the euclidean topology. For any integer 0s 1, we have defined in Section1.2the setC^s(⌦) of all real valueds times continuously di↵erentiable functions on ⌦, which is a real vector space w.r.t. pointwise addition and scalar multiplication. We have equipped this space with the C^s-topology (i.e. the topology of uniform convergence on compact sets of the functions and their derivatives up to orders) and showed that this turnsC^s(⌦) into a Fr´echet space.

Let K be a compact subset of ⌦, which means that it is bounded and closed inR^dand that its closure is contained in⌦. For any integer 0s 1, consider the subset Cc^k(K) ofC^s(⌦) consisting of all the functionsf 2C^s(⌦) whose support lies in K, i.e.

C^sc(K) :={f 2C^s(⌦) :supp(f)✓K},

wheresupp(f) denotes the support of the functionf on⌦, that is the closure in⌦ of the subset{x2⌦:f(x)6= 0}.

For any integer 0  s  1, Cc^s(K) is always a closed linear subspace of C^s(⌦). Indeed, for any f, g 2 C^sc(K) and any 2 R, we clearly have f+g2C^s(⌦) and f 2C^s(⌦) but alsosupp(f+g)✓supp(f)[supp(g)✓K and supp( f) = supp(f) ✓ K, which gives f +g, f 2 Cc^s(K). To show thatC^s(K) is closed inC^s(⌦), it suffices to prove that it is sequentially closed

as C^s(⌦) is a F-space. Consider a sequence (f_j)_j2N of functions in Cc^s(K) converging tof in theC^s topology. Then clearlyf 2C^s(⌦) and since all the fj vanish in the open set ⌦\K, obviously their limit f must also vanish in

⌦\K. Thus, regarded as a subspace of C^s(⌦), Cc^s(K) is also complete (see Proposition 2.5.8 in TVS-I) and so it is itself an F-space.

Let us now denote byCc^s(⌦) the union of the subspacesCc^s(K) asKvaries in all possible ways over the family of compact subsets of ⌦, i.e. Cc^s(⌦) is linear subspace of C^s(⌦) consisting of all the functions belonging to C^s(⌦) which have a compact support (this is what is actually encoded in the subscript c).

In particular, Cc¹(⌦) (smooth functions with compact support in ⌦) is called space of test functionsand plays an essential role in the theory of distributions.

We will not endow Cc^s(⌦) with the subspace topology induced by C^s(⌦), but we will consider a finer one, which will turnCc^s(⌦) into an LF-space. Let us consider a sequence (Kj)_j2N of compact subsets of ⌦ s.t. Kj ✓Kj+1,8j 2N and S₁

j=1K_j = ⌦. (Sometimes is even more advantageous to choose the K_j’s to be relatively compact i.e. the closures of open subsets of⌦such that Kj ✓K˚j+1,8j 2Nand S₁

j=1Kj =⌦.) Then Cc^s(⌦) = S₁

j=1Cc^s(K_j), as an arbitrary compact subset K of ⌦ is contained inKjfor some sufficiently largej. Because of our way of defining the F-spaces Cc^s(K_j), we have that Cc^s(K_j) ✓ Cc^s(K_j+1) and Cc^s(K_j+1) induces on the subset Cc^s(K_j) the same topology as the one originally given on it, i.e. the subspace topology induced onCc^s(K_j) byC^s(⌦). Thus we can equipCc^s(⌦) with the inductive topology ⌧_ind w.r.t. the sequence of F-spaces {C^sc(K_j), j 2N}, which makesCc^s(⌦) an LF-space. It is easy to check that⌧_inddoes not depend on the choice of the sequence of compact setsK_j’s provided they fill⌦.

Note that (Cc^s(⌦),⌧_ind) is not metrizable since it is not Baire (proof in Exercise Sheet 3).

Proposition 1.3.12. For any integer 0  s  1, consider Cc^s(⌦) endowed with the LF-topology ⌧_ind described above. Then we have the following contin- uous injections:

Cc¹(⌦)!Cc^s(⌦)!Cc^{s 1}(⌦), 80< s <1.

Proof. Let us just prove the first inclusion i:Cc¹(⌦) ! Cc^s(⌦) as the others follows in the same way. As Cc¹(⌦) = S₁

j=1Cc¹(K_j) is the inductive limit of the sequence of F-spaces (Cc¹(Kj))_j2N, where (Kj)j2N is a sequence of compact subsets of ⌦ such that Kj ✓ Kj+1,8j 2 N and S₁

j=1Kj = ⌦, by Proposition 1.3.7we know that i is continuous if and only if, for any j 2N, e_j := i Cc¹(K_j) is continuous. But from the definition we gave of the topology on each Cc^s(Kj) and Cc¹(Kj), it is clear that both the inclusions i_j : Cc¹(K_j) ! Cc^s(K_j) and s_j :Cc^s(K_j) ! Cc^s(⌦) are continuous. Hence, for each j2N,e_j =s_j i_j is indeed continuous.

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