COMPLETENESS OF REGULAR INDUCTIVE LIMITS

Regular LB-space is fast complete but may not be quasi-complete. Regular inductive limit of a sequence of fast complete, resp. weakly quasi-complete, resp. reflexive Banach, spaces is fast complete, resp. weakly quasi-complete, resp. reflexive complete, space.

iff it is regular, see [3], there is a natural question asked by Jorge Mujica in [4]: Is every regular LB-space complete?
Throughout the paper E c E 2 c is a sequence of locally convex spaces with continuous inclusions E E n e N. Their locally convex inductive limit is n R+I, denoted by E. The space E is called regular if every set bounded in E is bounded in some E n 2. MAIN RESULTS.
Let F be a locally convex space and A c F absolutely convex.We denote by F A the seminormed space U{nA;neN} whose topology is generated by the Minkowski functional of A. If F A is Banach space, A is called Banach disk.The space F is called fast complete if every set bounded in F is contained in a bounded Banach disk.Every sequentially complete space is fast complete and there are fast complete spaces which are sequentially incomplete, see [5].
EXAMPLE.For each neN and x NxN C, put llx n max {sup{j-ilxij.l; i n, j e N}, sup {Ixijl; i > n, j e N} En {x; llx lln En En+l' n e N, are continuous, E is regular and not quasi-complete.
CLAIM i.Each space E is Banach.n PROOF.Let {x(k)} be a Cauchy sequence in E For each i, j e N the sequence n {x(k)ij} is Cauchy in C and has a limit xij.Let x be the matrix with the entries xij" Given > 0, there is k such that p, rk implies ix(p) x(r) .<e.Hence, Take i > n and choose Ji so that Jx(k)ij < for j > Ji" x e U {En; n e N}.
follows from the inequalities llxll I 2 CLAIM 3. E is regular.
PROOF.Let D c E be not bounded in any E For each n e N choose x(n where =k >--0, Z a, k i, and y(k) e nr(k) Bm(k).To prove that ly(k)i(n)j(n) =< n for k e N, we have to distinguish three cases: (a) k > n: Then lY(k On the other hand Ix(n)i(n),j(n)l > n and x(n) cannot be a convex combination of y(k), k < s, i.e. x(n) nU.Since U is a 0-neighborhood in E, D is not bounded in E.
CLAIM 4. E is not quasi-complete.
PROOF.Let A {6 c NxN; {j e N; (i,j) e 6} is finite, i e N} be ordered by set inclusion.Denote by x(6) the set characteristic function of 6 e A. Then {x(6); 6 e A} c B 1 and the filter associated with 6 x( 6) is bounded in E 1 hence also bounded in E.

cNxN cNxN
Let P be the projection of an NxN matrix on its n-th row.Take a n a closed absolutely convex 0-neighborhood V in E. For each n EN choose m(n) e N and r(n) > 0 so that r(n)B n c V, m(n) 2r(n) -I/n and put o {(i j) e NxN; j < m(i)} If , 6 e A, , 6 o, then x()ij x(6)13.. 0 for j re(i) and Pn(X() x( 6)) n sup {j-n x()nj x(6)nj l;J > m(n)} < m(n) -n 2-nr(n).Hence 2nPn(X() x( 6)) 2+riP (x() x( 6)) k e N Yk n n=1 is contained in V.It is also contained in B 1 and converges coordinate-wise to x(y) x( 6) in E l Hence x(y) x( 6) is in the weak closure of V. Since V is closed and convex, it is also weakly closed and x() x(6) e V.So {x(6); 6 e A} is a base of a bounded Cauchy filter in E. If it had a limit x e E, then x.. 1 for all i, j e N.
This would imply x E for any n e N and x E, q.e.d.n LEMMA.Regular inductive limit of a sequence of semireflexive, resp.reflexive, spaces is semireflexive, resp.reflexive.
PROOF.Let each E be semlreflexive.Since E indlim E is regular, its strong n n equalsto projlim (En) and (E)' c U {((En))';neN U {En; n e N} E.

dual E b
Let each E be reflexive.By [7;IV, 5.6] it suffices to show that E is semlreflexive n and barreled.Take a barrel B in E. For each n e N, B 0 E is a barrel in E Since n n E is reflexive, the barrel B 0 E is a neighborhood in E which implies that B is a n n n neighborhood in E and E is barreled.CONSEQUENCE.Inductive limit of a sequence of reflexive Banach spaces is reflexive.
PROOF.By [6;Th. 4] the inductive limit of reflexive Banach spaces is regular.the proof is complete.
(b) Follows from Lemma since any locally convex space is weakly quasi-complete iff it is semireflexive, [7;IV, 5.5].
(c) Follows from (b) since every weakly quasi-complete space is quasi-complete.
(d) Letbe a Cauchy filter in E. Then as a filter of continuous linear The space E is regular, [6; Th 4], h is continuous iff h-l(0) is closed in E b.
hence E b' projlim E'n is Frechet.Take a sequence {Xn,.n 1,2...} c h-l(0) which converges to Xo in E b.' We have to show that h(x o) 0. Choose e > 0. The set hence there is F e T such that B {Xn; n 0,1,2 is bounded in Eb, sup {If(Xn) h(Xn)l; f e F, x n e B} < e. Fix an f e F and choose n e N so that If(x n) f(x o) < e.Then lh(x o) lffil h(x o) h(Xn) l&l h(Xo) f(Xo) + If(x o) f(x n) +If(x n) h(Xn)l < 3e which implies h(x 0 O CONJECTURE.Regular LB-spacemay not be sequentially complete.
Each E fast completeE fast complete.Each E weakly quasi-complete E weakly quasi-complete.
THEOREM.Let E indlim E be regular.Then: n (a) n(b)