MAT 578 Functonal Analyss John Qugg Fall 2008 Locally convex spaces revsed September 6, 2008 Ths secton establshes the fundamental propertes of locally convex spaces. Acknowledgment: although I wrote these notes to reflect how I thnk of these thngs, some of the development s essentally the same as n Jack Spelberg s notes, and n fact hs notes were a valuable source for me. Remnder: all vector spaces wll have scalar feld F = R or C. Defnton. Let X be a vector space and x, y X. The lne segment jonng x and y s the set {( t)x + ty : 0 t }. Defnton 2. A subset A of a vector space X s convex f t contans the lne segment jonng any two of ts elements Observaton 3. The ntersecton of any famly of convex sets s convex. Defnton 4. A local base of a topologcal vector space s convex f t conssts of convex sets. By a reasonable use of the Englsh language, a balanced convex local base wll mean a local base whch s both balanced and convex. Defnton 5. A locally convex space s a topologcal vector space whch has a convex local base. Defnton 6. Let X be a vector space. () A convex combnaton of x,..., x n X s any vector of the form n n t x wth t 0, t = ; (2) The convex hull of a subset A of X, denoted by co A, s the ntersecton of all convex sets contanng A. Lemma 7. If A X, then: () co A s the smallest convex set contanng A; (2) co A concdes wth the set of all convex combnatons of elements of A. Proof. For (), just note that co A s convex snce t s an ntersecton of convex sets. For (2), let B denote the set of all convex combnatons of elements of A. Then co A B, snce (by an easy nducton argument) every convex set s closed under convex combnatons. On the other hand, a straghtforward computaton shows that a convex combnaton of convex combnatons of elements of A s agan a convex combnaton of elements of A, so B s convex. Snce A B, we have co A B. Thus we must have co A = B, as desred.
2 Lemma 8. Every locally convex space has a balanced convex local base. Proof. Let B be the famly of all convex neghborhoods of 0 n a locally convex space X, so that B s a convex local base. It suffces to show that for all U B there exsts V B such that V s balanced and V U. Snce the balanced neghborhoods of 0 form a local base, we can choose a balanced neghborhood W of 0 such that W U. Put V = co W. Then W V U (because U s convex), so V s a convex neghborhood of 0. We fnsh by showng that V s balanced. Let s and x V. Then there exst t,..., t n 0 and y,..., y n W such that n t = and x = n t y. Snce W s balanced we have sy W for all. Thus n sx = t sy s a convex combnaton of elements of W, hence s an element of V. We now nvestgate how locally convex spaces can be characterzed by semnorms. Ths s n fact the reason why local convexty s so mportant. Defnton 9. Let p be a semnorm on a vector space X, and let a > 0. Put [p < a] = p (B a ) [p a] = p ( B a ) Thus [p < a] = {x X : p(x) < a} and [p a] = {x X : p(x) a}. Observaton 0. If p s a semnorm on a vector space X, then: () for every a > 0 the set [p < a] s balanced, convex, and absorbng; (2) [p < a] = a[p < ] = [ap < ]. The reason why semnorms are so closely ted to locally convex spaces s that, conversely, balanced convex absorbng sets gve rse to semnorms: Defnton. Let U be an absorbng set n X. The Mnkowsk functonal assocated to U s the functon p U : X R defned by p U (x) = nf{t > 0 : x tu}. Note that p U s well-defned snce U s absorbng. Theorem 2. Let U be a balanced, convex, and absorbng subset of a vector space X. Then the Mnkowsk functonal p U s a semnorm, and () [p U < ] U [p U ]. Proof. For ths proof let p = p U. It follows straght from the defnton that p s nonnegatve. Let x X and c F. We must show that p(cx) = c p(x). Frst suppose c = 0. Snce U s absorbng, we have 0 tu for some t > 0, and then n fact 0 tu for every t > 0. Thus p(0x) = p(0) = 0 = 0p(x).
Suppose next that c > 0. Then and takng the nf on both sdes gves {t > 0 : cx tu} = c{t > 0 : x tu}, p(cx) = cp(x). Fnally, f c 0 then c > 0, and for all t > 0 we have cx tu f and only f c x tu, snce U s balanced. Thus p(cx) = p( c x) = c p(x). For the trangle nequalty, let x, y X, and let p(x) < a and p(y) < b. Then there exst s (0, a) and t (0, b) such that Thus x su and y tu. ( s x + y su + tu = (s + t) s + t U + t ) s + t U (s + t)u because U s convex. Therefore p(x + y) s + t. Takng the nf over s and t gves p(x + y) p(x) p(y). For (), let x [p < ], so that p(x) <. Then there exsts t (0, ) such that x tu. Then t x U, hence x U snce U s convex and x s n the lne segment jonng two of ts elements 0 and t x. Fnally, f x U then p(x) by defnton. Later, for the proof of the Hahn-Banach Separaton Theorem, t wll be necessary to have the followng techncal varaton of Theorem 2: Lemma 3. In the above theorem, f we only assume U s convex and absorbng, then p U s stll a sublnear functonal, and () stll holds. Proof. Ths follows from a mldly careful examnaton of the above proof. Lemma 4. A semnorm on a topologcal vector space X s contnuous f t s contnuous at 0. Proof. Let p be a semnorm on X, and let {x } be a net convergng to x n X. Snce p satsfes the alternate form of the trangle nequalty, we have because x x 0. p(x ) p(x) p(x x) 0 Corollary 5. A semnorm p on a topologcal vector space X s contnuous f [p < ] s a neghborhood of 0. Proof. For all a > 0 the set [p < a] s a neghborhood of 0 (because multplcaton by a s a homeomorphsm). Thus p s contnuous at 0, hence s contnuous. 3
4 Lemma 6. If U s a convex open neghborhood of 0 n a topologcal vector space X, then U = [p U < ]. Proof. By Lemma 3, t suffces to show that f x U then p U (x) <. Snce U s open and scalar multplcaton s contnuous there exsts t (0, ) such that t x U, hence x tu. Thus p U (x) t <. Observaton 7. If p and q are semnorms on a vector space X, then p q f and only f [q < a] [p < a] for all a > 0. Corollary 8. Let p and q be semnorms on a topologcal vector space X. If p q and q s contnuous, then p s contnuous. Proof. Ths s mmedate from Observaton 7 and Corollary 5. Defnton 9. A famly P of semnorms on a vector space X s separatng f for all x X \ {0} there exsts p P such that p(x) 0. Lemma 20. The famly of all contnuous semnorms on a locally convex space s separatng. Proof. Let X be a locally convex space and x X \{0}. Then there exsts a convex balanced neghborhood U of 0 such that x / U. Then p U s a contnuous semnorm, and p U (x) snce x / U. We now show that, conversely, a separatng famly of semnorms gves a locally convex topology: Theorem 2. Let P be a separatng famly of semnorms on a vector space, and put { n } B = [p < a ] : p,..., p n P, a,..., a n > 0. Then B s a local base for a unque topology makng X nto a locally convex space. Proof. Note that B concdes wth the famly of all fnte ntersectons of sets n the famly S := {[p < a] : p P, a > 0}. S s a famly of convex balanced absorbng sets n X such that for all U S there exsts V S such that V + V U (f U = [p < a] take V = [p < a/2]). Thus the famly B s a local base for a unque topology makng X nto a pre-topologcal vector space. For each x X \ {0} there exsts U S such that x / U, snce the famly P s separatng. Thus X s Hausdorff. Snce every set n S s convex, so s every set n B. Therefore X s a locally convex space. Defnton 22. In the notaton of Theorem 2, P s a generatng famly of semnorms for the locally convex space X. Also, we say the unque locally convex topology on X wth local base B s generated by P. Thus, a famly P of semnorms of a locally convex space X s generatng f and only f the famly B assocated to P as n Theorem 2 s a local base.
Corollary 23. Let X be a locally convex space wth a generatng famly P of semnorms. Then a semnorm q on X s contnuous f and only f there exst p,..., p n P and a > 0 such that q a max p. Proof. If q a max p wth p,..., p n P, then by Lemma 8 q s contnuous because the semnorm a max p s. For the converse, assume that q s contnuous. Then the set U = [q < ] s a neghborhood of 0. Snce the famly B of Theorem 2 s a local base, there exst p,..., p n P and a,..., a n > 0 such that [p < a ] U. Put a = mn a. Then a > 0 and so [a max p < ] = [max p < a] [p < a ] U, q a max p. The followng lemma shows how a generatng famly of semnorms s used to characterze convergence: Lemma 24. Let X be a locally convex space wth a generatng famly P of semnorms. Then a net {x } I n X converges to x f and only f p(x x) 0 for all p P. Proof. Put y = x x. Then x x f and only f y 0. Snce each p P s contnuous, f y 0 then p(y ) 0. For the converse, assume that p(y ) 0 for all p P. Let U be a neghborhood of 0. We must show that there exsts k I such that y U for all k. We can choose p,..., p n P and a,..., a n > 0 such that n [p j < a j ] U. Snce lm I p j (y ) = 0 for all j, we can choose k I such that for all k we have p j (y ) < a j for every j =,..., n, hence y U. Next we characterze locally convex topologes n terms of maps nto normed spaces. Suppose p s a semnorm on a vector space X. Put N = p (0). Then N s a subspace of X by the propertes of semnorms. Let Q : X X/N be the quotent map. It s easy to see that there s a norm on X/N defned by Q(x) = p(x). Lemma 25. Wth the above notaton, f X s a topologcal vector space then p s contnuous f and only f Q s contnuous nto the normed space X/N. 5
6 Proof. Each of p and Q s contnuous f and only f t s contnuous at 0. Let x 0 n the topologcal vector space X. Then by constructon p(x ) 0 f and only f Q(x ) 0. The result follows. Now we play the above game wth a separatng famly P of semnorms: for each p P we have a quotent map Q p : X X/N p nto a normed space X/N p wth norm gven by Q p (x) = p(x). Lemma 26. Wth the above notaton, the topology on X generated by the famly P concdes wth the weakest topology makng every Q p contnuous. Proof. Let T be the topology on X generated by the famly P, and let T be the weakest topology makng every Q p contnuous. Both topologes make X nto a topologcal vector space, so t suffces to show that x 0 n T f and only f x 0 n T. For each p P we have p(x ) 0 f and only f Q p (x ) 0, equvalently Q p (x ) 0 n the normed space X/N p. By Lemma 24, x 0 n T f and only f p(x ) 0 for all p P, and by defnton x 0 n T f and only f Q p (x ) 0 for all p P. The result follows. Fnally, we characterze locally convex topologes n terms of contnuty of semnorms: Corollary 27. Let X be a vector space wth a separatng famly P of semnorms. Then the topology on X generated by P n the sense of Defnton 22 concdes wth the weakest topology on X makng each semnorm n P contnuous. Proof. By Lemma 26 the topology on X generated by P concdes wth the weakest topology makng each quotent map Q p : X X/N p contnuous (n the notaton of the above dscusson), whch by Lemma 25 concdes wth the weakest topology on X makng each semnorm n P contnuous.