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Automatic Continuity from a personal perspective Krzysztof Jarosz www.siue.edu/~kjarosz Southern Illinois University Edwardsville Krzysztof Jarosz (SIUE) Automatic Continuity 1 / 70 Questions Automatic continuity: T : X ! Y is such that ........ (an algebraic condition) + T is continuous Reversed question: T : X ! Y is continuous =) T and X , Y are such that ............ Krzysztof Jarosz (SIUE) Automatic Continuity 2 / 70 Examples Example If T : X ! Y is linear and dim X < ∞ then T is continuous. Example Any linear selfadjoint map T : H ! H on a Hilbert space H is continuous. Proof. Assume that xn ! 0, and Txn ! y . We have hTxn , y i k hxn , Ty i ! hy , y i = ky k2 ! h0, Ty i = 0. Hence y = 0, so by the Closed Graph Theorem T is continuous. kT k can be arbitrary, so while we automatically conclude that T is continuous, we can not conclude anything about the degree of continuity. Krzysztof Jarosz (SIUE) Automatic Continuity 3 / 70 Examples Example If F : A ! C is a linear-multiplicative functional on Banach algebra A then F is continuous and kF k = 1. Proof. Assume kb k < 1 but Fb = 1. We have b = (e ! b ) ∑n∞=0 b n so ∞ 1 = F (b ) = (Fe Fb ) F ∑ bn = (1 1) (...) = 0; n =0 the contradiction shows that kF k = 1. Remark Also true for almost multiplicative functionals: Krzysztof Jarosz (SIUE) jF (fg ) F (f ) F (g )j Automatic Continuity ε kf k kg k . 4 / 70 Examples Corollary If T : A ! B is a linear-multiplicative map from a Banach algebra A into a commutative semisimple Banach algebra B then T is continuous. De…nition A commutative Banach algebra B is semisimple if it has su¢ ciently many linear-multiplicative functionals: for any b 2 B n f0g there is a linear-multiplicative functional G on B such that G (b ) 6= 0. Proof. For any linear-multiplicative functional F : B ! C, F T is a linear and multiplicative functional, so based on the previous example it is continuous. Hence, from the Closed Graph Theorem, T is continuous. Krzysztof Jarosz (SIUE) Automatic Continuity 5 / 70 Michael’s Problem Problem (Open) Is any linear-multiplicative functional on a Fréchet algebra A automatically continuous? De…nition A is a Fréchet algebra if the complete topology on A is given by a sequence of submultiplicative seminorms k kn : kfg kn kf kn kg kn . Theorem Without loss of generality we may assume that A is equal to the algebra of continuous functions de…ned on l ∞ and generated by the projections π k (αj )j∞=1 = αk , with kf kn = sup Krzysztof Jarosz (SIUE) n f (αj )j∞=1 : (αj )j∞=1 Automatic Continuity o n . 6 / 70 Main Automatic Continuity Questions (Banach Algebras) 1 2 A, B - Banach (topological) algebras. Under what conditions any homomorphism T : A ! B is automatically continuous? In particular which Banach algebras have unique Banach algebra norm? A - Banach algebra, E - Banach A-module. Under what conditions any derivation D : A ! E is automatically continuous? De…nition (derivation) A linear map D : A ! E with D (fg ) = D (f ) g + fD (g ) . De…nition (point derivation) For a …xed F 2 M (A) a linear map D : A ! C s.t. D (fg ) = D (f ) F (g ) + F (f ) D (g ) . A point derivation is a derivation into a 1-dimensional bimodule C with df a z = F (a) z, for a 2 A, z 2 C. Krzysztof Jarosz (SIUE) Automatic Continuity 7 / 70 Derivations - examples Example A = A (∆) - disc algebra on the unit disc ∆; E = A ∆ 1 2 on the disc of radius 1 2; D (f ) = Example D : A ! A : Da (x ) = ax - disc algebra f 0. xa. Theorem If A is a commutative, semisimple Banach algebra then there is no non-zero derivation D : A ! A. Krzysztof Jarosz (SIUE) Automatic Continuity 8 / 70 Derivations & homomorphisms Theorem If there is a discontinuous derivation D from a Banach algebra A into a bimodule E then there is a Banach algebra B and a discontinuous homomorphism T : A ! B. Proof. df L df B = A E ; (a1 , x1 ) (a2 , x2 ) = (a1 a2 , a1 x2 + x1 a2 ); check that B is a Banach algebra and that T (a) = (a, Da) is an algebra homomorphism. Krzysztof Jarosz (SIUE) Automatic Continuity 9 / 70 Main Automatic Continuity Questions (non-Banach Algebras) 1 A, B - Banach spaces; R : A ! A, S : B ! B continuous linear maps. Under what conditions any linear map T intertwining with (R, S ) (meaning: TR = ST ) is automatically continuous? 2 Other algebraic conditions implying continuity. Krzysztof Jarosz (SIUE) Automatic Continuity 10 / 70 Examples - translation invariant maps Example If S 1 is a unit circle and T : Lp S 1 ! C , 1 < p < ∞ is a linear functional such that T (fα ) = T (f ) for f 2 Lp S 1 , α 2 S 1, where fα (x ) = f (α + x ) , then T is continuous. Remark Not True for L1 (G ) , L∞ (G ) , or C (X ) , open problem for A (D) , H ∞ , H p . We will later discuss this type of problems in a much greater details. Krzysztof Jarosz (SIUE) Automatic Continuity 11 / 70 Examples - separating maps Example If T : C (K ) ! C (L) is a linear bijection such that ab = 0 ) TaTb = 0, where K , L are compact, then T is continuous. Remark Also true if K , L are arbitrary subsets of the real line but open problem for general Hausdor¤ spaces. There have been a number of papers dealing with disjointedness preserving linear maps (separating maps) due to Y. Abramovich and A. Kitover, J. Araujo, G. Dolinar, J. Font and S. Hernandez, K. Jarosz, and many others addressing problems involving automatic continuity, when inverses are disjointedness preserving, stability of disjointedness preserving maps, etc. Krzysztof Jarosz (SIUE) Automatic Continuity 12 / 70 Reversed Question 1 Problem Assume T : V ! V is a linear map. Can we introduce a complete norm on V making T continuous? Obvious conditions: spectrum of T is compact dim V is …nite or uncountable. Example Let A be an in…nite dimensional vector space, let A0 be an in…nite, countably dimensional subspace of A and let B be a subspace of A such L that A = A0 B. Then the projection T onto B can not be continuous under any complete norm since its kernel is countably dimensional and consequently not closed. Obviously the spectrum of T is compact. Krzysztof Jarosz (SIUE) Automatic Continuity 13 / 70 Reversed Question 2 Problem When the solution to the previous Problem is unique? Assume V is a Banach space and T : V ! V is a continuous linear map. Can we introduce another inequivalent complete norm on V such that T is still continuous? T B (X , k k) \ B (X , j j) ? =) (X , k k) = (X , j j) . Fact It may be possible, e.g. if T = Id or if T is a projection onto an in…nite dimensional subspace. Krzysztof Jarosz (SIUE) Automatic Continuity 14 / 70 Classical Results - Kaplansky’s Theorem Theorem If T is an injective algebra homomorphism from C (X ) into a normed algebra B then kTf k kf k , for all f 2 C (X ) . Corollary Assume p ( ) is an algebra norm (meaning: p (fg ) C (X ) then p (f ) kf k for all f 2 C (X ) . p (f ) p (g ) ) on Corollary If (C (X ) , p ( )) is a Banach algebra then p ( ) ' k k∞ . Problem Must T in the Kaplansky’s Theorem be automatically continuous? Krzysztof Jarosz (SIUE) Automatic Continuity 15 / 70 Kaplansky’s Theorem - proof Given T : C (X ) ,! B. WLOG we assume that T (C (X )) is dense in B and B is a commutative Banach algebra. Put X0 = fF T : F 2 M (B )g ' fx 2 X : δx extends to a multiplicative functional on B g . Since M (B ) is compact in the weak topology of B and the topology of X coincides with the weak topology of X C (X ), X0 is a closed subset of X . Assume that X0 X and let f , g be non-zero elements of C (X ) s.t. supp (f ) , supp (g ) X nX0 and fg = f . Since B is commutative and F (Tg ) = 0 for all F 2 M (B ) it follows that Tg 2 radB, and consequently e Tg is an invertible element of B. However 0 = T (f fg ) = T (e g ) Tf = (e Tg ) Tf so Tf = 0, so T is not injective. Since X0 = X we have kTf k Krzysztof Jarosz (SIUE) max fjF (Tf )j : F 2 M (B )g = kf k . Automatic Continuity 16 / 70 Irving Kaplansky (March 22, 1917 – June 25, 2006). Born in Toronto; Ph.D. Harvard University, 1941; professor at the University of Chicago from 1945 until 1984, when he became the director of the Mathematical Sciences Research Institute in Berkeley; President of AMS: 1985-1986. Awarded the AMS Leroy P., Steele Prize for Lifetime Achievement in 1989. Krzysztof Jarosz (SIUE) Automatic Continuity 17 / 70 Reversed question Kaplansky: multiplication on C (X ) is given =) conclusion about the norm reversed question: ? the norm on C (X ) is given =) conclusion about the multiplication Krzysztof Jarosz (SIUE) Automatic Continuity 18 / 70 Reversed question Theorem Assume is a multiplication on C (X ) . Then 1 if (C (X ) , ) is a Banach algebra (meaning: kf with the same unit e = 1, then = . 2 if kf gk (1 + ε) kf k kg k and ke i k k = O (ε) and, ii (C (X ) , ) ' C (X ). gk kf k kg k ) 1k < ε then Remark Parts 1 & 2(i) remain valid for all uniform algebras; part 2(ii) is true for A (D) and H ∞ (D) , but false for many other algebras; an open question for any algebra of analytic functions of many variables. Krzysztof Jarosz (SIUE) Automatic Continuity 19 / 70 Johnson’s Theorem Any homomorphism from a Banach algebra into a commutative, semisimple Banach algebra B is continuous and consequently any two complete algebra norms on B are equivalent (as discussed earlier). The …rst part of the statement is not true for noncommutative algebras , but the second one is: Theorem (B.E. Johnson) An homomorphism from a Banach algebra onto a semisimple Banach algebra is continuous. Fact For general Banach algebras both properties obviously fail: Take a vector space with two inequivalent Banach norms (e.g. take a Hilbert space (H, k k2 ) , a linear bijection Φ : H ! C [0, 1] and put p (h) = kThk ) and de…ne zero multiplication (ab = 0 for all a, b). Krzysztof Jarosz (SIUE) Automatic Continuity 20 / 70 De…nition of a noncommutative semisimple Banach algebra De…nition Rad (A) = fa 2 A : map Ma : A ! A of multiplication by a is "small"g. A is semisimple if Rad (A) = f0g. For a unital Banach algebra Rad (A) is equal to the intersection of all left maximal ideals (same as: kernels of irreducible representations). Examples For f 2 C (X ) n f0g the Mf is never "small". For an algebra with zero multiplication any Mf is "small". Let A L (Cn ) be generated by T (z1 , ..., zn ) = (0, z1, ..., zn 1 ) then T 2 Rad (A) , however T 2 / Rad (L (Cn )); in fact n Rad (L (C )) = f0g , the left maximal ideals of L (Cn ) are of the form: fS 2 L (Cn ) : S (z) = 0g , z 2 Cn . Krzysztof Jarosz (SIUE) Automatic Continuity 21 / 70 Barry E. Johnson, born on August 1, 1937 in London; undergraduate at the University of Tasmania, Australia; 1961, Ph.D. from Cambridge University; professor at Newcastle University since 1965; Fellow of the Royal Society; President of the London Mathematical Society from 1980-82, died on May 5, 2002. Krzysztof Jarosz (SIUE) Automatic Continuity 22 / 70 Johnson’s Theorem - proof (part 1) A, B - unital Banach algebras, B - semisimple, T : A ! B; J - maximum left ideal in B, π J : B ! B/J irreducible representation; we have to show that π J T is continuous for all such J. Step 1: dim B/J < ∞ (in this case we just need to show that ker (π J T ) is closed) If ker (π J closed. T ) \ fx 2 A : ke Assume x 2 ker (π J We get T ) and ke e so so ker (π J x k < 1g = ∅ then ker (π J x =u T ) is x k < 1, put u = ∑n∞=1 (e (e x )n . x) u e = x + xu = x (e + u ) 2 ker (π J T) T ) = A. Krzysztof Jarosz (SIUE) Automatic Continuity 23 / 70 Johnson’s Theorem - proof (part 2) A, B - unital Banach algebras, B - semisimple, T : A ! B; J - maximum left ideal in B, π J : B ! B/J irreducible representation; we have to show that π J T is continuous for all such J. Step 2: dim B/J = ∞ B/J is an A-module: a (b + J ) = T (a) b + J; for any …xed a the map B/J 3 b + J ! T (a) b + J 2 B/J is continuous; B/J is irreducible A-module, it follows (not obvious) that B/J is a normed module: ka (b + J )k M ka k kb + J k . For b = e we get kπ J so π J T (a)k = kT (a) e + J k = ka (e + J )k M ka k T is continuous. Krzysztof Jarosz (SIUE) Automatic Continuity 24 / 70 Discontinuous Homomorphisms So far we mainly discussed situations where homomorphisms and other maps are automatically continuous. Now we ask for examples where it may not be true; we also ask how "bad" a discontinuous map could be. Example (trivial) A - Banach space, T : A ! A - discontinuous linear map; de…ne zero multiplication (ab = 0 for all a, b) on A then T is a discontinuous algebra homomorphism. Hence, we want to ask about homomorphism of Banach algebras for which ab is rarely equal to zero, e.g. about semisimple algebras or even better about algebras of continuous or analytic functions. Krzysztof Jarosz (SIUE) Automatic Continuity 25 / 70 Discontinuous Homomorphisms - "easy" example EXAMPLE. For a compact set K C we denote by R (K ) the closed subalgebra of C (K ) generated by the rational functions with poles outside K . For "nice" sets K (e.g. if intK = K is simply connected) df R (K ) = A (K ) = ff 2 C (K ) : f is analytic on intK g , and for any z 2 intK there is exactly one point derivation Dz at z (Dz (fg ) = f (z ) Dz (g ) + Dz (f ) g (z )) and there is no point derivation for any z 2 ∂K . However for a very "irregular" K there may be a discontinuous point derivation D0 at some point z0 2 ∂K . Put L df df B = R (K ) C and (f1 , w1 ) (f2 , w2 ) = (f1 f2 , f1 (z0 ) w2 + w1 f2 (z0 )). B is a Banach algebra and T : R (K ) ! R (K ) M df C : T (f ) = (f , D0 (f )) is a discontinuous algebra homomorphism. REMARK. This method is not available for "nice" algebras A; there is no discontinuous point derivation for A = A (D), or A = C (X ) . Krzysztof Jarosz (SIUE) Automatic Continuity 26 / 70 Discontinuous Homomorphisms - disc algebra Theorem There are discontinuous homomorphisms from the disc algebra A (D) into L1 (0, 1) , , into (C [0, 1] , ), and into any radical L1 (v ). De…nition For v : R+ ! R+ s.t. v (0) = 1, v (t ) > 0, v (s + t ) 1 L (v ) = EXAMPLE. v 1, v (t ) = exp t 2 . f : kf k = or Z ∞ 0 jf (t )j v (t ) dt < ∞ . v (t ) = exp ( t ) , Fact L1 (v ) is a commutative Banach algebra; it is radical i¤ limt !∞ v (t )1/t = 0. Krzysztof Jarosz (SIUE) v (s ) v (t ) Automatic Continuity or 27 / 70 Theorem (Functional Calculus) A - commutative Banach algebra, a = (a1 , ..., am ) 2 Am , then there is a unique continuous homomorphism from Hol (σ (a)) into A: Hol (σ (a)) 3 f 7 ! f (a) 2 A, s.t. Zj (a) = aj , 1 (a) = e, and f[ (a) = f (â1 , ..., âm ) . Theorem Functional Calculus is automatically continuous for semisimple algebras. Problem What about radical algebras and m = 1 ? If σ (a) = f0g then Hol (σ (a)) is the algebra of power series with positive radius of convergence; denote by C [[X ]] the algebra of all power series. Problem Can C [[X ]] be embedded in a Banach algebra? Krzysztof Jarosz (SIUE) Automatic Continuity 28 / 70 Discontinuous functional calculus Theorem A - commutative, unital Banach algebra. There exists a monomorphism Φ : C [[X ]] ! A with Φ (X ) = a i¤ a 2 radA and a has a …nite closed descent. If q 2 C [[X ]] is transcendental then Φ (q ) is not unique. De…nition a 2 A has …nite closed descent if 9m 2 N s.t. am +1 A is dense in am A. Example For A = C [0, 1] , f0 (t ) = t has …nite closed descent with m = 1 since f0k f : f 2 A is dense in ff 2 A : f (0) = 0g for all k 1 but f0 is not in the radical. Fact The algebras L1 (0, 1) , descent. Krzysztof Jarosz (SIUE) , (C [0, 1] , ), and L1 (v ) have …nite closed Automatic Continuity 29 / 70 Discontinuous Homomorphisms - disc algebra Theorem There are discontinuous homomorphisms from the disc algebra A (D) into L1 (0, 1) , , into (C [0, 1] , ), and into any radical L1 (v ). Proof. Let π Φ A (D) ,! C [[X ]] ! L1 (v ) M eC where π is the natural embedding and Φ is a discontinuous homomorphism (we may have Φ s.t. Φ π (exp) 6= 0 but Φ π (p ) = 0 for any polynomial p ). Krzysztof Jarosz (SIUE) Automatic Continuity 30 / 70 Discontinuous functional calculus Proof (general idea). Assume Φ has already been de…ned on a subalgebra of F1 of C [[X ]] and f 2 C [[X ]] nF1 ; we want to extend Φ to the algebra generated by F1 and f : fa0 + a1 f + ... + an f n : aj 2 F1 g . If f is transcendental over F1 then the powers of f are linearly independent and the task to select Φ (f ) is relatively easy (but not trivial since it has to allow for further extensions). If however a0 + a1 f + ... + an f n = 0 we have to select Φ (f ) 2 A very carefully so that Φ (a0 ) + Φ (a1 ) Φ (f ) + ... + Φ (an ) (Φ (f ))n = 0. Lemma If a 2 A has …nite closed descent then multiplication by a is a linear T bijection of fan A : n 2 Ng onto itself. Krzysztof Jarosz (SIUE) Automatic Continuity 31 / 70 Discontinuous homomorphisms - more questions Further questions Is there a discontinuous homomorphism from C (X ) ? How "bad" can a discontinuous homomorphism be? Could we describe a general structure of such homomorphisms? Krzysztof Jarosz (SIUE) Automatic Continuity 32 / 70 Structure of discontinuous homomorphisms Theorem (Bade & Curtis, 1960) Let T be a homomorphism from C (X ) into a Banach algebra B. Then T df is continuous on JF = ff 2 C (X ) : f = 0 on a neighborhood of F g for some …nite set F X . T is continuous i¤ F = ∅. William G. Bade, Berkeley Krzysztof Jarosz (SIUE) Automatic Continuity Philip C. Jr. Curtis, UCLA 33 / 70 Proof - part 1 Put F = fx 2 X : 9U 3 x 9K 8f U =) kTf k supp (f ) K kf kg . Step 1: Assume F = ∅. Let U = fUj : j = 1, ..., p g be a …nite cover of X s.t. 8 Uj 9 Kj 8f supp (f ) Uj =) kTf k Kj k f k , let ∑pj=1 χj = 1 be a …nite partition of unit inscribed in U (suppχj Then ! p kTf k = T ∑ χj f j =1 p ∑ Uj ). p T χj f j =1 ∑ Kj j =1 χj f max fKj g kf k . So T is continuous. Krzysztof Jarosz (SIUE) Automatic Continuity 34 / 70 Proof - part 2 Step 2: Assume jFj = ∞. Let x1 , x2 , ... be an in…nite sequence of points of discontinuity, let Vj be pairwise disjoint open neighborhoods of xj . Let hj 2 C (X ) , xj 2 int (fx : hj (x ) = 1g) supp (hj ) Vj . Put ∞ f = ∑ fj , supp (fj ) j =1 fx : hj (x ) = 1g , 1 k fj k = , j kTfj k > j kThj k . We have kTf k kThj k kT (fhj )k = kT (fj )k > j kThj k , so kTf k > j for each j; a contradiction. Krzysztof Jarosz (SIUE) Automatic Continuity 35 / 70 Discontinuous homomorphisms - general theory De…nition A, B - Banach spaces; T : A ! B linear, (X , τ ) - Hausdor¤ space JA , JB : τ ! closed subspaces of A (B, resp.); JA (U1 ) \ ... \ JA (Un 1 ) + JA T (JA (U )) JB ( U ) ; (Un ) = A, for all pairwise disjoint U j 2 τ; x0 2 X is a point of discontinuity of T i¤ π JB (V ) T : A ! B/JB (V ) is discontinuous for all V 3 x0 . Example A = C (X ) ; JA (U ) = ff 2 A : suppf X nU g , B - Banach algebra, T : A ! B homomorphism, JB (V ) = T (JA (V )). Then x0 2 X is a point of discontinuity i¤ T is discontinuous in any neighborhood of x0 . Theorem A linear map can only have …nitely many points of discontinuity. Krzysztof Jarosz (SIUE) Automatic Continuity 36 / 70 Discontinuous homomorphisms - separating maps The same idea works for other classes of spaces and maps. For example a separating map ab = 0 ) TaTb = 0 can have only …nitely many points of discontinuity. In 2004 L. Brown & N.G. Wong described all discontinuous separating functionals on C0 (X ). Such functionals arise from prime ideals in C0 (X ), in particular, every unbounded disjointedness preserving linear functional on c0 (which can be identi…ed with C0 (N )) can be constructed explicitly through an ultra…lter on N complementary to a cozero set ideal. This ultra…lter method can be extended to produce many, but in general not all, such functionals on C0 (X ) for arbitrary X . Krzysztof Jarosz (SIUE) Automatic Continuity 37 / 70 Almost separating maps De…nition A map Φ from C (X ) to C (Y ) is ε-disjointness preserving if, whenever f and g are in C (X ) with fg = 0, then kΦ(f )Φ(g )k εkf kkg k. Problem Is any ε-disjointness preserving linear map automatically near a disjointedness preserving linear map? When such a map is automatically continuous? Theorem For every continuous or surjective ε-disjointness preserving linear map Φ : C (X ) ! C (Y ) there is a disjointedness preserving linear map p Ψ : C (X ) ! C (Y ) such that kΦ Ψk 20 ε. Every unbounded ε-disjointness preserving linear functional on C (X ) must, in fact, be disjointedness preserving. Krzysztof Jarosz (SIUE) Automatic Continuity 38 / 70 Discontinuous homomorphisms on C(X) Theorem (Dales & Esterle) Let X be an in…nite compact set. Assuming the continuum hypothesis there is a discontinuous homomorphism from C (X ) into a Banach algebra. Theorem Let X be an in…nite compact set and A a commutative radical Banach algebra with bounded approximate identity. Assuming the continuum hypothesis there is a discontinuous algebra homomorphism from C (X ) into A, and there is an incomplete algebra norm on C (X ). Each non trivial prime ideal P of C (X ) s.t. jC (X ) /P j = 2@0 is the kernel of a discontinuous homomorphism. De…nition A proper ideal P is called prime if ab 2 P =) a 2 P or b 2 P, for all a, b. Krzysztof Jarosz (SIUE) Automatic Continuity 39 / 70 Discontinuous homomorphisms on Banach algebras Theorem Let B be a commutative Banach algebra and let A be a commutative radical Banach algebra with bounded approximate identity. Assuming the continuum hypothesis, each prime ideal P of B s.t. jB/P j = 2@0 and dim (B/P ) > 1 is the kernel of a discontinuous L homomorphism from B into A eC. such an ideal exists in B if B has in…nitely many characters, or if B is separable and the prime radical of B has in…nite codimension. Krzysztof Jarosz (SIUE) Automatic Continuity 40 / 70 Homomorphisms on C(X) in other models of ZFC set theory Theorem There is a model of ZFC set theory (Zermelo–Fraenkel set theory + axiom of choice) which satis…es the Martin’s Axiom and s.t. all homomorphisms from C (X ) are continuous. De…nition In axiomatic set theory, Martin’s axiom, named after Donald A. Martin, is a statement which is independent of the usual axioms of ZFC set theory. It is implied by the continuum hypothesis, so consistent with ZFC, but is also consistent with ZF + not CH. It can informally be considered to say that all cardinals less than the continuum behave roughly like @0 . Krzysztof Jarosz (SIUE) Automatic Continuity 41 / 70 H. Garth Dales University of Leeds Krzysztof Jarosz (SIUE) Jean Esterle Université de Bordeaux I Automatic Continuity 42 / 70 Proof of the Dales-EsterleTheorem The discontinuous homomorphism B ! A is constructed in several steps: Φ Φ2 Φ Φ Φ Φ B !1 c ,! l ∞ !3 R !4 F (R, G ) !5 A∞ !6 A, where = l = R = F (R, G ) = A∞ = c ∞ Krzysztof Jarosz (SIUE) C (N [ f∞g) = convergent sequences C ( βN) = bounded sequences nonstandard real numbers formal power series over group G functions analytic at in…nity. Automatic Continuity 43 / 70 Proof of the Dales-EsterleTheorem Assume Fn is a weak * convergent sequence of distinct characters on B. Then Φ1 B3b7 ! (Fn (b )) 2 c is a surjective homomorphism. The next map Φ2 is a natural embedding: Φ2 c ,! l ∞ Krzysztof Jarosz (SIUE) Automatic Continuity 44 / 70 Proof of the Dales-EsterleTheorem Select p 2 βNnN and put Jp = ff 2 C ( βN) : f = 0 on a neighborhood of pg (a prime ideal), R = C ( βN) /Jp (integral domain); R = quotient …eld of C ( βN) /Jp , Φ 3 C ( βN) 3 f 7 ! f + Jp 2 R . R is a totally ordered real-closed η 1 …eld, where the order is de…ned by [f ] df [g ] () fn 2 N : f (n) and η 1 means if card (T1 ) , card (T2 ) then 9s g (n)g is a neighborhood of p, @0 and T1 < T2 2 R T1 < f s g < T2 . Theorem (assuming continuum hypothesis) Any two totally ordered real-closed η 1 …elds are isomorphic. Krzysztof Jarosz (SIUE) Automatic Continuity 45 / 70 Proof of the Dales-EsterleTheorem Construct an increasing system Gξ : ξ < v 1 of totally ordered divisible groups s.t. each Gξ has a countable co…nal and coinitial subset, Put (1 ) Gv 1 = [ Gξ : ξ < v 1 , For a given group G de…ne the formal power series algebra over G by n o F (R, G ) = f 2 RG : supp (f ) is well ordered with f ? g (t ) = ∑ ff (r ) g (s ) : r + s = t g , Prove that all F R, Gξ are in fact totally ordered, real-closed …elds s.t. each subset has a countable co…nal and coinitial subset, Krzysztof Jarosz (SIUE) Automatic Continuity 46 / 70 Proof of the Dales-EsterleTheorem (1 ) Prove that F R, Gv1 is a totally ordered, real-closed η 1 …eld of cardinality 2@0 = @1 , so it is isomorphic with R , Put Bv 1 = Bv0 1 = n n (1 ) x 2 F R, Gv1 (1 ) x 2 F R, Gv1 o 0 , o : inf supp x > 0 : inf supp x and prove that Bv0 1 is a maximal ideal in Bv1 , (1 ) Show that the isomorphism Φ4 from R into F R, Gv1 maps C ( βN) /Jp into Bv1 , Krzysztof Jarosz (SIUE) Automatic Continuity 47 / 70 Proof of the Dales-EsterleTheorem Construct Φ5 from Bv0 1 into A∞ = [ fHol (fz 2 C : Re z > 1, jz j > ng)g . Show that the sequence ∞ En ( z ) = n ∏ pz 1 exp p z/n j =1 behaves like a bounded approximate identity A∞ . Use the above to construct Φ6 from A∞ into a radical Banach algebra with an approximate identity. Krzysztof Jarosz (SIUE) Automatic Continuity 48 / 70 Translation invariant maps & norms - 3 questions F (G ) = Banach space of functions on a group G (Lp (G , µ) , M (G , µ) , C (G ) , C0 (G ) , C (G , E ) , Hol (G ) , D (G ) , ...) df fg ( t ) = f ( t + g ) . 1 Automatic continuity of translation invariant functionals ? F : F (G ) ! C, F (fg ) = F (f ) =) F is continuous; 2 Automatic continuity of translation invariant operators ? T : F (G ) ! F (G ) , T (fg ) = (T (f ))g =) T is continuous; 3 Uniqueness of translation invariant norms ? j j on F (G ) s.t. translations f 7 ! fg are continuous =) j j ' k k . Krzysztof Jarosz (SIUE) Automatic Continuity 49 / 70 Translation invariant functionals: compact groups Automatic continuity of translation invariant functionals have been studied since early seventies by G.H. Meisters, W. Hebisch, J. Krawczyk, R. Nillsen, R.J. Loy, P. Ludvik, G.S. Woodward, S. Saeki, and more recently by A.R. Villena, K. Jarosz, and others. Theorem (G.H. Meisters, 1971 & 1972) For a compact abelian group G with a …nite number of components translation invariant functionals on L2 (G ) are continuous. Theorem (G. Woodward, 1974) For a compact abelian group G there are translation invariant discontinuous functionals on C (G ) , L1 (G ) , and on L∞ (G ) . Theorem (J. Extremera, J.F. Mena, A.R. Villena, 2002) For a compact connected abelian group G all translation invariant functionals on Lp (G ) ; 1 < p < ∞ are continuous. Krzysztof Jarosz (SIUE) Automatic Continuity 50 / 70 Translation invariant functionals: non-compact groups Theorem (G. Woodward, 1974) For a large class of noncompact, abelian groups G there are discontinuous translation invariant functionals on C0 (G ) , C (G ) , Lp (G ) , 1 < p ∞. Theorem (S. Seaki, 1985) For a noncompact, abelian group G there are discontinuous translation invariant functionals on M (G ), L1 (G ), and on Lp (G ) for 1 < p ∞. Krzysztof Jarosz (SIUE) Automatic Continuity 51 / 70 Translation invariant functionals - the basic approach A linear map F : F (G ) ! C is translation invariant i¤ span ff fg : f 2 F ( G ) , g 2 G g ker F . So the question about the existence and uniqueness of translation invariant linear functionals is directly related to the question about the codimension of span ff fg : f 2 F (G ) , g 2 G g in F (G ). Krzysztof Jarosz (SIUE) Automatic Continuity 52 / 70 Translation invariant functionals versus norms & operators Theorem (B.E. Johnson, 1985; J. Alaminos, J. Extremera & A.R. Villena, 2006) Assume G is a locally compact abelian group and X = Lp (G ) , C0 (G ) .Then if G is compact then the translation invariant linear operators on X are continuous i¤ the translation invariant functionals on X are continuous, and i¤ X has a unique translation invariant norm, if G is not compact there are no translation invariant discontinuous operators on X and X has a unique translation invariant norm. For noncommutative case see: [J. Alaminos, J. Extremera & A.R. Villena; Studia Math. (173) 2006] Krzysztof Jarosz (SIUE) Automatic Continuity 53 / 70 General setting for group operations X , Y - Banach spaces; G - topological group τ X : G ! L (X ) , τ Y : G ! B (Y ) representations of G on X & Y Φ : X ! Y s.t. Φ τX = τY Φ When Φ is automatically continuous? See recent papers by: J. Alaminos, J. Extremera, J.F. Mena, E. Moreno, A.R. Villena. Krzysztof Jarosz (SIUE) Automatic Continuity 54 / 70 Uniqueness of norm Problem V - Banach space, T : V ! V continuous linear map. Can we introduce another inequivalent complete norm on V s.t. T is still continuous? T ? B (X , k k) \ B (X , j j) =) (X , k k) = (X , j j) . Problem Which multipliers M determine the complete norm topology? A C (K ) , g 2 C (K ) , Mg : A ! A, Mg (f ) = gf Theorem (A. R. Villena, 1997) A - closed subalgebra of C (K ) , g 2 A s.t. int g 1 (λ) = ∅ for all λ, then Mg determines the complete norm topology of A. Krzysztof Jarosz (SIUE) Automatic Continuity 55 / 70 Separating Space De…nition For T : X ! Y the separating space S (T ) = fy 2 Y : 9xi ! 0, Txi ! y g . Theorem (Basic properties) T is continuous i¤ S (T ) = f0g , S (T ) is a closed linear subspace (ideal), for continuous S, S (S (T )) = S (S T ), in particular ST is continuos i¤ S (T ) ker S, for continuous R, S (T Krzysztof Jarosz (SIUE) R) S (T ) . Automatic Continuity 56 / 70 Stability Lemma Lemma Assume ... X #T ... Y R !3 S !3 X #T Y R !2 S !2 X #T Y R !1 S !1 X #T Y where Rn , Sn are continuous. Then there is an n 2 N s.t. for any m S (T R1 R2 ... Rn ) = S (T R1 R2 n ... Rm ) . Equivalently Sn Krzysztof Jarosz (SIUE) ... S1 (S (T )) = Sm Automatic Continuity ... S1 (S (T )). 57 / 70 Proof of Villena’s Theorem Proof. Let A = C ([0, 1]) , g (t ) = t and put A = (C ([0, 1]) , k.k∞ ) , X = (C ([0, 1]) , k.k) , Id : X ! A. Assume f0 2 S (Id ) , f0 6= 0, put Sn Rn : A ! A, : X ! X, Sn = Mt Rn = Mt tn tn where t1 , t2 , ... distinct points in [0, 1] s.t. f0 (tn ) 6= 0. Since Sn Id = Id Rn from the Stability Lemma, there is an n, s.t. for m Sn ... S1 (S (Id )) = Sm ... S1 (S (Id )) however (t t1 ) ... (t Krzysztof Jarosz (SIUE) tn ) f0 2 Sn n ff : f (t1 ) = ... = f (tm ) = 0g ... S1 (S (Id )) is not equal to 0 at tn +1 . Automatic Continuity 58 / 70 Uniqueness of norm & multipliers on Banach algebras Theorem (KJ, 2001) Let A be a unital, semisimple, commutative Banach algebra. Then multiplication by a 2 A determines the complete norm topology of A i¤ for each scalar λ such that (a + λe ) is a divisor of zero, the codimension of (a + λe ) A is …nite. Theorem (KJ, 2001) Let A be a unital, semisimple, commutative Banach algebra such that the maximal ideal space of A does not have any isolated point. Then a 2 A determines the complete norm topology of A i¤ for each scalar λ, the element a + λe is not a divisor of zero. Theorem (KJ, 2001) For any separable Banach space X there is a bounded linear map T on X that determines a complete norm topology of X . Krzysztof Jarosz (SIUE) Automatic Continuity 59 / 70 Uniqueness of norm & multipliers on function spaces Theorem (KJ, 2001) A C0 (X ), and M a …nite set of multipliers on A. Then M determines the complete norm topology of A i¤ 8x 2 X ! span [ M (A) is …nite-codim or M 2Mx \ M 2Mx ker M = f0g . Corollary Ω Cn , open, connected, A - Banach space of analytic functions on Ω separating the points of Ω. Then any nonconstant multiplier determines the complete norm topology of A. Krzysztof Jarosz (SIUE) Automatic Continuity 60 / 70 Uniqueness of norm & multipliers on function spaces Theorem (KJ, 2001) B T S B (A), span ( M 2B M (A)) in…nite codim and does not contain M 2B ker M, then B does not determine the complete norm topology of A. Theorem (KJ, 2001) C0 (X ), and M a set of multipliers on A, then A 1 If 9x 2 X such that span [ M 2Mx M (A) ! is in…nite-codim and \ M 2Mx ker M 6= f0g then M does not determine the complete norm topology of A 2 T If 8x 2 X , M 2Mx ker M = f0g, or 9M1 , ..., Mn 2 Mx such that S span nj=1 Mj (A) is …nite codim, then M determines the complete norm topology of A. Krzysztof Jarosz (SIUE) Automatic Continuity 61 / 70 Uniqueness of norm: multipliers versus translations Assume F (G ) consists of integrable functions on G (or measures, or distributions). The Fourier transform is de…ned by b ^ : F (G ) ! C0 G Hence any g 2 G we have \ T g (f ) ( χ ) = = Z ZG G :b f (χ) = f (t + g ) χ f (t ) χ 1 (t 1 Z G f (t ) χ 1 (t ) dt. (t ) dt g ) dt = χ (g ) b f (χ) cg is a multiplication by so Tg is a translation on F (G ) by g i¤ T df gb (χ) = χ (g ) on F[ (G ). Krzysztof Jarosz (SIUE) Automatic Continuity 62 / 70 Proof (For any separable Banach space X there is a bounded linear map T determining the topology of X) X - separable Banach space, let (xn , xn ) be an M-bounded biorthogonal system for X : xn (xk ) = δnk , ∞ \ j =1 ker xj = f0g , kxk k = 1, 1 x (x ) xn for x 2 X . n n n =1 2 ∑ So T is continuous, σ (T ) = f0g [ Krzysztof Jarosz (SIUE) (T for all n, k 2 N, ∞ T (x ) = xk M, span fxj : j 2 Ng is dense in X . and De…ne T : X ! X by kxn k 1 2n λId ) (x ) = : n 2 N , and 1 2k Automatic Continuity λ xk (x ) . 63 / 70 Proof j j - another complete norm on X s.t. T is also continuous on (X , j j). We have 1 σB (X ,j j) (T ) = σB (X ,k k) (T ) = f0g [ :n2N . 2n Indeed if λ is not in the spectrum of T 2 B (X , j j), then the linear map T λId has an inverse, since T λId is also k k-continuous, by the Open Mapping Theorem, its inverse is k k-continuous as well, so λ is not in the spectrum of T 2 B (X , k k) . Let Id : (X , j j) ! (X , k k) and df Sn = T 1 Id 2 T 1 Id 22 ... T 1 Id 2n (S) . The maps (T λId ) are continuous in both norms and commute with Id, so, by the Stability Lemma, there is an N 2 N such that Sn = SN for n Krzysztof Jarosz (SIUE) Automatic Continuity N. 64 / 70 Proof Let y0 2 S (Id ). Assume there is an n0 > N such that xn0 (y0 ) 6= 0. Then xn0 T1/2 T1/22 T1/23 ...T1/2N (y0 ) 1 1 1 1 xn0 (y0 ) 6= 0, = ... n 1 n 0 0 2 2 2 2N but xn0 T1/2n0 = 0, so that xn0 T1/2 T1/22 ...T1/2N ...T1/2n0 (y ) = 0 for any y 2 X . Hence Sn0 is contained in the kernel of xn0 while SN is not; this is a contradiction since xn0 is k k-continuous. Hence S (Id ) Krzysztof Jarosz (SIUE) span xj : j Automatic Continuity N , 65 / 70 Proof Let f 2 Hol (σ (T )) s.t. f f Put o 1, ..., 2 N , and n o 1 on a neighborhood of σ (T ) n 1, ..., 2 N . 0 on a neighborhood of n N df P = f (T ) = Id ∑ xj ( ) xj , j =1 P is well de…ned and continuous with respect to both topologies, and it is T a linear projection onto N j =1 ker xj . Since S (Id ) span fxj : j Ng , P is continuous also as a map from (X , j j) into (X , k k). Hence PX is a closed, complemented, …nite-codimensional subspace of both (X , j j) and (X , k k), and S is continuousnon PX . Since o S is also continuous on the …nite-dimensional space span xj : j X. Krzysztof Jarosz (SIUE) N , the map S is continuous on Automatic Continuity 66 / 70 Another proof Theorem If B is a set of continuous linear maps on a Banach space A s.t. T S span ( M 2B M (A)) in…nite codim and does not contain M 2B ker M, then B does not determine the complete norm topology of A. T S Proof. a0 2 ( M 2B ker M ) nspan ( M 2B M (A)), A0 = span fa0 g , S B a linear complement of span (A0 [ M 2B M (A)) , F discontinuous linear functional on B. Then ! A = A0 M span [ M (A) M 2B M B. Put P : A ! A, P (a1 , a2 , a3 ) = (a1 + F (a3 ) a0 , 0, 0) , jaj = ka/A0 k + kPak , for a 2 A. Check that j j is a complete norm. Krzysztof Jarosz (SIUE) Automatic Continuity 67 / 70 Another proof To show that any M 2 B is continuous on (A, j j) …x a 2 A and let λ be a scalar such that ka/A 0 k = ka + λa0 k . By our assumption P M = 0 and Ma0 = 0 so jMaj = (Ma)/A0 + kPMak = (Ma)/A0 kMak = kM (a + λa0 )k kM k ka + λa0 k = kM k ka/A0 k kM k jaj , where kM k is the norm of M in (A, k k) . The above shows that the norm of M in (A, j j) is not greater than kM k. Krzysztof Jarosz (SIUE) Automatic Continuity 68 / 70 Questions Problem Assume X is a nonseparable Banach space. Does there exist a bounded linear map on X that determines a complete norm topology of that space? Problem Do multipliers determine the complete norm topology on Lp spaces? Problem Does the multiplication by f (t ) = t determine the complete norm topology of Lp [0, 1] ? Krzysztof Jarosz (SIUE) Automatic Continuity 69 / 70 THANK YOU Krzysztof Jarosz (SIUE) Automatic Continuity 70 / 70