Download Automatic Continuity from a personal perspective Krzysztof Jarosz www.siue.edu/~kjarosz

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