Large sublattices (subalgebras) of subsets of Banach lattices - - PowerPoint PPT Presentation

Large sublattices (subalgebras) of subsets of Banach lattices (algebras)

Transfinite Methods in Banach Spaces and Algebras of Operators, Banach Center, Bedlewo, July 2016

Main question

Question Suppose A is a “large” subset of a Banach lattice (resp. algebra)

• X. Does A ∪ {0} contain large (closed) sublatices (resp.

subalgebras)? Convention: All spaces, lattices, algebras etc. are infinite dimensional, unless specified otherwise. “Large” may mean that a sublattice (subalgebra) is: Infinite dimensional. Dense in A ∪ {0}. Has “many” generators, in the lattice (resp. algebraic) sense (not in the topological sense). If S is a minimal set of generators of Z, S′ is another set of generators, and S is infinite, then |S| |S′|.

Main question

Question Suppose A is a “large” subset of a Banach lattice (resp. algebra)

• X. Does A ∪ {0} contain large (closed) sublatices (resp.

subalgebras)? Convention: All spaces, lattices, algebras etc. are infinite dimensional, unless specified otherwise. “Large” may mean that a sublattice (subalgebra) is: Infinite dimensional. Dense in A ∪ {0}. Has “many” generators, in the lattice (resp. algebraic) sense (not in the topological sense). If S is a minimal set of generators of Z, S′ is another set of generators, and S is infinite, then |S| |S′|.

Main question

Question Suppose A is a “large” subset of a Banach lattice (resp. algebra)

• X. Does A ∪ {0} contain large (closed) sublatices (resp.

subalgebras)? Convention: All spaces, lattices, algebras etc. are infinite dimensional, unless specified otherwise. “Large” may mean that a sublattice (subalgebra) is: Infinite dimensional. Dense in A ∪ {0}. Has “many” generators, in the lattice (resp. algebraic) sense (not in the topological sense). If S is a minimal set of generators of Z, S′ is another set of generators, and S is infinite, then |S| |S′|.

Main question

Question Suppose A is a “large” subset of a Banach lattice (resp. algebra)

• X. Does A ∪ {0} contain large (closed) sublatices (resp.

subalgebras)? Convention: All spaces, lattices, algebras etc. are infinite dimensional, unless specified otherwise. “Large” may mean that a sublattice (subalgebra) is: Infinite dimensional. Dense in A ∪ {0}. Has “many” generators, in the lattice (resp. algebraic) sense (not in the topological sense). If S is a minimal set of generators of Z, S′ is another set of generators, and S is infinite, then |S| |S′|.

Main question

Question Suppose A is a “large” subset of a Banach lattice (resp. algebra)

• X. Does A ∪ {0} contain large (closed) sublatices (resp.

subalgebras)? Convention: All spaces, lattices, algebras etc. are infinite dimensional, unless specified otherwise. “Large” may mean that a sublattice (subalgebra) is: Infinite dimensional. Dense in A ∪ {0}. Has “many” generators, in the lattice (resp. algebraic) sense (not in the topological sense). If S is a minimal set of generators of Z, S′ is another set of generators, and S is infinite, then |S| |S′|.

Some history: Banach space case

Definition (Lineability and spaceability) For a Banach space X, A ⊂ X is: Lineable if A ∪ {0} contains a linear subspace. Spaceable if A ∪ {0} contains a closed linear subspace. Dense lineable if A ∪ {0} contains a linear subspace dense in A ∪ {0}.

Some history: Banach space case

Definition (Lineability and spaceability) For a Banach space X, A ⊂ X is: Lineable if A ∪ {0} contains a linear subspace. Spaceable if A ∪ {0} contains a closed linear subspace. Dense lineable if A ∪ {0} contains a linear subspace dense in A ∪ {0}.

Some history: Banach space case

Definition (Lineability and spaceability) For a Banach space X, A ⊂ X is: Lineable if A ∪ {0} contains a linear subspace. Spaceable if A ∪ {0} contains a closed linear subspace. Dense lineable if A ∪ {0} contains a linear subspace dense in A ∪ {0}.

Some history: Banach space case

• N. Kalton and A. Wilansky 1975: if A is a closed subspace of X,

with dim X/A = ∞, then X\A is spaceable (that is, X\A ∪ {0} contains a closed infinite dimensional subspace).

• L. Drewnowski 1984 (generalized by D. Kitson and R. Timoney

2011): if A is a non-closed operator range in X, then X\A is spaceable. Let ND[0, 1] be the space of nowhere differentiable functions in C[0, 1]. (i) V. Fonf, V. Gurarii, and M. Kadets 1966-1999: ND[0, 1] is spaceable. (ii) L. Bernal-Gonzalez 2008: ND[0, 1] is densely lineable (contains a dense subspace).

Some history: Banach space case

• N. Kalton and A. Wilansky 1975: if A is a closed subspace of X,

with dim X/A = ∞, then X\A is spaceable (that is, X\A ∪ {0} contains a closed infinite dimensional subspace).

• L. Drewnowski 1984 (generalized by D. Kitson and R. Timoney

2011): if A is a non-closed operator range in X, then X\A is spaceable. Let ND[0, 1] be the space of nowhere differentiable functions in C[0, 1]. (i) V. Fonf, V. Gurarii, and M. Kadets 1966-1999: ND[0, 1] is spaceable. (ii) L. Bernal-Gonzalez 2008: ND[0, 1] is densely lineable (contains a dense subspace).

Some history: Banach space case

• N. Kalton and A. Wilansky 1975: if A is a closed subspace of X,

with dim X/A = ∞, then X\A is spaceable (that is, X\A ∪ {0} contains a closed infinite dimensional subspace).

• L. Drewnowski 1984 (generalized by D. Kitson and R. Timoney

2011): if A is a non-closed operator range in X, then X\A is spaceable. Let ND[0, 1] be the space of nowhere differentiable functions in C[0, 1]. (i) V. Fonf, V. Gurarii, and M. Kadets 1966-1999: ND[0, 1] is spaceable. (ii) L. Bernal-Gonzalez 2008: ND[0, 1] is densely lineable (contains a dense subspace).

Banach lattices

Examples of Banach lattice: Lp(µ), C(K). Notation: X+ = {x ∈ X : x 0}. Definition (Latticeability) Suppose X is a Banach lattice. A subset A ⊂ X is (completely) latticeable if X contains a (closed) infinite dimensional sublattice Z so that Z ⊂ A ∪ {0}. Latticeability ∼ lineability Closed latticeability ∼ spaceability

Banach lattices

Examples of Banach lattice: Lp(µ), C(K). Notation: X+ = {x ∈ X : x 0}. Definition (Latticeability) Suppose X is a Banach lattice. A subset A ⊂ X is (completely) latticeable if X contains a (closed) infinite dimensional sublattice Z so that Z ⊂ A ∪ {0}. Latticeability ∼ lineability Closed latticeability ∼ spaceability

Banach lattices

Examples of Banach lattice: Lp(µ), C(K). Notation: X+ = {x ∈ X : x 0}. Definition (Latticeability) Suppose X is a Banach lattice. A subset A ⊂ X is (completely) latticeable if X contains a (closed) infinite dimensional sublattice Z so that Z ⊂ A ∪ {0}. Latticeability ∼ lineability Closed latticeability ∼ spaceability

Complements of closed subspaces

Suppose Y is a closed subspace of X. What kind of sublattices does (X\Y ) ∪ {0} contain? Theorem (a) If Y is a closed subspace of a Banach lattice X with dim X/Y n ∈ N, then ∃ an n-dimensional sublattice Z ⊂ X so that Z ∩ Y = {0}. (b) Consequently, if Y is a closed subspace of a Banach lattice X with dim X/Y = ∞, then ∀n ∈ N ∃ an n-dimensional sublattice Z ⊂ X s.t. Z ∩ Y = {0}. Question In (b), can Z be infinite dimensional?

Complements of closed subspaces

Suppose Y is a closed subspace of X. What kind of sublattices does (X\Y ) ∪ {0} contain? Theorem (a) If Y is a closed subspace of a Banach lattice X with dim X/Y n ∈ N, then ∃ an n-dimensional sublattice Z ⊂ X so that Z ∩ Y = {0}. (b) Consequently, if Y is a closed subspace of a Banach lattice X with dim X/Y = ∞, then ∀n ∈ N ∃ an n-dimensional sublattice Z ⊂ X s.t. Z ∩ Y = {0}. Question In (b), can Z be infinite dimensional?

Complements of closed subspaces

Suppose Y is a closed subspace of X. What kind of sublattices does (X\Y ) ∪ {0} contain? Theorem (a) If Y is a closed subspace of a Banach lattice X with dim X/Y n ∈ N, then ∃ an n-dimensional sublattice Z ⊂ X so that Z ∩ Y = {0}. (b) Consequently, if Y is a closed subspace of a Banach lattice X with dim X/Y = ∞, then ∀n ∈ N ∃ an n-dimensional sublattice Z ⊂ X s.t. Z ∩ Y = {0}. Question In (b), can Z be infinite dimensional?

Complements of closed subspaces

Suppose Y is a closed subspace of X. What kind of sublattices does (X\Y ) ∪ {0} contain? Theorem (a) If Y is a closed subspace of a Banach lattice X with dim X/Y n ∈ N, then ∃ an n-dimensional sublattice Z ⊂ X so that Z ∩ Y = {0}. (b) Consequently, if Y is a closed subspace of a Banach lattice X with dim X/Y = ∞, then ∀n ∈ N ∃ an n-dimensional sublattice Z ⊂ X s.t. Z ∩ Y = {0}. Question In (b), can Z be infinite dimensional?

Complements of closed ideals

An subspace Y of a Banach lattice X is called an ideal if, for any y ∈ Y , and any x ∈ X satisfying |x| |y|, we have x ∈ Y . Theorem Suppose Y is a closed ideal in X, with dim X/Y = ∞. Then X+ contains disjoint non-zero elements (xi)i∈N so that Y ∩ Z = {0}, where Z = span[(xi)i∈N]. In particular, X\Y is completely latticeable.

Complements of closed ideals

An subspace Y of a Banach lattice X is called an ideal if, for any y ∈ Y , and any x ∈ X satisfying |x| |y|, we have x ∈ Y . Theorem Suppose Y is a closed ideal in X, with dim X/Y = ∞. Then X+ contains disjoint non-zero elements (xi)i∈N so that Y ∩ Z = {0}, where Z = span[(xi)i∈N]. In particular, X\Y is completely latticeable.

Complements of closed subspaces

Theorem Suppose Y is a fin. dim. subspace of a Banach lattice X. Then X+ contains disjoint non-zero elements (xi)i∈N so that Y ∩ Z = {0}, where Z is the closed ideal generated (xi)i∈N: Z =

• z ∈ X : |z| |x| for some x ∈ span[(xi)i∈N]
• Proposition

C[0, 1] contains a closed inf. codim. subspace Y so that (C[0, 1]\Y ) ∪ {0} contains no non-trivial ideals.

Complements of closed subspaces

Theorem Suppose Y is a fin. dim. subspace of a Banach lattice X. Then X+ contains disjoint non-zero elements (xi)i∈N so that Y ∩ Z = {0}, where Z is the closed ideal generated (xi)i∈N: Z =

• z ∈ X : |z| |x| for some x ∈ span[(xi)i∈N]
• Proposition

C[0, 1] contains a closed inf. codim. subspace Y so that (C[0, 1]\Y ) ∪ {0} contains no non-trivial ideals.

Complements of closed subspaces: atomic lattices

Suppose X is a sequence space – that is, the order structure is determined by a 1-unconditional basis (σi)i∈N. Theorem Suppose Y is a closed subspace of X, with dim X/Y = ∞. Then there exist k1 < k2 < . . . so that Y ∩ span

• (σki)i∈N
• = {0}.

In particular, X\Y is completely latticeable.

Complements of closed subspaces: atomic lattices

Suppose X is a sequence space – that is, the order structure is determined by a 1-unconditional basis (σi)i∈N. Theorem Suppose Y is a closed subspace of X, with dim X/Y = ∞. Then there exist k1 < k2 < . . . so that Y ∩ span

• (σki)i∈N
• = {0}.

In particular, X\Y is completely latticeable.

Complements of closed subspaces: atomic lattices

Proposition Suppose X is either ℓp (1 < p < ∞) or c0, and Y is a closed subspace of X, with dim X/Y = ∞. Then X\Y is completely

• latticeable. Moreover, there exists a closed sublattice Z ⊂ X and a

constant c so that z + y cz for any z ∈ Z and y ∈ Y . Remark Proposition fails for X = ℓ1.

Complements of closed subspaces: atomic lattices

Proposition Suppose X is either ℓp (1 < p < ∞) or c0, and Y is a closed subspace of X, with dim X/Y = ∞. Then X\Y is completely

• latticeable. Moreover, there exists a closed sublattice Z ⊂ X and a

constant c so that z + y cz for any z ∈ Z and y ∈ Y . Remark Proposition fails for X = ℓ1.

Definition of K¨

• the function spaces

Definition (K¨

• the function space)

Suppose (Ω, µ) is a non-atomic measure space. A Banach space X = X(Ω, µ) of (equivalence classes of) locally integrable functions (modulo a.e. equality) is a K¨

• the function space if:

1 If S ⊂ Ω has finite measure, then χS ∈ X. 2 If f ∈ X, g is measurable, and |g| |f | a.e., then g ∈ X, and

g f . Examples: Lp(µ) (1 p ∞), but not C[0, 1].

Definition of K¨

• the function spaces

Definition (K¨

• the function space)

Suppose (Ω, µ) is a non-atomic measure space. A Banach space X = X(Ω, µ) of (equivalence classes of) locally integrable functions (modulo a.e. equality) is a K¨

• the function space if:

1 If S ⊂ Ω has finite measure, then χS ∈ X. 2 If f ∈ X, g is measurable, and |g| |f | a.e., then g ∈ X, and

g f . Examples: Lp(µ) (1 p ∞), but not C[0, 1].

Definition of K¨

• the function spaces

Definition (K¨

• the function space)

Suppose (Ω, µ) is a non-atomic measure space. A Banach space X = X(Ω, µ) of (equivalence classes of) locally integrable functions (modulo a.e. equality) is a K¨

• the function space if:

1 If S ⊂ Ω has finite measure, then χS ∈ X. 2 If f ∈ X, g is measurable, and |g| |f | a.e., then g ∈ X, and

g f . Examples: Lp(µ) (1 p ∞), but not C[0, 1].

Complements of closed subspaces: K¨

• the function spaces

Definition (Equiintegrability) Z ⊂ X is X-equiintegrable if ∀ ε > 0 ∃ δ > 0 s.t. z1AX < ε if z ∈ Z and µ(A) < δ. Notation: B(Y ) = {y ∈ Y : y 1}. Proposition Suppose X(0, 1) is a reflexive K¨

• the function space on (0, 1), and

a subspace G ⊂ X ∗ is such that B(G) is not X ∗-equiintegrable. Then X\G ⊥ is completely latticeable. In fact, one can can find pairwise disjoint f1, f2, . . . ∈ X+ so that span

• (fi)i∈N
• ∩ G ⊥ = {0}.

Complements of closed subspaces: K¨

• the function spaces

Definition (Equiintegrability) Z ⊂ X is X-equiintegrable if ∀ ε > 0 ∃ δ > 0 s.t. z1AX < ε if z ∈ Z and µ(A) < δ. Notation: B(Y ) = {y ∈ Y : y 1}. Proposition Suppose X(0, 1) is a reflexive K¨

• the function space on (0, 1), and

a subspace G ⊂ X ∗ is such that B(G) is not X ∗-equiintegrable. Then X\G ⊥ is completely latticeable. In fact, one can can find pairwise disjoint f1, f2, . . . ∈ X+ so that span

• (fi)i∈N
• ∩ G ⊥ = {0}.

Complements of closed subspaces: K¨

• the function spaces

Definition (Equiintegrability) Z ⊂ X is X-equiintegrable if ∀ ε > 0 ∃ δ > 0 s.t. z1AX < ε if z ∈ Z and µ(A) < δ. Notation: B(Y ) = {y ∈ Y : y 1}. Proposition Suppose X(0, 1) is a reflexive K¨

• the function space on (0, 1), and

a subspace G ⊂ X ∗ is such that B(G) is not X ∗-equiintegrable. Then X\G ⊥ is completely latticeable. In fact, one can can find pairwise disjoint f1, f2, . . . ∈ X+ so that span

• (fi)i∈N
• ∩ G ⊥ = {0}.

Complements of closed subspaces: K¨

• the function spaces

Definition (Equiintegrability) Z ⊂ X is X-equiintegrable if ∀ ε > 0 ∃ δ > 0 s.t. z1AX < ε if z ∈ Z and µ(A) < δ. Notation: B(Y ) = {y ∈ Y : y 1}. Proposition Suppose X(0, 1) is a reflexive K¨

• the function space on (0, 1), and

a subspace G ⊂ X ∗ is such that B(G) is not X ∗-equiintegrable. Then X\G ⊥ is completely latticeable. In fact, one can can find pairwise disjoint f1, f2, . . . ∈ X+ so that span

• (fi)i∈N
• ∩ G ⊥ = {0}.

Complements of closed subspaces: K¨

• the function spaces

Corollary Suppose 1 < p < 2, 1/p + 1/q = 1, and the subspace G ⊂ Lq contains an isomorphic copy of ℓq. Then Lp\G ⊥ is completely latticeable. For Λ ⊂ Z, set LΛ

p(T) = {f ∈ Lp(T) : ∀n /

∈ Λ, f (n) = 0}. Λ ⊂ Z contains arbitrarily long arithmetic sequences if if for any k ∈ N there exist a, n ∈ Z so that a, a + n, . . . , a + kn ∈ Λ. Corollary Suppose Λ ⊂ Z is such that Λc contains arbitrarily long arithmetic

• sequences. For 1 < p < ∞, Lp(T)\LΛ

p(T) is completely latticeable.

Complements of closed subspaces: K¨

• the function spaces

Corollary Suppose 1 < p < 2, 1/p + 1/q = 1, and the subspace G ⊂ Lq contains an isomorphic copy of ℓq. Then Lp\G ⊥ is completely latticeable. For Λ ⊂ Z, set LΛ

p(T) = {f ∈ Lp(T) : ∀n /

∈ Λ, f (n) = 0}. Λ ⊂ Z contains arbitrarily long arithmetic sequences if if for any k ∈ N there exist a, n ∈ Z so that a, a + n, . . . , a + kn ∈ Λ. Corollary Suppose Λ ⊂ Z is such that Λc contains arbitrarily long arithmetic

• sequences. For 1 < p < ∞, Lp(T)\LΛ

p(T) is completely latticeable.

Complements of closed subspaces: K¨

• the function spaces

Corollary Suppose 1 < p < 2, 1/p + 1/q = 1, and the subspace G ⊂ Lq contains an isomorphic copy of ℓq. Then Lp\G ⊥ is completely latticeable. For Λ ⊂ Z, set LΛ

p(T) = {f ∈ Lp(T) : ∀n /

∈ Λ, f (n) = 0}. Λ ⊂ Z contains arbitrarily long arithmetic sequences if if for any k ∈ N there exist a, n ∈ Z so that a, a + n, . . . , a + kn ∈ Λ. Corollary Suppose Λ ⊂ Z is such that Λc contains arbitrarily long arithmetic

• sequences. For 1 < p < ∞, Lp(T)\LΛ

p(T) is completely latticeable.

Complements of closed subspaces: K¨

• the function spaces

Corollary Suppose 1 < p < 2, 1/p + 1/q = 1, and the subspace G ⊂ Lq contains an isomorphic copy of ℓq. Then Lp\G ⊥ is completely latticeable. For Λ ⊂ Z, set LΛ

p(T) = {f ∈ Lp(T) : ∀n /

∈ Λ, f (n) = 0}. Λ ⊂ Z contains arbitrarily long arithmetic sequences if if for any k ∈ N there exist a, n ∈ Z so that a, a + n, . . . , a + kn ∈ Λ. Corollary Suppose Λ ⊂ Z is such that Λc contains arbitrarily long arithmetic

• sequences. For 1 < p < ∞, Lp(T)\LΛ

p(T) is completely latticeable.

Complements of closed subspaces: Lp spaces

Proposition If X = X(0, 1) is a K¨

• the function space, and a subspace Y ⊂ X is

such that B(Y ) is X-equiintegrable, then X\Y is completely latticeable. Corollary If Y is a reflexive subspace of L1(µ), then L1(µ)\Y is completely latticeable.

Complements of closed subspaces: Lp spaces

Proposition If X = X(0, 1) is a K¨

• the function space, and a subspace Y ⊂ X is

such that B(Y ) is X-equiintegrable, then X\Y is completely latticeable. Corollary If Y is a reflexive subspace of L1(µ), then L1(µ)\Y is completely latticeable.

Complements of dense subspaces

Theorem For 0 < p ∞, there exists a vector lattice Z ⊂ ℓp\(∪q<pℓq) ∪ {0} (or Z ⊂ c0\(∪q<∞ℓq) ∪ {0} if p = ∞) so that:

1 Z = ℓp (c0 if p = ∞). 2 If a set S generates Z as a vector lattice, then |S| 2ℵ0.

Algebrability: known results

f ∈ C[0, 1] is nowhere H¨

• lder if, for any x ∈ [0, 1], and for any

α > 0, we have sup

y∈[0,1]

|f (x) − f (y)| |x − y|α = ∞.

• F. Bayart and L. Quarta 2007: Let Y ⊂ C[0, 1] be the set of

nowhere H¨

• lder functions. Then Y contains a dense subalgebra

with infinitely many generators.

Algebrability: known results

f ∈ C[0, 1] is nowhere H¨

• lder if, for any x ∈ [0, 1], and for any

α > 0, we have sup

y∈[0,1]

|f (x) − f (y)| |x − y|α = ∞.

• F. Bayart and L. Quarta 2007: Let Y ⊂ C[0, 1] be the set of

nowhere H¨

• lder functions. Then Y contains a dense subalgebra

with infinitely many generators.

Subalgebras of complements of dense subspaces

Theorem Each of the following sets contains a dense subalgebra whose minimal set of generators has the cardinality of continuum. c0\(∪q<∞ℓq) ∪ {0} (pointwise multiplication). C0(R)\(∪p<∞Lp(R)) ∪ {0} (pointwise multiplication). L1(T)\(∪p>1Lp(T)) ∪ {0} (convolution). S∞\

• ∪p<∞ Sp
• , where Sp is the Schatten p-space on ℓ2, and

S∞ is the space of compact operators on ℓ2.

Subalgebras of complements of dense subspaces

Theorem Each of the following sets contains a dense subalgebra whose minimal set of generators has the cardinality of continuum. c0\(∪q<∞ℓq) ∪ {0} (pointwise multiplication). C0(R)\(∪p<∞Lp(R)) ∪ {0} (pointwise multiplication). L1(T)\(∪p>1Lp(T)) ∪ {0} (convolution). S∞\

• ∪p<∞ Sp
• , where Sp is the Schatten p-space on ℓ2, and

S∞ is the space of compact operators on ℓ2.

Subalgebras of complements of dense subspaces

Theorem Each of the following sets contains a dense subalgebra whose minimal set of generators has the cardinality of continuum. c0\(∪q<∞ℓq) ∪ {0} (pointwise multiplication). C0(R)\(∪p<∞Lp(R)) ∪ {0} (pointwise multiplication). L1(T)\(∪p>1Lp(T)) ∪ {0} (convolution). S∞\

• ∪p<∞ Sp
• , where Sp is the Schatten p-space on ℓ2, and

S∞ is the space of compact operators on ℓ2.

Subalgebras of complements of dense subspaces

Theorem Each of the following sets contains a dense subalgebra whose minimal set of generators has the cardinality of continuum. c0\(∪q<∞ℓq) ∪ {0} (pointwise multiplication). C0(R)\(∪p<∞Lp(R)) ∪ {0} (pointwise multiplication). L1(T)\(∪p>1Lp(T)) ∪ {0} (convolution). S∞\

• ∪p<∞ Sp
• , where Sp is the Schatten p-space on ℓ2, and

S∞ is the space of compact operators on ℓ2.

Back to spaceability: non-regular operators

T = (tij)n

i,j=1 ∈ B(ℓp) (1 < p < ∞) is regular if |T| = (|tij|)n i,j=1 is

a bounded operator. Notation: K(X) = {T ∈ B(X) : T is compact}. Kr(X) = {T ∈ K(X) : T is regular}. Theorem For 1 < p < ∞, K(ℓp)\Kr(ℓp) is: Spaceable (contains a closed inf. dim. subspace). Densely lineable (contains a dense subspace).

Back to spaceability: non-regular operators

T = (tij)n

i,j=1 ∈ B(ℓp) (1 < p < ∞) is regular if |T| = (|tij|)n i,j=1 is

a bounded operator. Notation: K(X) = {T ∈ B(X) : T is compact}. Kr(X) = {T ∈ K(X) : T is regular}. Theorem For 1 < p < ∞, K(ℓp)\Kr(ℓp) is: Spaceable (contains a closed inf. dim. subspace). Densely lineable (contains a dense subspace).

Back to spaceability: non-regular operators

T = (tij)n

i,j=1 ∈ B(ℓp) (1 < p < ∞) is regular if |T| = (|tij|)n i,j=1 is

a bounded operator. Notation: K(X) = {T ∈ B(X) : T is compact}. Kr(X) = {T ∈ K(X) : T is regular}. Theorem For 1 < p < ∞, K(ℓp)\Kr(ℓp) is: Spaceable (contains a closed inf. dim. subspace). Densely lineable (contains a dense subspace).

A proof

Theorem (to be proved) If Y is a closed subspace of a Banach lattice X with dim X/Y n ∈ N, then there exists an n-dimensional sublattice Z ⊂ X so that Z ∩ Y = {0}. Definition A Banach lattice X is Dedekind (or order) complete if any subset

• f X, which has an upper bound, has a supremum.

Examples of Dedekind complete lattices: Lp (1 p ∞), dual Banach lattices. C[0, 1] is not Dedekind complete.

A proof

Theorem (to be proved) If Y is a closed subspace of a Banach lattice X with dim X/Y n ∈ N, then there exists an n-dimensional sublattice Z ⊂ X so that Z ∩ Y = {0}. Definition A Banach lattice X is Dedekind (or order) complete if any subset

• f X, which has an upper bound, has a supremum.

Examples of Dedekind complete lattices: Lp (1 p ∞), dual Banach lattices. C[0, 1] is not Dedekind complete.

A proof

Theorem (to be proved) If Y is a closed subspace of a Banach lattice X with dim X/Y n ∈ N, then there exists an n-dimensional sublattice Z ⊂ X so that Z ∩ Y = {0}. Definition A Banach lattice X is Dedekind (or order) complete if any subset

• f X, which has an upper bound, has a supremum.

Examples of Dedekind complete lattices: Lp (1 p ∞), dual Banach lattices. C[0, 1] is not Dedekind complete.

A proof

Theorem (to be proved) If Y is a closed subspace of a Banach lattice X with dim X/Y n ∈ N, then there exists an n-dimensional sublattice Z ⊂ X so that Z ∩ Y = {0}. Definition A Banach lattice X is Dedekind (or order) complete if any subset

• f X, which has an upper bound, has a supremum.

Examples of Dedekind complete lattices: Lp (1 p ∞), dual Banach lattices. C[0, 1] is not Dedekind complete.

Proof: complements of subspaces

Theorem (to be proved) If Y is a closed subspace of a Banach lattice X with dim X/Y n ∈ N, then there exists an n-dimensional sublattice Z ⊂ X so that Z ∩ Y = {0}. Proof when X is Dedekind complete. (1) If G is a subspace of a fin. dim. Banach lattice F, then ∃ a sublattice Z ⊂ F, s.t. Z ∩ G = {0}, dim Z + dim G = dim F. Proof: identify F with Fm (m = dim F), use linear algebra.

Proof: complements of subspaces

Theorem (to be proved) If Y is a closed subspace of a Banach lattice X with dim X/Y n ∈ N, then there exists an n-dimensional sublattice Z ⊂ X so that Z ∩ Y = {0}. Proof when X is Dedekind complete. (1) If G is a subspace of a fin. dim. Banach lattice F, then ∃ a sublattice Z ⊂ F, s.t. Z ∩ G = {0}, dim Z + dim G = dim F. Proof: identify F with Fm (m = dim F), use linear algebra.

Proof: complements of subspaces

Theorem (to be proved) If Y is a closed subspace of a Banach lattice X with dim X/Y n ∈ N, then there exists an n-dimensional sublattice Z ⊂ X so that Z ∩ Y = {0}. Proof when X is Dedekind complete. (1) If G is a subspace of a fin. dim. Banach lattice F, then ∃ a sublattice Z ⊂ F, s.t. Z ∩ G = {0}, dim Z + dim G = dim F. Proof: identify F with Fm (m = dim F), use linear algebra.

Proof: complements of subspaces

Theorem (to be proved) If Y is a closed subspace of a Banach lattice X with dim X/Y n ∈ N, then there exists an n-dimensional sublattice Z ⊂ X so that Z ∩ Y = {0}. Proof when X is Dedekind complete. (2) Fact: if E is a fin. dim. subspace of a Dedekind complete Banach lattice X, and ε > 0, then ∃ a fin. dim. sublattice F ⊂ X and an automorphism T : X → X s.t. TE ⊂ F, I − T < ε. Find a subspace E ⊂ X s.t. dim E = n, E ∩ Y = {0}. Use (2) to find a fin. dim. sublattice F ⊂ X and T ∈ B(X) s.t. T(E) ∩ Y = {0}. Then dim F/G n, where G = Y ∩ F. Use (1) to find a sublattice Z ⊂ F s.t. dim Z = n, Z ∩ G = {0}. Then Z ∩ Y = {0}.

Proof: complements of subspaces

Theorem (to be proved) If Y is a closed subspace of a Banach lattice X with dim X/Y n ∈ N, then there exists an n-dimensional sublattice Z ⊂ X so that Z ∩ Y = {0}. Proof when X is Dedekind complete. (2) Fact: if E is a fin. dim. subspace of a Dedekind complete Banach lattice X, and ε > 0, then ∃ a fin. dim. sublattice F ⊂ X and an automorphism T : X → X s.t. TE ⊂ F, I − T < ε. Find a subspace E ⊂ X s.t. dim E = n, E ∩ Y = {0}. Use (2) to find a fin. dim. sublattice F ⊂ X and T ∈ B(X) s.t. T(E) ∩ Y = {0}. Then dim F/G n, where G = Y ∩ F. Use (1) to find a sublattice Z ⊂ F s.t. dim Z = n, Z ∩ G = {0}. Then Z ∩ Y = {0}.

Proof: complements of subspaces

Theorem (to be proved) If Y is a closed subspace of a Banach lattice X with dim X/Y n ∈ N, then there exists an n-dimensional sublattice Z ⊂ X so that Z ∩ Y = {0}. Proof when X is Dedekind complete. (2) Fact: if E is a fin. dim. subspace of a Dedekind complete Banach lattice X, and ε > 0, then ∃ a fin. dim. sublattice F ⊂ X and an automorphism T : X → X s.t. TE ⊂ F, I − T < ε. Find a subspace E ⊂ X s.t. dim E = n, E ∩ Y = {0}. Use (2) to find a fin. dim. sublattice F ⊂ X and T ∈ B(X) s.t. T(E) ∩ Y = {0}. Then dim F/G n, where G = Y ∩ F. Use (1) to find a sublattice Z ⊂ F s.t. dim Z = n, Z ∩ G = {0}. Then Z ∩ Y = {0}.

Proof: complements of subspaces

Theorem (to be proved) If Y is a closed subspace of a Banach lattice X with dim X/Y n ∈ N, then there exists an n-dimensional sublattice Z ⊂ X so that Z ∩ Y = {0}. Proof for general X. X ∗∗ is Dedekind complete, hence ∃ n-dimensional sublattice W ⊂ X ∗∗ s.t. W ∩ Y ⊥⊥ = {0}. Find c ∈ (0, 1/9) s.t. dist(w, Y ⊥⊥) 3cw ∀ w ∈ W . Find x∗

1, . . . , x∗ N ∈ B(Y ⊥) ⊂ X ∗ s.t.

max

1iN

• x∗

i , w

• 2cw ∀ w ∈ W .

Let V =

• x∗∗ ∈ X ∗∗ : max1iN
• x∗

i , w

• < c
• .

Local reflexivity: ∃ lattice homomorphism T : W → Z ⊂ X s.t. T, T −1 < 1 + ε, and (I − T)B(W ) ⊂ εV . ⇒ Z ∩ Y = {0}.

Proof: complements of subspaces

Theorem (to be proved) If Y is a closed subspace of a Banach lattice X with dim X/Y n ∈ N, then there exists an n-dimensional sublattice Z ⊂ X so that Z ∩ Y = {0}. Proof for general X. X ∗∗ is Dedekind complete, hence ∃ n-dimensional sublattice W ⊂ X ∗∗ s.t. W ∩ Y ⊥⊥ = {0}. Find c ∈ (0, 1/9) s.t. dist(w, Y ⊥⊥) 3cw ∀ w ∈ W . Find x∗

1, . . . , x∗ N ∈ B(Y ⊥) ⊂ X ∗ s.t.

max

1iN

• x∗

i , w

• 2cw ∀ w ∈ W .

Let V =

• x∗∗ ∈ X ∗∗ : max1iN
• x∗

i , w

• < c
• .

Local reflexivity: ∃ lattice homomorphism T : W → Z ⊂ X s.t. T, T −1 < 1 + ε, and (I − T)B(W ) ⊂ εV . ⇒ Z ∩ Y = {0}.

Proof: complements of subspaces

Theorem (to be proved) If Y is a closed subspace of a Banach lattice X with dim X/Y n ∈ N, then there exists an n-dimensional sublattice Z ⊂ X so that Z ∩ Y = {0}. Proof for general X. X ∗∗ is Dedekind complete, hence ∃ n-dimensional sublattice W ⊂ X ∗∗ s.t. W ∩ Y ⊥⊥ = {0}. Find c ∈ (0, 1/9) s.t. dist(w, Y ⊥⊥) 3cw ∀ w ∈ W . Find x∗

1, . . . , x∗ N ∈ B(Y ⊥) ⊂ X ∗ s.t.

max

1iN

• x∗

i , w

• 2cw ∀ w ∈ W .

Let V =

• x∗∗ ∈ X ∗∗ : max1iN
• x∗

i , w

• < c
• .

Local reflexivity: ∃ lattice homomorphism T : W → Z ⊂ X s.t. T, T −1 < 1 + ε, and (I − T)B(W ) ⊂ εV . ⇒ Z ∩ Y = {0}.

Proof: complements of subspaces

Theorem (to be proved) If Y is a closed subspace of a Banach lattice X with dim X/Y n ∈ N, then there exists an n-dimensional sublattice Z ⊂ X so that Z ∩ Y = {0}. Proof for general X. X ∗∗ is Dedekind complete, hence ∃ n-dimensional sublattice W ⊂ X ∗∗ s.t. W ∩ Y ⊥⊥ = {0}. Find c ∈ (0, 1/9) s.t. dist(w, Y ⊥⊥) 3cw ∀ w ∈ W . Find x∗

1, . . . , x∗ N ∈ B(Y ⊥) ⊂ X ∗ s.t.

max

1iN

• x∗

i , w

• 2cw ∀ w ∈ W .

Let V =

• x∗∗ ∈ X ∗∗ : max1iN
• x∗

i , w

• < c
• .

Local reflexivity: ∃ lattice homomorphism T : W → Z ⊂ X s.t. T, T −1 < 1 + ε, and (I − T)B(W ) ⊂ εV . ⇒ Z ∩ Y = {0}.

Proof: c0\(∪q<∞ℓq) is densely latticeable

Theorem There exists a vector lattice Z ⊂ c0\(∪q<∞ℓq) ∪ {0} s.t.:

1 Z = c0. 2 If a set S generates Z as a vector lattice, then |S| 2ℵ0.

In fact, Z has a set of free generators, with cardinality 2ℵ0.

• 1. Combinatorial lemma. ∃ set I, |I| = 2ℵ0, and functions

fi : N → N (i ∈ I) s.t. ∀n, ∀k1, . . . , kn ∈ N, ∀ distinct i1, . . . , in ∈ I, ∃m ∈ N s.t. fij(m) = kj (1 j n).

Proof: c0\(∪q<∞ℓq) is densely latticeable

Theorem There exists a vector lattice Z ⊂ c0\(∪q<∞ℓq) ∪ {0} s.t.:

1 Z = c0. 2 If a set S generates Z as a vector lattice, then |S| 2ℵ0.

In fact, Z has a set of free generators, with cardinality 2ℵ0.

• 1. Combinatorial lemma. ∃ set I, |I| = 2ℵ0, and functions

fi : N → N (i ∈ I) s.t. ∀n, ∀k1, . . . , kn ∈ N, ∀ distinct i1, . . . , in ∈ I, ∃m ∈ N s.t. fij(m) = kj (1 j n).

Proof: c0\(∪q<∞ℓq) is densely latticeable

Theorem There exists a vector lattice Z ⊂ c0\(∪q<∞ℓq) ∪ {0} s.t.:

1 Z = c0. 2 If a set S generates Z as a vector lattice, then |S| 2ℵ0.

In fact, Z has a set of free generators, with cardinality 2ℵ0.

• 1. Combinatorial lemma. ∃ set I, |I| = 2ℵ0, and functions

fi : N → N (i ∈ I) s.t. ∀n, ∀k1, . . . , kn ∈ N, ∀ distinct i1, . . . , in ∈ I, ∃m ∈ N s.t. fij(m) = kj (1 j n).

Proof: c0(N × N)\(∪q<∞ℓq(N × N))

• 2. Construction.

Let Sn be ℓ∞({1, . . . , n} × {1, . . . , n}) ⊂ c0(N × N). Let S = ∪n>10Sn. By Lemma, ∃ functions fu : N → K = Q ∩ (−1, 1) (u ∈ S) s.t. ∀n, ∀k1, . . . , kn ∈ K, ∀ distinct u1, . . . , un ∈ S, ∃m ∈ N with fuj(m) = kj for 1 j n. Define Φ : S → c0(N × N): for u = (uij)1i,jn ∈ Sn set

• Φ(u)
• ij =

   uij 1 i, j n fu(i)/ ln j j > n i > n, j n . Let Z be the vector lattice generated by Φ(S).

Proof: c0(N × N)\(∪q<∞ℓq(N × N))

• 2. Construction.

Let Sn be ℓ∞({1, . . . , n} × {1, . . . , n}) ⊂ c0(N × N). Let S = ∪n>10Sn. By Lemma, ∃ functions fu : N → K = Q ∩ (−1, 1) (u ∈ S) s.t. ∀n, ∀k1, . . . , kn ∈ K, ∀ distinct u1, . . . , un ∈ S, ∃m ∈ N with fuj(m) = kj for 1 j n. Define Φ : S → c0(N × N): for u = (uij)1i,jn ∈ Sn set

• Φ(u)
• ij =

   uij 1 i, j n fu(i)/ ln j j > n i > n, j n . Let Z be the vector lattice generated by Φ(S).

Proof: c0(N × N)\(∪q<∞ℓq(N × N))

• 2. Construction.

Let Sn be ℓ∞({1, . . . , n} × {1, . . . , n}) ⊂ c0(N × N). Let S = ∪n>10Sn. By Lemma, ∃ functions fu : N → K = Q ∩ (−1, 1) (u ∈ S) s.t. ∀n, ∀k1, . . . , kn ∈ K, ∀ distinct u1, . . . , un ∈ S, ∃m ∈ N with fuj(m) = kj for 1 j n. Define Φ : S → c0(N × N): for u = (uij)1i,jn ∈ Sn set

• Φ(u)
• ij =

   uij 1 i, j n fu(i)/ ln j j > n i > n, j n . Let Z be the vector lattice generated by Φ(S).

Proof: c0(N × N)\(∪q<∞ℓq(N × N))

• 3. Density. Z is dense in c0: for u ∈ S, u − Φ(u) 1/ ln(n + 1).
• 4. Z ∩ (∪qℓq) = {0}; (Φ(u))u∈S is a countable set of free

generators of Z. Elements of Z are of the form Ψ(Φ(u1), . . . , Φ(un)), where u1, . . . , un are distinct, and Ψ : Rn → R is a finite composition of +, −, c·, ∨, ∧. Need: either Ψ(Φ(u1), . . . , Φ(un)) = 0, or Ψ(Φ(u1), . . . , Φ(un)) / ∈ ∪q<∞ℓq. Ψ ≡ 0 ⇒ Ψ(Φ(u1), . . . , Φ(un)) = 0. Otherwise ∃ k1, . . . , kn ∈ K s.t. c := Ψ(k1, . . . , kn) = 0. Find m s.t. fuℓ(m) = kℓ for 1 ℓ m. For j large enough

• Ψ(Φ(u1), . . . , Φ(un))
• mj = (ln j)−1Ψ
• fu1(m), . . . , fun(m)
• = c

ln j . Thus Ψ(Φ(u1), . . . , Φ(un)) / ∈ ∪q<∞ℓq.

Proof: c0(N × N)\(∪q<∞ℓq(N × N))

• 3. Density. Z is dense in c0: for u ∈ S, u − Φ(u) 1/ ln(n + 1).
• 4. Z ∩ (∪qℓq) = {0}; (Φ(u))u∈S is a countable set of free

generators of Z. Elements of Z are of the form Ψ(Φ(u1), . . . , Φ(un)), where u1, . . . , un are distinct, and Ψ : Rn → R is a finite composition of +, −, c·, ∨, ∧. Need: either Ψ(Φ(u1), . . . , Φ(un)) = 0, or Ψ(Φ(u1), . . . , Φ(un)) / ∈ ∪q<∞ℓq. Ψ ≡ 0 ⇒ Ψ(Φ(u1), . . . , Φ(un)) = 0. Otherwise ∃ k1, . . . , kn ∈ K s.t. c := Ψ(k1, . . . , kn) = 0. Find m s.t. fuℓ(m) = kℓ for 1 ℓ m. For j large enough

• Ψ(Φ(u1), . . . , Φ(un))
• mj = (ln j)−1Ψ
• fu1(m), . . . , fun(m)
• = c

ln j . Thus Ψ(Φ(u1), . . . , Φ(un)) / ∈ ∪q<∞ℓq.

Proof: c0(N × N)\(∪q<∞ℓq(N × N))

• 3. Density. Z is dense in c0: for u ∈ S, u − Φ(u) 1/ ln(n + 1).
• 4. Z ∩ (∪qℓq) = {0}; (Φ(u))u∈S is a countable set of free

generators of Z. Elements of Z are of the form Ψ(Φ(u1), . . . , Φ(un)), where u1, . . . , un are distinct, and Ψ : Rn → R is a finite composition of +, −, c·, ∨, ∧. Need: either Ψ(Φ(u1), . . . , Φ(un)) = 0, or Ψ(Φ(u1), . . . , Φ(un)) / ∈ ∪q<∞ℓq. Ψ ≡ 0 ⇒ Ψ(Φ(u1), . . . , Φ(un)) = 0. Otherwise ∃ k1, . . . , kn ∈ K s.t. c := Ψ(k1, . . . , kn) = 0. Find m s.t. fuℓ(m) = kℓ for 1 ℓ m. For j large enough

• Ψ(Φ(u1), . . . , Φ(un))
• mj = (ln j)−1Ψ
• fu1(m), . . . , fun(m)
• = c

ln j . Thus Ψ(Φ(u1), . . . , Φ(un)) / ∈ ∪q<∞ℓq.

Proof: complements of finite dimensional subspaces

Theorem (to be proved) Suppose Y is a fin. dim. subspace of a Banach lattice X. Then X+ contains disjoint non-zero elements (xi)i∈N so that Y ∩ Z = {0}, where Z is the closed ideal generated (xi)i∈N: Z =

• z ∈ X : |z| |x| for some x ∈ span[(xi)i∈N]
• Fact: Any inf. dim. Banach lattice contains an infinite disjoint
• sequence. Find mutually disjoint non-zero (uij)n

i,j=1 in X+.

Claim: ∃ j ∈ N s.t. span[(zi)∞

i=1] ∩ Y = {0} whenever ∀i ∃ci > 0

s.t. |zi| ciuij. Once claim is established: set xi = uij.

Proof: complements of finite dimensional subspaces

Theorem (to be proved) Suppose Y is a fin. dim. subspace of a Banach lattice X. Then X+ contains disjoint non-zero elements (xi)i∈N so that Y ∩ Z = {0}, where Z is the closed ideal generated (xi)i∈N: Z =

• z ∈ X : |z| |x| for some x ∈ span[(xi)i∈N]
• Fact: Any inf. dim. Banach lattice contains an infinite disjoint
• sequence. Find mutually disjoint non-zero (uij)n

i,j=1 in X+.

Claim: ∃ j ∈ N s.t. span[(zi)∞

i=1] ∩ Y = {0} whenever ∀i ∃ci > 0

s.t. |zi| ciuij. Once claim is established: set xi = uij.

Proof: complements of finite dimensional subspaces

Theorem (to be proved) Suppose Y is a fin. dim. subspace of a Banach lattice X. Then X+ contains disjoint non-zero elements (xi)i∈N so that Y ∩ Z = {0}, where Z is the closed ideal generated (xi)i∈N: Z =

• z ∈ X : |z| |x| for some x ∈ span[(xi)i∈N]
• Fact: Any inf. dim. Banach lattice contains an infinite disjoint
• sequence. Find mutually disjoint non-zero (uij)n

i,j=1 in X+.

Claim: ∃ j ∈ N s.t. span[(zi)∞

i=1] ∩ Y = {0} whenever ∀i ∃ci > 0

s.t. |zi| ciuij. Once claim is established: set xi = uij.

Proof: complements of finite dimensional subspaces

Theorem (to be proved) Suppose Y is a fin. dim. subspace of a Banach lattice X. Then X+ contains disjoint non-zero elements (xi)i∈N so that Y ∩ Z = {0}, where Z is the closed ideal generated (xi)i∈N: Z =

• z ∈ X : |z| |x| for some x ∈ span[(xi)i∈N]
• (uij) ∈ X+\{0} mutually disjoint.

Claim: ∃ j ∈ N s.t. span[(zi)∞

i=1] ∩ Y = {0} whenever ∀i ∃ci > 0

s.t. |zi| ciuij. Proof of Claim: Otherwise, ∀j ∃yj ∈ span[(zi)∞

i=1] ∩ Y with zi’s

as above, and yj = 1. All yj’s are disjoint, hence

• lin. independent. But Y is fin. dim..

Proof: complements of finite dimensional subspaces

Theorem (to be proved) Suppose Y is a fin. dim. subspace of a Banach lattice X. Then X+ contains disjoint non-zero elements (xi)i∈N so that Y ∩ Z = {0}, where Z is the closed ideal generated (xi)i∈N: Z =

• z ∈ X : |z| |x| for some x ∈ span[(xi)i∈N]
• (uij) ∈ X+\{0} mutually disjoint.

Claim: ∃ j ∈ N s.t. span[(zi)∞

i=1] ∩ Y = {0} whenever ∀i ∃ci > 0

s.t. |zi| ciuij. Proof of Claim: Otherwise, ∀j ∃yj ∈ span[(zi)∞

i=1] ∩ Y with zi’s

as above, and yj = 1. All yj’s are disjoint, hence

• lin. independent. But Y is fin. dim..

Proof: complements of finite dimensional subspaces

Theorem (to be proved) Suppose Y is a fin. dim. subspace of a Banach lattice X. Then X+ contains disjoint non-zero elements (xi)i∈N so that Y ∩ Z = {0}, where Z is the closed ideal generated (xi)i∈N: Z =

• z ∈ X : |z| |x| for some x ∈ span[(xi)i∈N]
• (uij) ∈ X+\{0} mutually disjoint.

Claim: ∃ j ∈ N s.t. span[(zi)∞

i=1] ∩ Y = {0} whenever ∀i ∃ci > 0

s.t. |zi| ciuij. Proof of Claim: Otherwise, ∀j ∃yj ∈ span[(zi)∞

i=1] ∩ Y with zi’s

as above, and yj = 1. All yj’s are disjoint, hence

• lin. independent. But Y is fin. dim..

Last slide

Based on: T.O., A note on latticeability and algebrability, JMAA 2016.

Thank you for your attention!

Last slide

Based on: T.O., A note on latticeability and algebrability, JMAA 2016.

Thank you for your attention!