Hypercyclic N-linear operators Preliminary work, with Alberto - - PowerPoint PPT Presentation

Hypercyclic N-linear operators

Preliminary work, with Alberto Conejero (Univ. Polit´ ecnica de Valencia) Juan B` es Bowling Green State University

Hypercyclic N-linear operators

Greetings from Alberto Conejero

Hypercyclic N-linear operators

Theorem (N. Bernardes, 1998)

No homogeneous polynomial of degree d ≥ 2 on a Banach space can be hypercyclic. Recall that a homogenous polynomial P of degree 2 is given by P(x) = L(x, x), x ∈ X, where L : X × X → X is a continuous bilinear map. And a polynomial P is said to be hypercyclic if it has a dense orbit {Pn(x) : n ≥ 0}.

Hypercyclic N-linear operators

If we drop the homogeneity requirement,

• F. Mart´

ınez-Gim´ enez, A. Peris (2009): Every separable, infinite dimensional Fr´ echet space supports non-homogeneous mixing polynomials of any positive degree.

• N. Bernardes, A. Peris (2014): If the Fr´

echet space has an unconditional basis, then it has non-homogeneous polynomials that are frequently hypercyclic and chaotic.

Hypercyclic N-linear operators

We are interested in the new direction started by K.-G. Grosse-Erdmann and S. G. Kim (2013): Instead of looking at the homogeneous polynomial P, they considered the underlying bilinear mapping (x, y) → L(x, y)

Question: Can a bilinear mapping be ’hypercyclic’, that is, have a dense ’orbit’? What is the ’orbit’ of a bilinear mapping?

Hypercyclic N-linear operators

We are interested in the new direction started by K.-G. Grosse-Erdmann and S. G. Kim (2013): Instead of looking at the homogeneous polynomial P, they considered the underlying bilinear mapping (x, y) → L(x, y)

Question: Can a bilinear mapping be ’hypercyclic’, that is, have a dense ’orbit’? What is the ’orbit’ of a bilinear mapping?

Hypercyclic N-linear operators

For a linear map T : X → X and x ∈ X, Orb(x, T) = {x, Tx, T 2x, . . . } = {x} ∪ {x, Tx} ∪ {x, Tx, T 2x} ∪ . . .

• Definition. (K.-G. Grosse-Erdmann, S. G. Kim, ’13)

For a bilinear map L : X × X → X and (x, y) ∈ X × X, let Orb((x, y), L) = {x, y} ∪ {x, y, L(x, x), L(x, y), L(y, x), L(y, y)} ∪ . . . = ∪∞

n=0Un,

where U0 = {x, y} and Un := Un−1 ∪ {L(u, v) : u, v ∈ Un−1}, n ≥ 1. L is bihypercyclic provided Orb((x, y), L) = X for some (x, y) ∈ X × X. Such (x, y) is called a bihypercyclic pair for L.

Hypercyclic N-linear operators

• Example. Is L : C × C → C, L(x, y) = xy, bihypercyclic?

For x, y ∈ C Orb((x, y), L) = {x, y, x2, xy, y 2, x3, x2y, xy 2, y 3, . . . , y 4, . . . } = {xky m}k,m≥1 So the hypercyclicy follows by (N. Feldman, 2008) For each n ≥ 1, there are complex numbers a1, . . . , an, b1, . . . , bn such that the set of vectors               ak

1bk1 1

ak

2bk2 2

. . . ak

nbkn n

             

k,k1,k2,...,kn≥1

is dense in Cn. In general,

Grosse-Erdmann & Kim, 2013: Each Kn has bihypercyclic

• perators.

Hypercyclic N-linear operators

• Example. Is L : C × C → C, L(x, y) = xy, bihypercyclic?

For x, y ∈ C Orb((x, y), L) = {x, y, x2, xy, y 2, x3, x2y, xy 2, y 3, . . . , y 4, . . . } = {xky m}k,m≥1 So the hypercyclicy follows by (N. Feldman, 2008) For each n ≥ 1, there are complex numbers a1, . . . , an, b1, . . . , bn such that the set of vectors               ak

1bk1 1

ak

2bk2 2

. . . ak

nbkn n

             

k,k1,k2,...,kn≥1

is dense in Cn. In general,

Grosse-Erdmann & Kim, 2013: Each Kn has bihypercyclic

• perators.

Hypercyclic N-linear operators

• Example. Is L : C × C → C, L(x, y) = xy, bihypercyclic?

For x, y ∈ C Orb((x, y), L) = {x, y, x2, xy, y 2, x3, x2y, xy 2, y 3, . . . , y 4, . . . } = {xky m}k,m≥1 So the hypercyclicy follows by (N. Feldman, 2008) For each n ≥ 1, there are complex numbers a1, . . . , an, b1, . . . , bn such that the set of vectors               ak

1bk1 1

ak

2bk2 2

. . . ak

nbkn n

             

k,k1,k2,...,kn≥1

is dense in Cn. In general,

Grosse-Erdmann & Kim, 2013: Each Kn has bihypercyclic

• perators.

Hypercyclic N-linear operators

• Example. Is L : C × C → C, L(x, y) = xy, bihypercyclic?

For x, y ∈ C Orb((x, y), L) = {x, y, x2, xy, y 2, x3, x2y, xy 2, y 3, . . . , y 4, . . . } = {xky m}k,m≥1 So the hypercyclicy follows by (N. Feldman, 2008) For each n ≥ 1, there are complex numbers a1, . . . , an, b1, . . . , bn such that the set of vectors               ak

1bk1 1

ak

2bk2 2

. . . ak

nbkn n

             

k,k1,k2,...,kn≥1

is dense in Cn. In general,

Grosse-Erdmann & Kim, 2013: Each Kn has bihypercyclic

• perators.

Hypercyclic N-linear operators

• Example. Is L : C × C → C, L(x, y) = xy, bihypercyclic?

For x, y ∈ C Orb((x, y), L) = {x, y, x2, xy, y 2, x3, x2y, xy 2, y 3, . . . , y 4, . . . } = {xky m}k,m≥1 So the hypercyclicy follows by (N. Feldman, 2008) For each n ≥ 1, there are complex numbers a1, . . . , an, b1, . . . , bn such that the set of vectors               ak

1bk1 1

ak

2bk2 2

. . . ak

nbkn n

             

k,k1,k2,...,kn≥1

is dense in Cn. In general,

Grosse-Erdmann & Kim, 2013: Each Kn has bihypercyclic

• perators.

Hypercyclic N-linear operators

• Example. Is L : C × C → C, L(x, y) = xy, bihypercyclic?

For x, y ∈ C Orb((x, y), L) = {x, y, x2, xy, y 2, x3, x2y, xy 2, y 3, . . . , y 4, . . . } = {xky m}k,m≥1 So the hypercyclicy follows by (N. Feldman, 2008) For each n ≥ 1, there are complex numbers a1, . . . , an, b1, . . . , bn such that the set of vectors               ak

1bk1 1

ak

2bk2 2

. . . ak

nbkn n

             

k,k1,k2,...,kn≥1

is dense in Cn. In general,

Grosse-Erdmann & Kim, 2013: Each Kn has bihypercyclic

• perators.

Hypercyclic N-linear operators

• Example. Is L : C × C → C, L(x, y) = xy, bihypercyclic?

For x, y ∈ C Orb((x, y), L) = {x, y, x2, xy, y 2, x3, x2y, xy 2, y 3, . . . , y 4, . . . } = {xky m}k,m≥1 So the hypercyclicy follows by (N. Feldman, 2008) For each n ≥ 1, there are complex numbers a1, . . . , an, b1, . . . , bn such that the set of vectors               ak

1bk1 1

ak

2bk2 2

. . . ak

nbkn n

             

k,k1,k2,...,kn≥1

is dense in Cn. In general,

Grosse-Erdmann & Kim, 2013: Each Kn has bihypercyclic

• perators.

Hypercyclic N-linear operators

• Example. Is L : C × C → C, L(x, y) = xy, bihypercyclic?

For x, y ∈ C Orb((x, y), L) = {x, y, x2, xy, y 2, x3, x2y, xy 2, y 3, . . . , y 4, . . . } = {xky m}k,m≥1 So the hypercyclicy follows by (N. Feldman, 2008) For each n ≥ 1, there are complex numbers a1, . . . , an, b1, . . . , bn such that the set of vectors               ak

1bk1 1

ak

2bk2 2

. . . ak

nbkn n

             

k,k1,k2,...,kn≥1

is dense in Cn. In general,

Grosse-Erdmann & Kim, 2013: Each Kn has bihypercyclic

• perators.

Hypercyclic N-linear operators

• Example. Is L : C × C → C, L(x, y) = xy, bihypercyclic?

For x, y ∈ C Orb((x, y), L) = {x, y, x2, xy, y 2, x3, x2y, xy 2, y 3, . . . , y 4, . . . } = {xky m}k,m≥1 So the hypercyclicy follows by (N. Feldman, 2008) For each n ≥ 1, there are complex numbers a1, . . . , an, b1, . . . , bn such that the set of vectors               ak

1bk1 1

ak

2bk2 2

. . . ak

nbkn n

             

k,k1,k2,...,kn≥1

is dense in Cn. In general,

Grosse-Erdmann & Kim, 2013: Each Kn has bihypercyclic

• perators.

Hypercyclic N-linear operators

Example (Grosse-Erdmann and Kim)

Suppose there exists x ∈ X so that T := L(x, ·) : X → X is hypercyclic, with a hypercyclic vector y. Then (x, y) is a bihypercyclic pair for L. In fact, Ty = L(x, y), T 2y = L(x, Ty) = L(x, L(x, y)), T 3y = L(x, T 2y) = L(x, L(x, L(x, y))), . . . So Orb(T, y) ⊂ Orb((x, y), L)

• Theorem (Grosse-Erdmann and Kim) Every separable Banach

space suports a bihypercyclic bilinear operator.

If dim(X) = ∞, get T a hypercyclic operator on X, and any 0 = x∗ ∈ X ∗. Then L := x∗ ⊗ T, (x, y) → x∗(x)T(y) is bihypercyclic.

Hypercyclic N-linear operators

Example (Grosse-Erdmann and Kim)

Suppose there exists x ∈ X so that T := L(x, ·) : X → X is hypercyclic, with a hypercyclic vector y. Then (x, y) is a bihypercyclic pair for L. In fact, Ty = L(x, y), T 2y = L(x, Ty) = L(x, L(x, y)), T 3y = L(x, T 2y) = L(x, L(x, L(x, y))), . . . So Orb(T, y) ⊂ Orb((x, y), L)

• Theorem (Grosse-Erdmann and Kim) Every separable Banach

space suports a bihypercyclic bilinear operator.

If dim(X) = ∞, get T a hypercyclic operator on X, and any 0 = x∗ ∈ X ∗. Then L := x∗ ⊗ T, (x, y) → x∗(x)T(y) is bihypercyclic.

Hypercyclic N-linear operators

Example (Grosse-Erdmann and Kim)

Suppose there exists x ∈ X so that T := L(x, ·) : X → X is hypercyclic, with a hypercyclic vector y. Then (x, y) is a bihypercyclic pair for L. In fact, Ty = L(x, y), T 2y = L(x, Ty) = L(x, L(x, y)), T 3y = L(x, T 2y) = L(x, L(x, L(x, y))), . . . So Orb(T, y) ⊂ Orb((x, y), L)

• Theorem (Grosse-Erdmann and Kim) Every separable Banach

space suports a bihypercyclic bilinear operator.

If dim(X) = ∞, get T a hypercyclic operator on X, and any 0 = x∗ ∈ X ∗. Then L := x∗ ⊗ T, (x, y) → x∗(x)T(y) is bihypercyclic.

Hypercyclic N-linear operators

Example (Grosse-Erdmann and Kim)

Suppose there exists x ∈ X so that T := L(x, ·) : X → X is hypercyclic, with a hypercyclic vector y. Then (x, y) is a bihypercyclic pair for L. In fact, Ty = L(x, y), T 2y = L(x, Ty) = L(x, L(x, y)), T 3y = L(x, T 2y) = L(x, L(x, L(x, y))), . . . So Orb(T, y) ⊂ Orb((x, y), L)

• Theorem (Grosse-Erdmann and Kim) Every separable Banach

space suports a bihypercyclic bilinear operator.

If dim(X) = ∞, get T a hypercyclic operator on X, and any 0 = x∗ ∈ X ∗. Then L := x∗ ⊗ T, (x, y) → x∗(x)T(y) is bihypercyclic.

Hypercyclic N-linear operators

Example (Grosse-Erdmann and Kim)

Suppose there exists x ∈ X so that T := L(x, ·) : X → X is hypercyclic, with a hypercyclic vector y. Then (x, y) is a bihypercyclic pair for L. In fact, Ty = L(x, y), T 2y = L(x, Ty) = L(x, L(x, y)), T 3y = L(x, T 2y) = L(x, L(x, L(x, y))), . . . So Orb(T, y) ⊂ Orb((x, y), L)

• Theorem (Grosse-Erdmann and Kim) Every separable Banach

space suports a bihypercyclic bilinear operator.

If dim(X) = ∞, get T a hypercyclic operator on X, and any 0 = x∗ ∈ X ∗. Then L := x∗ ⊗ T, (x, y) → x∗(x)T(y) is bihypercyclic.

Hypercyclic N-linear operators

Example (Grosse-Erdmann and Kim)

Suppose there exists x ∈ X so that T := L(x, ·) : X → X is hypercyclic, with a hypercyclic vector y. Then (x, y) is a bihypercyclic pair for L. In fact, Ty = L(x, y), T 2y = L(x, Ty) = L(x, L(x, y)), T 3y = L(x, T 2y) = L(x, L(x, L(x, y))), . . . So Orb(T, y) ⊂ Orb((x, y), L)

• Theorem (Grosse-Erdmann and Kim) Every separable Banach

space suports a bihypercyclic bilinear operator.

If dim(X) = ∞, get T a hypercyclic operator on X, and any 0 = x∗ ∈ X ∗. Then L := x∗ ⊗ T, (x, y) → x∗(x)T(y) is bihypercyclic.

Hypercyclic N-linear operators

Example (Grosse-Erdmann and Kim)

Suppose there exists x ∈ X so that T := L(x, ·) : X → X is hypercyclic, with a hypercyclic vector y. Then (x, y) is a bihypercyclic pair for L. In fact, Ty = L(x, y), T 2y = L(x, Ty) = L(x, L(x, y)), T 3y = L(x, T 2y) = L(x, L(x, L(x, y))), . . . So Orb(T, y) ⊂ Orb((x, y), L)

• Theorem (Grosse-Erdmann and Kim) Every separable Banach

space suports a bihypercyclic bilinear operator.

If dim(X) = ∞, get T a hypercyclic operator on X, and any 0 = x∗ ∈ X ∗. Then L := x∗ ⊗ T, (x, y) → x∗(x)T(y) is bihypercyclic.

Hypercyclic N-linear operators

Example (Grosse-Erdmann and Kim)

Suppose there exists x ∈ X so that T := L(x, ·) : X → X is hypercyclic, with a hypercyclic vector y. Then (x, y) is a bihypercyclic pair for L. In fact, Ty = L(x, y), T 2y = L(x, Ty) = L(x, L(x, y)), T 3y = L(x, T 2y) = L(x, L(x, L(x, y))), . . . So Orb(T, y) ⊂ Orb((x, y), L)

• Theorem (Grosse-Erdmann and Kim) Every separable Banach

space suports a bihypercyclic bilinear operator.

If dim(X) = ∞, get T a hypercyclic operator on X, and any 0 = x∗ ∈ X ∗. Then L := x∗ ⊗ T, (x, y) → x∗(x)T(y) is bihypercyclic.

Hypercyclic N-linear operators

Example (Grosse-Erdmann and Kim)

Suppose there exists x ∈ X so that T := L(x, ·) : X → X is hypercyclic, with a hypercyclic vector y. Then (x, y) is a bihypercyclic pair for L. In fact, Ty = L(x, y), T 2y = L(x, Ty) = L(x, L(x, y)), T 3y = L(x, T 2y) = L(x, L(x, L(x, y))), . . . So Orb(T, y) ⊂ Orb((x, y), L)

• Theorem (Grosse-Erdmann and Kim) Every separable Banach

space suports a bihypercyclic bilinear operator.

If dim(X) = ∞, get T a hypercyclic operator on X, and any 0 = x∗ ∈ X ∗. Then L := x∗ ⊗ T, (x, y) → x∗(x)T(y) is bihypercyclic.

Hypercyclic N-linear operators

Recall that the set of hypercyclic vectors of a hypercyclic linear operator is residual and contains (but zero) a dense linear manifold. What about the set of bihypercyclic pairs of a bihypercyclic bilinear map? (Grosse-Erdmann and Kim) This set is a Gδ. However, it is not dense when X is a Banach space. Indeed, if x, y ≤ 1 L, then L(x, y) ≤ 1 L and Orb((x, y), L) is bounded.

Nilson’s

Notice also that the set { bihypercyclic pairs } ∪ {(0, 0)} will contain no line in this case, either.

d-hyp(T1, T2) Hypercyclic N-linear operators

Recall that the set of hypercyclic vectors of a hypercyclic linear operator is residual and contains (but zero) a dense linear manifold. What about the set of bihypercyclic pairs of a bihypercyclic bilinear map? (Grosse-Erdmann and Kim) This set is a Gδ. However, it is not dense when X is a Banach space. Indeed, if x, y ≤ 1 L, then L(x, y) ≤ 1 L and Orb((x, y), L) is bounded.

Nilson’s

Notice also that the set { bihypercyclic pairs } ∪ {(0, 0)} will contain no line in this case, either.

d-hyp(T1, T2) Hypercyclic N-linear operators

Recall that the set of hypercyclic vectors of a hypercyclic linear operator is residual and contains (but zero) a dense linear manifold. What about the set of bihypercyclic pairs of a bihypercyclic bilinear map? (Grosse-Erdmann and Kim) This set is a Gδ. However, it is not dense when X is a Banach space. Indeed, if x, y ≤ 1 L, then L(x, y) ≤ 1 L and Orb((x, y), L) is bounded.

Nilson’s

Notice also that the set { bihypercyclic pairs } ∪ {(0, 0)} will contain no line in this case, either.

d-hyp(T1, T2) Hypercyclic N-linear operators

Here’s a curious fact about bihypercyclic operators: For a hypercyclic operator T we know that λT is hypercyclic if |λ| = 1 (Le´

• n, M¨

uller, 2004), but not necessarily if |λ| = 1. However,

(Grosse-Erdmann and Kim, 2013) If L is bihypercyclic, so is λL, for each λ = 0.

Indeed, since (λL)( 1 λx, 1 λy) = 1 λL(x, y) we have that Orb(( 1 λx, 1 λy), λL) = 1 λOrb((x, y), L).

• Hypercyclic N-linear operators

Here’s a curious fact about bihypercyclic operators: For a hypercyclic operator T we know that λT is hypercyclic if |λ| = 1 (Le´

• n, M¨

uller, 2004), but not necessarily if |λ| = 1. However,

(Grosse-Erdmann and Kim, 2013) If L is bihypercyclic, so is λL, for each λ = 0.

Indeed, since (λL)( 1 λx, 1 λy) = 1 λL(x, y) we have that Orb(( 1 λx, 1 λy), λL) = 1 λOrb((x, y), L).

• Hypercyclic N-linear operators

Here’s a curious fact about bihypercyclic operators: For a hypercyclic operator T we know that λT is hypercyclic if |λ| = 1 (Le´

• n, M¨

uller, 2004), but not necessarily if |λ| = 1. However,

(Grosse-Erdmann and Kim, 2013) If L is bihypercyclic, so is λL, for each λ = 0.

Indeed, since (λL)( 1 λx, 1 λy) = 1 λL(x, y) we have that Orb(( 1 λx, 1 λy), λL) = 1 λOrb((x, y), L).

• Hypercyclic N-linear operators

We propose here an alternative orbit for bilinear maps. Motivation: For a linear operator T : X → X and x ∈ X, Orb(x, T) = {x, Tx, T 2x, . . . } = {x0, x1, x2, . . . } where x0 = x and xn = Txn−1.

Definition (B.- Conejero) Given a bilinear operator L : X × X → X and (x, y) ∈ X × X, let

• rb((x, y), L) = {un : n ≥ 1}

where (u1, u2) = (x, y), and u2+n = L(un, un+1), n ≥ 1. Also, (x, y) is a hypercyclic vector for L provided orb((x, y), L) = X.

That is, u1 = x, u2 = y, u3 = L(x, y), u4 = L(y, L(x, y)), u5 = L(L(x, y), L(y, L(x, y))), etc, and

• rb((x, y), L) ⊂ Orb((x, y), L).

So if (x, y) is hypercyclic vector for L, it is also a bihypercyclic pair for L.

Hypercyclic N-linear operators

We propose here an alternative orbit for bilinear maps. Motivation: For a linear operator T : X → X and x ∈ X, Orb(x, T) = {x, Tx, T 2x, . . . } = {x0, x1, x2, . . . } where x0 = x and xn = Txn−1.

Definition (B.- Conejero) Given a bilinear operator L : X × X → X and (x, y) ∈ X × X, let

• rb((x, y), L) = {un : n ≥ 1}

where (u1, u2) = (x, y), and u2+n = L(un, un+1), n ≥ 1. Also, (x, y) is a hypercyclic vector for L provided orb((x, y), L) = X.

That is, u1 = x, u2 = y, u3 = L(x, y), u4 = L(y, L(x, y)), u5 = L(L(x, y), L(y, L(x, y))), etc, and

• rb((x, y), L) ⊂ Orb((x, y), L).

So if (x, y) is hypercyclic vector for L, it is also a bihypercyclic pair for L.

Hypercyclic N-linear operators

We propose here an alternative orbit for bilinear maps. Motivation: For a linear operator T : X → X and x ∈ X, Orb(x, T) = {x, Tx, T 2x, . . . } = {x0, x1, x2, . . . } where x0 = x and xn = Txn−1.

Definition (B.- Conejero) Given a bilinear operator L : X × X → X and (x, y) ∈ X × X, let

• rb((x, y), L) = {un : n ≥ 1}

where (u1, u2) = (x, y), and u2+n = L(un, un+1), n ≥ 1. Also, (x, y) is a hypercyclic vector for L provided orb((x, y), L) = X.

That is, u1 = x, u2 = y, u3 = L(x, y), u4 = L(y, L(x, y)), u5 = L(L(x, y), L(y, L(x, y))), etc, and

• rb((x, y), L) ⊂ Orb((x, y), L).

So if (x, y) is hypercyclic vector for L, it is also a bihypercyclic pair for L.

Hypercyclic N-linear operators

We propose here an alternative orbit for bilinear maps. Motivation: For a linear operator T : X → X and x ∈ X, Orb(x, T) = {x, Tx, T 2x, . . . } = {x0, x1, x2, . . . } where x0 = x and xn = Txn−1.

Definition (B.- Conejero) Given a bilinear operator L : X × X → X and (x, y) ∈ X × X, let

• rb((x, y), L) = {un : n ≥ 1}

where (u1, u2) = (x, y), and u2+n = L(un, un+1), n ≥ 1. Also, (x, y) is a hypercyclic vector for L provided orb((x, y), L) = X.

That is, u1 = x, u2 = y, u3 = L(x, y), u4 = L(y, L(x, y)), u5 = L(L(x, y), L(y, L(x, y))), etc, and

• rb((x, y), L) ⊂ Orb((x, y), L).

So if (x, y) is hypercyclic vector for L, it is also a bihypercyclic pair for L.

Hypercyclic N-linear operators

We propose here an alternative orbit for bilinear maps. Motivation: For a linear operator T : X → X and x ∈ X, Orb(x, T) = {x, Tx, T 2x, . . . } = {x0, x1, x2, . . . } where x0 = x and xn = Txn−1.

Definition (B.- Conejero) Given a bilinear operator L : X × X → X and (x, y) ∈ X × X, let

• rb((x, y), L) = {un : n ≥ 1}

where (u1, u2) = (x, y), and u2+n = L(un, un+1), n ≥ 1. Also, (x, y) is a hypercyclic vector for L provided orb((x, y), L) = X.

That is, u1 = x, u2 = y, u3 = L(x, y), u4 = L(y, L(x, y)), u5 = L(L(x, y), L(y, L(x, y))), etc, and

• rb((x, y), L) ⊂ Orb((x, y), L).

So if (x, y) is hypercyclic vector for L, it is also a bihypercyclic pair for L.

Hypercyclic N-linear operators

We propose here an alternative orbit for bilinear maps. Motivation: For a linear operator T : X → X and x ∈ X, Orb(x, T) = {x, Tx, T 2x, . . . } = {x0, x1, x2, . . . } where x0 = x and xn = Txn−1.

Definition (B.- Conejero) Given a bilinear operator L : X × X → X and (x, y) ∈ X × X, let

• rb((x, y), L) = {un : n ≥ 1}

where (u1, u2) = (x, y), and u2+n = L(un, un+1), n ≥ 1. Also, (x, y) is a hypercyclic vector for L provided orb((x, y), L) = X.

That is, u1 = x, u2 = y, u3 = L(x, y), u4 = L(y, L(x, y)), u5 = L(L(x, y), L(y, L(x, y))), etc, and

• rb((x, y), L) ⊂ Orb((x, y), L).

So if (x, y) is hypercyclic vector for L, it is also a bihypercyclic pair for L.

Hypercyclic N-linear operators

We propose here an alternative orbit for bilinear maps. Motivation: For a linear operator T : X → X and x ∈ X, Orb(x, T) = {x, Tx, T 2x, . . . } = {x0, x1, x2, . . . } where x0 = x and xn = Txn−1.

Definition (B.- Conejero) Given a bilinear operator L : X × X → X and (x, y) ∈ X × X, let

• rb((x, y), L) = {un : n ≥ 1}

where (u1, u2) = (x, y), and u2+n = L(un, un+1), n ≥ 1. Also, (x, y) is a hypercyclic vector for L provided orb((x, y), L) = X.

That is, u1 = x, u2 = y, u3 = L(x, y), u4 = L(y, L(x, y)), u5 = L(L(x, y), L(y, L(x, y))), etc, and

• rb((x, y), L) ⊂ Orb((x, y), L).

So if (x, y) is hypercyclic vector for L, it is also a bihypercyclic pair for L.

Hypercyclic N-linear operators

We saw L : C × C → C, L(x, y) = xy is bihypercyclic. Is it hypercyclic?

Notice that orb((x, y), L) = {un : n ≥ 1} is given by u1 = x u2 = y u3 = L(u1, u2) = xy, u4 = L(u2, u3) = y(xy) = xy 2, u5 = L(u3, u4) = x2y 3, . . . etc. That is, u2+n = xFny Fn+1 for each n ≥ 1, where (Fk) is the Fibonacci sequence F1 = F2 = 1, Fk = Fk−1 + Fk−2. Recall that lim

n→∞

Fn+1 Fn = ϕ (Keppler) Fn = ϕn − ψn √ 5 (Binet’s Formula), where ϕ = 1+

√ 5 2

≈ 1.61 and ψ = 1−

√ 5 2

≈ −.61.

Hypercyclic N-linear operators

We saw L : C × C → C, L(x, y) = xy is bihypercyclic. Is it hypercyclic?

Notice that orb((x, y), L) = {un : n ≥ 1} is given by u1 = x u2 = y u3 = L(u1, u2) = xy, u4 = L(u2, u3) = y(xy) = xy 2, u5 = L(u3, u4) = x2y 3, . . . etc. That is, u2+n = xFny Fn+1 for each n ≥ 1, where (Fk) is the Fibonacci sequence F1 = F2 = 1, Fk = Fk−1 + Fk−2. Recall that lim

n→∞

Fn+1 Fn = ϕ (Keppler) Fn = ϕn − ψn √ 5 (Binet’s Formula), where ϕ = 1+

√ 5 2

≈ 1.61 and ψ = 1−

√ 5 2

≈ −.61.

Hypercyclic N-linear operators

We saw L : C × C → C, L(x, y) = xy is bihypercyclic. Is it hypercyclic?

Notice that orb((x, y), L) = {un : n ≥ 1} is given by u1 = x u2 = y u3 = L(u1, u2) = xy, u4 = L(u2, u3) = y(xy) = xy 2, u5 = L(u3, u4) = x2y 3, . . . etc. That is, u2+n = xFny Fn+1 for each n ≥ 1, where (Fk) is the Fibonacci sequence F1 = F2 = 1, Fk = Fk−1 + Fk−2. Recall that lim

n→∞

Fn+1 Fn = ϕ (Keppler) Fn = ϕn − ψn √ 5 (Binet’s Formula), where ϕ = 1+

√ 5 2

≈ 1.61 and ψ = 1−

√ 5 2

≈ −.61.

Hypercyclic N-linear operators

We saw L : C × C → C, L(x, y) = xy is bihypercyclic. Is it hypercyclic?

Notice that orb((x, y), L) = {un : n ≥ 1} is given by u1 = x u2 = y u3 = L(u1, u2) = xy, u4 = L(u2, u3) = y(xy) = xy 2, u5 = L(u3, u4) = x2y 3, . . . etc. That is, u2+n = xFny Fn+1 for each n ≥ 1, where (Fk) is the Fibonacci sequence F1 = F2 = 1, Fk = Fk−1 + Fk−2. Recall that lim

n→∞

Fn+1 Fn = ϕ (Keppler) Fn = ϕn − ψn √ 5 (Binet’s Formula), where ϕ = 1+

√ 5 2

≈ 1.61 and ψ = 1−

√ 5 2

≈ −.61.

Hypercyclic N-linear operators

We saw L : C × C → C, L(x, y) = xy is bihypercyclic. Is it hypercyclic?

Notice that orb((x, y), L) = {un : n ≥ 1} is given by u1 = x u2 = y u3 = L(u1, u2) = xy, u4 = L(u2, u3) = y(xy) = xy 2, u5 = L(u3, u4) = x2y 3, . . . etc. That is, u2+n = xFny Fn+1 for each n ≥ 1, where (Fk) is the Fibonacci sequence F1 = F2 = 1, Fk = Fk−1 + Fk−2. Recall that lim

n→∞

Fn+1 Fn = ϕ (Keppler) Fn = ϕn − ψn √ 5 (Binet’s Formula), where ϕ = 1+

√ 5 2

≈ 1.61 and ψ = 1−

√ 5 2

≈ −.61.

Hypercyclic N-linear operators

If {u2+n = xFny Fn+1}n≥1 is dense in C, then

• |u2+n| = |x|Fn|yn|Fn+1 =
• |x| |y|

Fn+1 Fn

Fn

n≥1

is dense in (0, ∞). This forces |x||y|ϕ = 1. But then log |u2+n| = Fn log |x| + Fn+1 log |y| = (Fn+1 − ϕFn) log |y|, and Fn+1 − ϕFn = ϕn+1 − ψn+1 √ 5 − ϕϕn − ψn √ 5 = ψn(ϕ − ψ √ 5 ) →

n→∞ 0,

forcing |u2+n| →

n→∞ 1, a contradiction.

So L is bihypercyclic but not hypercyclic In general, there exists no hypercyclic bilinear map L : C × C → C.

Hypercyclic N-linear operators

If {u2+n = xFny Fn+1}n≥1 is dense in C, then

• |u2+n| = |x|Fn|yn|Fn+1 =
• |x| |y|

Fn+1 Fn

Fn

n≥1

is dense in (0, ∞). This forces |x||y|ϕ = 1. But then log |u2+n| = Fn log |x| + Fn+1 log |y| = (Fn+1 − ϕFn) log |y|, and Fn+1 − ϕFn = ϕn+1 − ψn+1 √ 5 − ϕϕn − ψn √ 5 = ψn(ϕ − ψ √ 5 ) →

n→∞ 0,

forcing |u2+n| →

n→∞ 1, a contradiction.

So L is bihypercyclic but not hypercyclic In general, there exists no hypercyclic bilinear map L : C × C → C.

Hypercyclic N-linear operators

If {u2+n = xFny Fn+1}n≥1 is dense in C, then

• |u2+n| = |x|Fn|yn|Fn+1 =
• |x| |y|

Fn+1 Fn

Fn

n≥1

is dense in (0, ∞). This forces |x||y|ϕ = 1. But then log |u2+n| = Fn log |x| + Fn+1 log |y| = (Fn+1 − ϕFn) log |y|, and Fn+1 − ϕFn = ϕn+1 − ψn+1 √ 5 − ϕϕn − ψn √ 5 = ψn(ϕ − ψ √ 5 ) →

n→∞ 0,

forcing |u2+n| →

n→∞ 1, a contradiction.

So L is bihypercyclic but not hypercyclic In general, there exists no hypercyclic bilinear map L : C × C → C.

Hypercyclic N-linear operators

If {u2+n = xFny Fn+1}n≥1 is dense in C, then

• |u2+n| = |x|Fn|yn|Fn+1 =
• |x| |y|

Fn+1 Fn

Fn

n≥1

is dense in (0, ∞). This forces |x||y|ϕ = 1. But then log |u2+n| = Fn log |x| + Fn+1 log |y| = (Fn+1 − ϕFn) log |y|, and Fn+1 − ϕFn = ϕn+1 − ψn+1 √ 5 − ϕϕn − ψn √ 5 = ψn(ϕ − ψ √ 5 ) →

n→∞ 0,

forcing |u2+n| →

n→∞ 1, a contradiction.

So L is bihypercyclic but not hypercyclic In general, there exists no hypercyclic bilinear map L : C × C → C.

Hypercyclic N-linear operators

Do hypercyclic bilinear operators exist?

Example.

Let ω = KN (K = R or C), endowed with the product topology, and let (x1, x2, . . . )

B

→ (x2, x3, . . . ) (x1, x2, . . . )

e∗

1

→ x1 Then L = e∗

1 ⊗ B : ω × ω −

→ ω is hypercyclic. (x, y) → e∗

1 (x)By

Hypercyclic N-linear operators

Do hypercyclic bilinear operators exist?

Example.

Let ω = KN (K = R or C), endowed with the product topology, and let (x1, x2, . . . )

B

→ (x2, x3, . . . ) (x1, x2, . . . )

e∗

1

→ x1 Then L = e∗

1 ⊗ B : ω × ω −

→ ω is hypercyclic. (x, y) → e∗

1 (x)By

Hypercyclic N-linear operators

L = e∗

1 ⊗ B : ω × ω −

→ ω, (x, y) → e∗

1 (x)By,

is hypercyclic. Proof: For x, y ∈ ω, the orbit orb((x, y), L) = {un : n ≥ 1} is given by u1 = x u2 = y u3 = L(x, y) = x1By u4 = L(u2, u3) = y1 x1B2y = x1 y1B2y u5 = L(u3, u4) = (x1y2) y1x1B3y = x2

1 y1 y2 B3y

u6 = L(u4, u5) = (x1y1y3) x2

1y1y2 B4y = x3 1 y 2 1 y2 y3 B4y

In general, u2+ℓ = xFℓ

1 y Fℓ−1 1

y Fℓ−2

2

. . . y F2

ℓ−2y F1 ℓ−1

• Cℓ=Cℓ(x1,y1,...yℓ−1)

Bℓy (ℓ ≥ 1), where (Fn)≥1 is the Fibonacci sequence F1 = F2 = 1, Fn = Fn−1 + Fn−2. Thus orb((e1, y), L) is dense when y = (g1,1, 1, . . . , 1, g2,1 Cn2−2 , g2,2 Cn2−2 , 1, . . . , 1, g3,1 Cn3−2 , . . . ), for some dense subset ∆ = {gn = (gn,k) : n ≥ 1} of ω where gn,k = 0 ⇔ 1 ≤ k ≤ n

Hypercyclic N-linear operators

L = e∗

1 ⊗ B : ω × ω −

→ ω, (x, y) → e∗

1 (x)By,

is hypercyclic. Proof: For x, y ∈ ω, the orbit orb((x, y), L) = {un : n ≥ 1} is given by u1 = x u2 = y u3 = L(x, y) = x1By u4 = L(u2, u3) = y1 x1B2y = x1 y1B2y u5 = L(u3, u4) = (x1y2) y1x1B3y = x2

1 y1 y2 B3y

u6 = L(u4, u5) = (x1y1y3) x2

1y1y2 B4y = x3 1 y 2 1 y2 y3 B4y

In general, u2+ℓ = xFℓ

1 y Fℓ−1 1

y Fℓ−2

2

. . . y F2

ℓ−2y F1 ℓ−1

• Cℓ=Cℓ(x1,y1,...yℓ−1)

Bℓy (ℓ ≥ 1), where (Fn)≥1 is the Fibonacci sequence F1 = F2 = 1, Fn = Fn−1 + Fn−2. Thus orb((e1, y), L) is dense when y = (g1,1, 1, . . . , 1, g2,1 Cn2−2 , g2,2 Cn2−2 , 1, . . . , 1, g3,1 Cn3−2 , . . . ), for some dense subset ∆ = {gn = (gn,k) : n ≥ 1} of ω where gn,k = 0 ⇔ 1 ≤ k ≤ n

Hypercyclic N-linear operators

L = e∗

1 ⊗ B : ω × ω −

→ ω, (x, y) → e∗

1 (x)By,

is hypercyclic. Proof: For x, y ∈ ω, the orbit orb((x, y), L) = {un : n ≥ 1} is given by u1 = x u2 = y u3 = L(x, y) = x1By u4 = L(u2, u3) = y1 x1B2y = x1 y1B2y u5 = L(u3, u4) = (x1y2) y1x1B3y = x2

1 y1 y2 B3y

u6 = L(u4, u5) = (x1y1y3) x2

1y1y2 B4y = x3 1 y 2 1 y2 y3 B4y

In general, u2+ℓ = xFℓ

1 y Fℓ−1 1

y Fℓ−2

2

. . . y F2

ℓ−2y F1 ℓ−1

• Cℓ=Cℓ(x1,y1,...yℓ−1)

Bℓy (ℓ ≥ 1), where (Fn)≥1 is the Fibonacci sequence F1 = F2 = 1, Fn = Fn−1 + Fn−2. Thus orb((e1, y), L) is dense when y = (g1,1, 1, . . . , 1, g2,1 Cn2−2 , g2,2 Cn2−2 , 1, . . . , 1, g3,1 Cn3−2 , . . . ), for some dense subset ∆ = {gn = (gn,k) : n ≥ 1} of ω where gn,k = 0 ⇔ 1 ≤ k ≤ n

Hypercyclic N-linear operators

L = e∗

1 ⊗ B : ω × ω −

→ ω, (x, y) → e∗

1 (x)By,

is hypercyclic. Proof: For x, y ∈ ω, the orbit orb((x, y), L) = {un : n ≥ 1} is given by u1 = x u2 = y u3 = L(x, y) = x1By u4 = L(u2, u3) = y1 x1B2y = x1 y1B2y u5 = L(u3, u4) = (x1y2) y1x1B3y = x2

1 y1 y2 B3y

u6 = L(u4, u5) = (x1y1y3) x2

1y1y2 B4y = x3 1 y 2 1 y2 y3 B4y

In general, u2+ℓ = xFℓ

1 y Fℓ−1 1

y Fℓ−2

2

. . . y F2

ℓ−2y F1 ℓ−1

• Cℓ=Cℓ(x1,y1,...yℓ−1)

Bℓy (ℓ ≥ 1), where (Fn)≥1 is the Fibonacci sequence F1 = F2 = 1, Fn = Fn−1 + Fn−2. Thus orb((e1, y), L) is dense when y = (g1,1, 1, . . . , 1, g2,1 Cn2−2 , g2,2 Cn2−2 , 1, . . . , 1, g3,1 Cn3−2 , . . . ), for some dense subset ∆ = {gn = (gn,k) : n ≥ 1} of ω where gn,k = 0 ⇔ 1 ≤ k ≤ n

Hypercyclic N-linear operators

L = e∗

1 ⊗ B : ω × ω −

→ ω, (x, y) → e∗

1 (x)By,

is hypercyclic. Proof: For x, y ∈ ω, the orbit orb((x, y), L) = {un : n ≥ 1} is given by u1 = x u2 = y u3 = L(x, y) = x1By u4 = L(u2, u3) = y1 x1B2y = x1 y1B2y u5 = L(u3, u4) = (x1y2) y1x1B3y = x2

1 y1 y2 B3y

u6 = L(u4, u5) = (x1y1y3) x2

1y1y2 B4y = x3 1 y 2 1 y2 y3 B4y

In general, u2+ℓ = xFℓ

1 y Fℓ−1 1

y Fℓ−2

2

. . . y F2

ℓ−2y F1 ℓ−1

• Cℓ=Cℓ(x1,y1,...yℓ−1)

Bℓy (ℓ ≥ 1), where (Fn)≥1 is the Fibonacci sequence F1 = F2 = 1, Fn = Fn−1 + Fn−2. Thus orb((e1, y), L) is dense when y = (g1,1, 1, . . . , 1, g2,1 Cn2−2 , g2,2 Cn2−2 , 1, . . . , 1, g3,1 Cn3−2 , . . . ), for some dense subset ∆ = {gn = (gn,k) : n ≥ 1} of ω where gn,k = 0 ⇔ 1 ≤ k ≤ n

Hypercyclic N-linear operators

L = e∗

1 ⊗ B : ω × ω −

→ ω, (x, y) → e∗

1 (x)By,

is hypercyclic. Proof: For x, y ∈ ω, the orbit orb((x, y), L) = {un : n ≥ 1} is given by u1 = x u2 = y u3 = L(x, y) = x1By u4 = L(u2, u3) = y1 x1B2y = x1 y1B2y u5 = L(u3, u4) = (x1y2) y1x1B3y = x2

1 y1 y2 B3y

u6 = L(u4, u5) = (x1y1y3) x2

1y1y2 B4y = x3 1 y 2 1 y2 y3 B4y

In general, u2+ℓ = xFℓ

1 y Fℓ−1 1

y Fℓ−2

2

. . . y F2

ℓ−2y F1 ℓ−1

• Cℓ=Cℓ(x1,y1,...yℓ−1)

Bℓy (ℓ ≥ 1), where (Fn)≥1 is the Fibonacci sequence F1 = F2 = 1, Fn = Fn−1 + Fn−2. Thus orb((e1, y), L) is dense when y = (g1,1, 1, . . . , 1, g2,1 Cn2−2 , g2,2 Cn2−2 , 1, . . . , 1, g3,1 Cn3−2 , . . . ), for some dense subset ∆ = {gn = (gn,k) : n ≥ 1} of ω where gn,k = 0 ⇔ 1 ≤ k ≤ n

Hypercyclic N-linear operators

L = e∗

1 ⊗ B : ω × ω −

→ ω, (x, y) → e∗

1 (x)By,

is hypercyclic. Proof: For x, y ∈ ω, the orbit orb((x, y), L) = {un : n ≥ 1} is given by u1 = x u2 = y u3 = L(x, y) = x1By u4 = L(u2, u3) = y1 x1B2y = x1 y1B2y u5 = L(u3, u4) = (x1y2) y1x1B3y = x2

1 y1 y2 B3y

u6 = L(u4, u5) = (x1y1y3) x2

1y1y2 B4y = x3 1 y 2 1 y2 y3 B4y

In general, u2+ℓ = xFℓ

1 y Fℓ−1 1

y Fℓ−2

2

. . . y F2

ℓ−2y F1 ℓ−1

• Cℓ=Cℓ(x1,y1,...yℓ−1)

Bℓy (ℓ ≥ 1), where (Fn)≥1 is the Fibonacci sequence F1 = F2 = 1, Fn = Fn−1 + Fn−2. Thus orb((e1, y), L) is dense when y = (g1,1, 1, . . . , 1, g2,1 Cn2−2 , g2,2 Cn2−2 , 1, . . . , 1, g3,1 Cn3−2 , . . . ), for some dense subset ∆ = {gn = (gn,k) : n ≥ 1} of ω where gn,k = 0 ⇔ 1 ≤ k ≤ n

Hypercyclic N-linear operators

L = e∗

1 ⊗ B : ω × ω −

→ ω, (x, y) → e∗

1 (x)By,

is hypercyclic. Proof: For x, y ∈ ω, the orbit orb((x, y), L) = {un : n ≥ 1} is given by u1 = x u2 = y u3 = L(x, y) = x1By u4 = L(u2, u3) = y1 x1B2y = x1 y1B2y u5 = L(u3, u4) = (x1y2) y1x1B3y = x2

1 y1 y2 B3y

u6 = L(u4, u5) = (x1y1y3) x2

1y1y2 B4y = x3 1 y 2 1 y2 y3 B4y

In general, u2+ℓ = xFℓ

1 y Fℓ−1 1

y Fℓ−2

2

. . . y F2

ℓ−2y F1 ℓ−1

• Cℓ=Cℓ(x1,y1,...yℓ−1)

Bℓy (ℓ ≥ 1), where (Fn)≥1 is the Fibonacci sequence F1 = F2 = 1, Fn = Fn−1 + Fn−2. Thus orb((e1, y), L) is dense when y = (g1,1, 1, . . . , 1, g2,1 Cn2−2 , g2,2 Cn2−2 , 1, . . . , 1, g3,1 Cn3−2 , . . . ), for some dense subset ∆ = {gn = (gn,k) : n ≥ 1} of ω where gn,k = 0 ⇔ 1 ≤ k ≤ n

Hypercyclic N-linear operators

In general,

Theorem 1.

For each N ≥ 2, the set of hypercyclic vectors of L = e∗

1 ⊗ e∗ 1 ⊗ · · · ⊗ e∗ 1

• N−1

⊗ B : ωN → ω is residual in ωN.

Theorem 2. (July 2014?)

Let H(C) = {entire functions}, and N ≥ 2. Then L(f1, . . . , fN) = f1(0)f2(0) · · · fN−1(0) f ′

N(z)

has a residual set of hypercyclic vectors.

Hypercyclic N-linear operators

In general,

Theorem 1.

For each N ≥ 2, the set of hypercyclic vectors of L = e∗

1 ⊗ e∗ 1 ⊗ · · · ⊗ e∗ 1

• N−1

⊗ B : ωN → ω is residual in ωN.

Theorem 2. (July 2014?)

Let H(C) = {entire functions}, and N ≥ 2. Then L(f1, . . . , fN) = f1(0)f2(0) · · · fN−1(0) f ′

N(z)

has a residual set of hypercyclic vectors.

Hypercyclic N-linear operators

• Question. Is there any hypercyclic bilinear operator on a

Banach space?

We don’t know. As in the case of homogeneous polynomials,

Theorem 3.

Each separable, infinite-dimensional Fr´ echet space X supports a bilinear

• perator whose set of supercyclic vectors is residual in X × X.

(Recall (x1, x2) ∈ X × X is supercyclic for L provided C · orb((x1, x2), L) is dense in X) For X = w, there exist sequences (xn) in X and (fn) in X ∗ so that fn(xk) = 0 for n = k, fn(xn) ∈ (0, 1], and span{x′

ns} is dense in X.

Then L = f1 ⊗ R is supercyclic, where Ry =

n≥1 1 2n fn+1(y)yn (y ∈ X).

Hypercyclic N-linear operators

• Question. Is there any hypercyclic bilinear operator on a

Banach space?

We don’t know. As in the case of homogeneous polynomials,

Theorem 3.

Each separable, infinite-dimensional Fr´ echet space X supports a bilinear

• perator whose set of supercyclic vectors is residual in X × X.

(Recall (x1, x2) ∈ X × X is supercyclic for L provided C · orb((x1, x2), L) is dense in X) For X = w, there exist sequences (xn) in X and (fn) in X ∗ so that fn(xk) = 0 for n = k, fn(xn) ∈ (0, 1], and span{x′

ns} is dense in X.

Then L = f1 ⊗ R is supercyclic, where Ry =

n≥1 1 2n fn+1(y)yn (y ∈ X).

Hypercyclic N-linear operators

• Question. Is there any hypercyclic bilinear operator on a

Banach space?

We don’t know. As in the case of homogeneous polynomials,

Theorem 3.

Each separable, infinite-dimensional Fr´ echet space X supports a bilinear

• perator whose set of supercyclic vectors is residual in X × X.

(Recall (x1, x2) ∈ X × X is supercyclic for L provided C · orb((x1, x2), L) is dense in X) For X = w, there exist sequences (xn) in X and (fn) in X ∗ so that fn(xk) = 0 for n = k, fn(xn) ∈ (0, 1], and span{x′

ns} is dense in X.

Then L = f1 ⊗ R is supercyclic, where Ry =

n≥1 1 2n fn+1(y)yn (y ∈ X).

Hypercyclic N-linear operators

THANK YOU!

Hypercyclic N-linear operators