[PPT] - A z k -invariant subspace without the wandering property Daniel PowerPoint Presentation

SLIDE 1

A zk-invariant subspace without the wandering property

Daniel Seco

Universidad Carlos III de Madrid and Instituto de Ciencias Matemáticas

Workshop on Banach spaces and Banach lattices, ICMAT 12th September 2019

Seco (UC3M/ICMAT) Wandering property ICMAT 1 / 13

SLIDE 2

Spaces over the disc

Definition

Dirichlet-type space, Dα, is: {f ∈ Hol(D) : f(z) =

k∈N

akzk, ||f||2

α = ∞

k=0

|ak|2(k + 1)α < ∞}

Seco (UC3M/ICMAT) Wandering property ICMAT 2 / 13

SLIDE 3

Spaces over the disc

Definition

Dirichlet-type space, Dα, is: {f ∈ Hol(D) : f(z) =

k∈N

akzk, ||f||2

α = ∞

k=0

|ak|2(k + 1)α < ∞}

Examples

α = −1, A2 = Hol(D) ∩ L2(D)

Seco (UC3M/ICMAT) Wandering property ICMAT 2 / 13

SLIDE 4

Spaces over the disc

Definition

Dirichlet-type space, Dα, is: {f ∈ Hol(D) : f(z) =

k∈N

akzk, ||f||2

α = ∞

k=0

|ak|2(k + 1)α < ∞}

Examples

α = −1, A2 = Hol(D) ∩ L2(D) α = 0, H2 = Hol(D) ∩ L2(T)

Seco (UC3M/ICMAT) Wandering property ICMAT 2 / 13

SLIDE 5

Spaces over the disc

Definition

Dirichlet-type space, Dα, is: {f ∈ Hol(D) : f(z) =

k∈N

akzk, ||f||2

α = ∞

k=0

|ak|2(k + 1)α < ∞}

Examples

α = −1, A2 = Hol(D) ∩ L2(D) α = 0, H2 = Hol(D) ∩ L2(T) α = 1, D = Hol(D) ∩ {A(f(D)) < ∞}

Seco (UC3M/ICMAT) Wandering property ICMAT 2 / 13

SLIDE 6

Spaces over the disc

Definition

Dirichlet-type space, Dα, is: {f ∈ Hol(D) : f(z) =

k∈N

akzk, ||f||2

α = ∞

k=0

|ak|2(k + 1)α < ∞}

Examples

α = −1, A2 = Hol(D) ∩ L2(D) α = 0, H2 = Hol(D) ∩ L2(T) α = 1, D = Hol(D) ∩ {A(f(D)) < ∞} α > α′ ⇒ Dα ⊂ Dα′

Seco (UC3M/ICMAT) Wandering property ICMAT 2 / 13

SLIDE 7

Spaces over the disc

Definition

Dirichlet-type space, Dα, is: {f ∈ Hol(D) : f(z) =

k∈N

akzk, ||f||2

α = ∞

k=0

|ak|2(k + 1)α < ∞}

Examples

α = −1, A2 = Hol(D) ∩ L2(D) α = 0, H2 = Hol(D) ∩ L2(T) α = 1, D = Hol(D) ∩ {A(f(D)) < ∞} α > α′ ⇒ Dα ⊂ Dα′ f ∈ Dα ⇔ f ′ ∈ Dα−2

Seco (UC3M/ICMAT) Wandering property ICMAT 2 / 13

SLIDE 8

Spaces over the disc

Definition

Dirichlet-type space, Dα, is: {f ∈ Hol(D) : f(z) =

k∈N

akzk, ||f||2

α = ∞

k=0

|ak|2(k + 1)α < ∞}

Examples

α = −1, A2 = Hol(D) ∩ L2(D) α = 0, H2 = Hol(D) ∩ L2(T) α = 1, D = Hol(D) ∩ {A(f(D)) < ∞} α > α′ ⇒ Dα ⊂ Dα′ f ∈ Dα ⇔ f ′ ∈ Dα−2 Hilbert spaces with monomials as an orthogonal basis

Seco (UC3M/ICMAT) Wandering property ICMAT 2 / 13

SLIDE 9

Invariant subspaces

The (forward) shift operator is bdd: S : Dα → Dα : Sf(z) = zf(z). A closed subspace V of Dα is zk- invariant if SkV ⊂ V.

Seco (UC3M/ICMAT) Wandering property ICMAT 3 / 13

SLIDE 10

Invariant subspaces

The (forward) shift operator is bdd: S : Dα → Dα : Sf(z) = zf(z). A closed subspace V of Dα is zk- invariant if SkV ⊂ V. [f]zk(= [f]) = span{ztkf : t = 0, 1, 2, . . .}.

Seco (UC3M/ICMAT) Wandering property ICMAT 3 / 13

SLIDE 11

Invariant subspaces

The (forward) shift operator is bdd: S : Dα → Dα : Sf(z) = zf(z). A closed subspace V of Dα is zk- invariant if SkV ⊂ V. [f]zk(= [f]) = span{ztkf : t = 0, 1, 2, . . .}.

Theorem (Beurling, ’49)

For H2 (α = 0), M z-inv. subsp. ⇔ M = ϕH2 = [ϕ] with span(ϕ) = M ⊖ SM.

Seco (UC3M/ICMAT) Wandering property ICMAT 3 / 13

SLIDE 12

Invariant subspaces

The (forward) shift operator is bdd: S : Dα → Dα : Sf(z) = zf(z). A closed subspace V of Dα is zk- invariant if SkV ⊂ V. [f]zk(= [f]) = span{ztkf : t = 0, 1, 2, . . .}.

Theorem (Beurling, ’49)

For H2 (α = 0), M z-inv. subsp. ⇔ M = ϕH2 = [ϕ] with span(ϕ) = M ⊖ SM.

Theorem (Aleman, Richter, Sundberg, ’96)

For A2 (α = −1), M z-inv. ⇔ [M ⊖ SM] = M.

Seco (UC3M/ICMAT) Wandering property ICMAT 3 / 13

SLIDE 13

Wandering subspaces

Shimorin (’11) extended ARS’96 to α ∈ [−1, 1] (different ideas for α < 0 and α > 0). Hedenmalm-Zhu (’92) and Nowak et al. (’17) showed that the analogous fails for α ≤ −5 in some sense.

Definition

M has the zk wandering property if [M ⊖ SkM]zk = M.

Seco (UC3M/ICMAT) Wandering property ICMAT 4 / 13

SLIDE 14

Wandering subspaces

Shimorin (’11) extended ARS’96 to α ∈ [−1, 1] (different ideas for α < 0 and α > 0). Hedenmalm-Zhu (’92) and Nowak et al. (’17) showed that the analogous fails for α ≤ −5 in some sense.

Definition

M has the zk wandering property if [M ⊖ SkM]zk = M. If ∀M zk-inv. subsp. of H, M has the wandering property, then the wandering subspace property (WSP) holds for zk in H.

Seco (UC3M/ICMAT) Wandering property ICMAT 4 / 13

SLIDE 15

Wandering subspaces

Shimorin (’11) extended ARS’96 to α ∈ [−1, 1] (different ideas for α < 0 and α > 0). Hedenmalm-Zhu (’92) and Nowak et al. (’17) showed that the analogous fails for α ≤ −5 in some sense.

Definition

M has the zk wandering property if [M ⊖ SkM]zk = M. If ∀M zk-inv. subsp. of H, M has the wandering property, then the wandering subspace property (WSP) holds for zk in H. Norm dependent!

Seco (UC3M/ICMAT) Wandering property ICMAT 4 / 13

SLIDE 16

Wandering subspaces

Shimorin (’11) extended ARS’96 to α ∈ [−1, 1] (different ideas for α < 0 and α > 0). Hedenmalm-Zhu (’92) and Nowak et al. (’17) showed that the analogous fails for α ≤ −5 in some sense.

Definition

M has the zk wandering property if [M ⊖ SkM]zk = M. If ∀M zk-inv. subsp. of H, M has the wandering property, then the wandering subspace property (WSP) holds for zk in H. Norm dependent! Really!

Seco (UC3M/ICMAT) Wandering property ICMAT 4 / 13

SLIDE 17

The Problem

More work on WSP for other multipliers (i.e. mult. by an inner function) by Carswell, Duren, Khavinson, Shapiro, Stessin, Sundberg, Weir...

Seco (UC3M/ICMAT) Wandering property ICMAT 5 / 13

SLIDE 18

The Problem

More work on WSP for other multipliers (i.e. mult. by an inner function) by Carswell, Duren, Khavinson, Shapiro, Stessin, Sundberg, Weir... CDS’02: Not known whether always WSP for mult. by a finite Blaschke product in A2.

Seco (UC3M/ICMAT) Wandering property ICMAT 5 / 13

SLIDE 19

The Problem

More work on WSP for other multipliers (i.e. mult. by an inner function) by Carswell, Duren, Khavinson, Shapiro, Stessin, Sundberg, Weir... CDS’02: Not known whether always WSP for mult. by a finite Blaschke product in A2.

Conjecture

k > 1. For zk in A2, the wandering prop. holds.

Seco (UC3M/ICMAT) Wandering property ICMAT 5 / 13

SLIDE 20

The Problem

More work on WSP for other multipliers (i.e. mult. by an inner function) by Carswell, Duren, Khavinson, Shapiro, Stessin, Sundberg, Weir... CDS’02: Not known whether always WSP for mult. by a finite Blaschke product in A2.

Conjecture

k > 1. For zk in A2, the wandering prop. holds. This is the problem we study (but do not solve) today.

Seco (UC3M/ICMAT) Wandering property ICMAT 5 / 13

SLIDE 21

Results

Theorem

∀k ≥ 6, ∀α ∈ R, Dα admits equiv. norm · : zk-WSP fails.

Seco (UC3M/ICMAT) Wandering property ICMAT 6 / 13

SLIDE 22

Results

Theorem

∀k ≥ 6, ∀α ∈ R, Dα admits equiv. norm · : zk-WSP fails.

Theorem (Gallardo-Gutiérrez, Partington, Seco, ’19)

∀k ≥ 1, ∀α ∈ [−1, 1], Dα admits equiv. norm · : B-WSP holds (any B finite Blaschke product).

Seco (UC3M/ICMAT) Wandering property ICMAT 6 / 13

SLIDE 23

Results

Theorem

∀k ≥ 6, ∀α ∈ R, Dα admits equiv. norm · : zk-WSP fails.

Theorem (Gallardo-Gutiérrez, Partington, Seco, ’19)

∀k ≥ 1, ∀α ∈ [−1, 1], Dα admits equiv. norm · : B-WSP holds (any B finite Blaschke product). What about the usual norms in Dα?

Seco (UC3M/ICMAT) Wandering property ICMAT 6 / 13

SLIDE 24

Results

Theorem

∀k ≥ 6, ∀α ∈ R, Dα admits equiv. norm · : zk-WSP fails.

Theorem (Gallardo-Gutiérrez, Partington, Seco, ’19)

∀k ≥ 1, ∀α ∈ [−1, 1], Dα admits equiv. norm · : B-WSP holds (any B finite Blaschke product). What about the usual norms in Dα? k ≥ 10, α < −(5k +

700 (k−9)2 ), zk-WSP fails.

Seco (UC3M/ICMAT) Wandering property ICMAT 6 / 13

SLIDE 25

Results

Theorem

∀k ≥ 6, ∀α ∈ R, Dα admits equiv. norm · : zk-WSP fails.

Theorem (Gallardo-Gutiérrez, Partington, Seco, ’19)

∀k ≥ 1, ∀α ∈ [−1, 1], Dα admits equiv. norm · : B-WSP holds (any B finite Blaschke product). What about the usual norms in Dα? k ≥ 10, α < −(5k +

700 (k−9)2 ), zk-WSP fails.

−

log 2 log(k+1) ≤ α ≤ 1, zk-WSP holds.

Seco (UC3M/ICMAT) Wandering property ICMAT 6 / 13

SLIDE 26

Results

Theorem

∀k ≥ 6, ∀α ∈ R, Dα admits equiv. norm · : zk-WSP fails.

Theorem (Gallardo-Gutiérrez, Partington, Seco, ’19)

∀k ≥ 1, ∀α ∈ [−1, 1], Dα admits equiv. norm · : B-WSP holds (any B finite Blaschke product). What about the usual norms in Dα? k ≥ 10, α < −(5k +

700 (k−9)2 ), zk-WSP fails.

−

log 2 log(k+1) ≤ α ≤ 1, zk-WSP holds.

In z2A2, z2-WSP holds but in A2 we still DO NOT KNOW.

Seco (UC3M/ICMAT) Wandering property ICMAT 6 / 13

SLIDE 27

Results

Theorem

∀k ≥ 6, ∀α ∈ R, Dα admits equiv. norm · : zk-WSP fails.

Theorem (Gallardo-Gutiérrez, Partington, Seco, ’19)

∀k ≥ 1, ∀α ∈ [−1, 1], Dα admits equiv. norm · : B-WSP holds (any B finite Blaschke product). What about the usual norms in Dα? k ≥ 10, α < −(5k +

700 (k−9)2 ), zk-WSP fails.

−

log 2 log(k+1) ≤ α ≤ 1, zk-WSP holds.

In z2A2, z2-WSP holds but in A2 we still DO NOT KNOW. Today focus: α = −16, z6 wandering fails.

Seco (UC3M/ICMAT) Wandering property ICMAT 6 / 13

SLIDE 28

How does M look

α = −16, k = 6.

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How does M look

α = −16, k = 6. M = [F1, F2]z6

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How does M look

α = −16, k = 6. M = [F1, F2]z6 F1(z) = a0 + a1z + a2z2 + a3z3 + a4z4 + a6z6 + a7z7 + a8z8 + a9z9 F2(z) = b0 + b1z + b2z2 + b3z3 + b5z5

Seco (UC3M/ICMAT) Wandering property ICMAT 7 / 13

SLIDE 31

How does M look

α = −16, k = 6. M = [F1, F2]z6 F1(z) = a0 + a1z + a2z2 + a3z3 + a4z4 + a6z6 + a7z7 + a8z8 + a9z9 F2(z) = b0 + b1z + b2z2 + b3z3 + b5z5 a4, b5 different role: (2z6 + 5)F1(z) + (9z6 + 8)F2(z) = = (c0, ..., c3, 5a4, 8b5, c6, ..., c9, 2a4, 9b5, c12, ..., c15, 0, 0, c18, ...)

Seco (UC3M/ICMAT) Wandering property ICMAT 7 / 13

SLIDE 32

How does M look

α = −16, k = 6. M = [F1, F2]z6 F1(z) = a0 + a1z + a2z2 + a3z3 + a4z4 + a6z6 + a7z7 + a8z8 + a9z9 F2(z) = b0 + b1z + b2z2 + b3z3 + b5z5 a4, b5 different role: (2z6 + 5)F1(z) + (9z6 + 8)F2(z) = = (c0, ..., c3, 5a4, 8b5, c6, ..., c9, 2a4, 9b5, c12, ..., c15, 0, 0, c18, ...) ⇒ unique expression ⇒ “Fourier analysis”

Seco (UC3M/ICMAT) Wandering property ICMAT 7 / 13

SLIDE 33

How does M look

α = −16, k = 6. M = [F1, F2]z6 F1(z) = a0 + a1z + a2z2 + a3z3 + a4z4 + a6z6 + a7z7 + a8z8 + a9z9 F2(z) = b0 + b1z + b2z2 + b3z3 + b5z5 a4, b5 different role: (2z6 + 5)F1(z) + (9z6 + 8)F2(z) = = (c0, ..., c3, 5a4, 8b5, c6, ..., c9, 2a4, 9b5, c12, ..., c15, 0, 0, c18, ...) ⇒ unique expression ⇒ “Fourier analysis” ⇒ M = {f1(z6)F1(z) + f2(z6)F2(z) : f1, f2 ∈ Dα}

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Linear relations

f ∈ M ⊖ z6M ⇔ f, z6sFj = 0 (s ≥ 1, j = 1, 2).

Definition

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Linear relations

f ∈ M ⊖ z6M ⇔ f, z6sFj = 0 (s ≥ 1, j = 1, 2).

Definition

As,1 = z6(s−1)F1, z6sF1 = 3

h=0 aha6+hω6s+h

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SLIDE 36

Linear relations

f ∈ M ⊖ z6M ⇔ f, z6sFj = 0 (s ≥ 1, j = 1, 2).

Definition

As,1 = z6(s−1)F1, z6sF1 = 3

h=0 aha6+hω6s+h

As,2 = z6sF1, z6sF2

Seco (UC3M/ICMAT) Wandering property ICMAT 8 / 13

SLIDE 37

Linear relations

f ∈ M ⊖ z6M ⇔ f, z6sFj = 0 (s ≥ 1, j = 1, 2).

Definition

As,1 = z6(s−1)F1, z6sF1 = 3

h=0 aha6+hω6s+h

As,2 = z6sF1, z6sF2 As,3 = z6sF12, As,4 = z6sF22

Seco (UC3M/ICMAT) Wandering property ICMAT 8 / 13

SLIDE 38

Linear relations

f ∈ M ⊖ z6M ⇔ f, z6sFj = 0 (s ≥ 1, j = 1, 2).

Definition

As,1 = z6(s−1)F1, z6sF1 = 3

h=0 aha6+hω6s+h

As,2 = z6sF1, z6sF2 As,3 = z6sF12, As,4 = z6sF22 As,5 = z6(s−1)F1, z6sF2

Seco (UC3M/ICMAT) Wandering property ICMAT 8 / 13

SLIDE 39

Linear relations

f ∈ M ⊖ z6M ⇔ f, z6sFj = 0 (s ≥ 1, j = 1, 2).

Definition

As,1 = z6(s−1)F1, z6sF1 = 3

h=0 aha6+hω6s+h

As,2 = z6sF1, z6sF2 As,3 = z6sF12, As,4 = z6sF22 As,5 = z6(s−1)F1, z6sF2

Lemma

f ∈ M. Then f ⊥ z6M ⇔ ∀s ≥ 1 both 0 = ˆ f1(s+1)As+1,1+ ˆ f2(s+1)As+1,5+ ˆ f1(s)As,3+ ˆ f2(s)As,2+ ˆ f1(s−1)As,1 and 0 = ˆ f1(s)As,2 + ˆ f2(s)As,4 + ˆ f1(s − 1)As,5.

Seco (UC3M/ICMAT) Wandering property ICMAT 8 / 13

SLIDE 40

Linear conditions

Suppose f ∈ M ⊖ z6M. If A2,1 = A3,1 = A2,5 = A3,5 = 0 ⇒ ˆ f1(2) = ˆ f2(2) = 0.

Seco (UC3M/ICMAT) Wandering property ICMAT 9 / 13

SLIDE 41

Linear conditions

Suppose f ∈ M ⊖ z6M. If A2,1 = A3,1 = A2,5 = A3,5 = 0 ⇒ ˆ f1(2) = ˆ f2(2) = 0. If (previous) + A1,1 = 0, then M ⊥ z6M spanned by F2 and F4 where F4(z) = F1(z)(1 − z6A1,5A1,2 |A1,2|2 − A1,3A1,4 ) = (1 + z6/c)F1(z).

Seco (UC3M/ICMAT) Wandering property ICMAT 9 / 13

SLIDE 42

Linear conditions

Suppose f ∈ M ⊖ z6M. If A2,1 = A3,1 = A2,5 = A3,5 = 0 ⇒ ˆ f1(2) = ˆ f2(2) = 0. If (previous) + A1,1 = 0, then M ⊥ z6M spanned by F2 and F4 where F4(z) = F1(z)(1 − z6A1,5A1,2 |A1,2|2 − A1,3A1,4 ) = (1 + z6/c)F1(z). Notice then F1 ∈ [M ⊖ z6M] ⇔ 1 + z/c cyclic ⇔ c ≥ 1.

Seco (UC3M/ICMAT) Wandering property ICMAT 9 / 13

SLIDE 43

Linear conditions

Suppose f ∈ M ⊖ z6M. If A2,1 = A3,1 = A2,5 = A3,5 = 0 ⇒ ˆ f1(2) = ˆ f2(2) = 0. If (previous) + A1,1 = 0, then M ⊥ z6M spanned by F2 and F4 where F4(z) = F1(z)(1 − z6A1,5A1,2 |A1,2|2 − A1,3A1,4 ) = (1 + z6/c)F1(z). Notice then F1 ∈ [M ⊖ z6M] ⇔ 1 + z/c cyclic ⇔ c ≥ 1. Optimization problem on 12 variables, with 5 restrictions to show c < 1.

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SLIDE 44

The optimization problem

B0 := inf A1,3A1,4 − |A1,2|2 |A1,2A1,5| < 1

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SLIDE 45

The optimization problem

B0 := inf A1,3A1,4 − |A1,2|2 |A1,2A1,5| < 1 inf on (a0, ..., a3, a6, ..., a9, b0, ..., b3) ∈ C12:

Seco (UC3M/ICMAT) Wandering property ICMAT 10 / 13

SLIDE 46

The optimization problem

B0 := inf A1,3A1,4 − |A1,2|2 |A1,2A1,5| < 1 inf on (a0, ..., a3, a6, ..., a9, b0, ..., b3) ∈ C12: N     a0a6 a1a7 a2a8 a3a9     =     , N     b0a6 b1a7 b2a8 b3a9     =   A1,5   ,

Seco (UC3M/ICMAT) Wandering property ICMAT 10 / 13

SLIDE 47

The optimization problem

B0 := inf A1,3A1,4 − |A1,2|2 |A1,2A1,5| < 1 inf on (a0, ..., a3, a6, ..., a9, b0, ..., b3) ∈ C12: N     a0a6 a1a7 a2a8 a3a9     =     , N     b0a6 b1a7 b2a8 b3a9     =   A1,5   , and N is the 3 × 4 matrix given by N =   ω6 ω7 ω8 ω9 ω12 ω13 ω14 ω15 ω18 ω19 ω20 ω21   N0 :=     1 ω6 ω7 ω8 ω9 ω12 ω13 ω14 ω15 ω18 ω19 ω20 ω21    

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SLIDE 48

Further and further reductions...

Restrictions give a6, ..., a9 and b0, ..., b3 in terms of a0, ..., a3 and 3 mute variables.

Seco (UC3M/ICMAT) Wandering property ICMAT 11 / 13

SLIDE 49

Further and further reductions...

Restrictions give a6, ..., a9 and b0, ..., b3 in terms of a0, ..., a3 and 3 mute variables. Objective function, homogeneous ⇒ a0 = 1.

Seco (UC3M/ICMAT) Wandering property ICMAT 11 / 13

SLIDE 50

Further and further reductions...

Restrictions give a6, ..., a9 and b0, ..., b3 in terms of a0, ..., a3 and 3 mute variables. Objective function, homogeneous ⇒ a0 = 1. Classical calculus techniques and “good luck” reduce from 6 complex to 3 real positive variables: Find d1, d2, d3 ∈ R+ such that: 4C2C4 < 1, where

Seco (UC3M/ICMAT) Wandering property ICMAT 11 / 13

SLIDE 51

Further and further reductions...

Restrictions give a6, ..., a9 and b0, ..., b3 in terms of a0, ..., a3 and 3 mute variables. Objective function, homogeneous ⇒ a0 = 1. Classical calculus techniques and “good luck” reduce from 6 complex to 3 real positive variables: Find d1, d2, d3 ∈ R+ such that: 4C2C4 < 1, where C2 = 1 +

3

i=1

E2

i ω2k+i

di , C4 =

3

i=1

G2

i ωk+idi

E2

i

,

Seco (UC3M/ICMAT) Wandering property ICMAT 11 / 13

SLIDE 52

Further and further reductions...

Restrictions give a6, ..., a9 and b0, ..., b3 in terms of a0, ..., a3 and 3 mute variables. Objective function, homogeneous ⇒ a0 = 1. Classical calculus techniques and “good luck” reduce from 6 complex to 3 real positive variables: Find d1, d2, d3 ∈ R+ such that: 4C2C4 < 1, where C2 = 1 +

3

i=1

E2

i ω2k+i

di , C4 =

3

i=1

G2

i ωk+idi

E2

i

, N−1 =     1 E1 G1 ... ... E2 G2 ... ... E3 G3 ... ...     .

Seco (UC3M/ICMAT) Wandering property ICMAT 11 / 13

SLIDE 53

The solution to the problem

For us, this becomes something like d1, d2, d3: 4 · 10−6 · (1 + 82 d1 + 440 d2 + 194 d3 ) · (14d1 + 4d2 + d3) < 1.

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SLIDE 54

The solution to the problem

For us, this becomes something like d1, d2, d3: 4 · 10−6 · (1 + 82 d1 + 440 d2 + 194 d3 ) · (14d1 + 4d2 + d3) < 1. A simple educated guess d = (1, 4, 6) gives a good enough result (lhs < 0.033), and then B0 < 0.22.

Seco (UC3M/ICMAT) Wandering property ICMAT 12 / 13

SLIDE 55

The solution to the problem

For us, this becomes something like d1, d2, d3: 4 · 10−6 · (1 + 82 d1 + 440 d2 + 194 d3 ) · (14d1 + 4d2 + d3) < 1. A simple educated guess d = (1, 4, 6) gives a good enough result (lhs < 0.033), and then B0 < 0.22.

Remark

Changing those 12 values in the adequate place of the sequence ω will give an equiv. norm in any Dα with the same result for any k ≥ 6.

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SLIDE 56

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