Polynomials on Banach spaces

DefinitionContinuous polynomial functions on a complex normed space#✓fully proved

Let \(E\) be a complex normed space. A function \(F : E \to \mathbb{C}\) is a continuous polynomial function if there are a natural number \(n\), natural numbers \(m_1, \dots, m_n\) and continuous \(\mathbb{C}\)-multilinear maps \(A_i : E^{m_i} \to \mathbb{C}\), with \(A_i\) taking \(m_i\) arguments, such that \(F(x) = \sum_{i=1}^{n} A_i(x, \dots, x)\) for every \(x \in E\).

A \(0\)-multilinear map is a constant, so constant functions are polynomial functions.

def IsPolynomialOn (E : Type) [NormedAddCommGroup E] [NormedSpace ℂ E]
    (F : E → ℂ) : Prop :=
  ∃ (n : ℕ) (m : Fin n → ℕ)
    (A : (i : Fin n) → (E [× m i]→L[ℂ] ℂ)),
    ∀ x, F x = ∑ i, A i fun _ => x

Used by 1

Used byBezout identity for polynomials without a common zero

TheoremBezout identity for polynomials without a common zero#●ready to prove

Let \(E\) be a complex Banach space and let \(F, G : E \to \mathbb{C}\) be continuous polynomial functions with no common zero, that is, \(G(x) \neq 0\) whenever \(F(x) = 0\). Then there are continuous polynomial functions \(\varphi, \psi : E \to \mathbb{C}\) such that \(F(x) \varphi(x) + G(x) \psi(x) = 1\) for every \(x \in E\).

This is the identity recorded in the addendum to Problem 27 of the Scottish Book (Mazur-Orlicz, 4 April 1939). The problem asks whether two complex polynomials on a complex Banach space that both tend to \(0\) along a sequence \((x_n)\) in the unit ball must have a common zero. The answer is positive and follows from the identity by contradiction: continuous polynomials are bounded on the unit ball, so \(F(x_n) \varphi(x_n) + G(x_n) \psi(x_n)\) would tend to \(0\), yet it equals \(1\).

theorem_wanted bezout_identity_for_polynomials_without_common_zero :
    ∀ (E : Type) [NormedAddCommGroup E] [NormedSpace ℂ E] [CompleteSpace E]
      (F G : E → ℂ),
      IsPolynomialOn E F → IsPolynomialOn E G →
      (∀ x, F x = 0 → G x ≠ 0) →
      ∃ (φ ψ : E → ℂ), IsPolynomialOn E φ ∧ IsPolynomialOn E ψ ∧
        ∀ x, F x * φ x + G x * ψ x = 1

Sources

• S. Mazur and W. Orlicz, Problem 27 of the Scottish Book, with the positive answer via the Bezout identity in the addendum; see the commentary in R. D. Mauldin (ed.), The Scottish Book, 2nd ed. (Birkhäuser, 2015).

Uses 1

Statement usesContinuous polynomial functions on a complex normed space

DefinitionMazur's factorisation claim (Scottish Book Problem 55)#✓fully proved

The claim that for every real Banach space \(E\), every unbounded set \(R \subseteq E\) and every continuous bilinear form \(B : E \times E \to \mathbb{R}\), if the quadratic form \(W(x) = B(x, x)\) is bounded on the set of points at distance less than \(\varepsilon\) from \(R\) for some \(\varepsilon > 0\), then there are a continuous linear functional \(L : E \to \mathbb{R}\), a constant \(c \in \mathbb{R}\) and a real polynomial \(V\) such that \(W(x) = V(L(x) + c)\) for all \(x \in E\) and \(|L(r) + c|\) is bounded for \(r \in R\).

This is Mazur's question in the general Banach-space form for scalar-valued 2-polynomials: \(W\) factors as \(V \circ \varphi\) through the first-degree polynomial \(\varphi(x) = L(x) + c\), which maps \(R\) to a bounded set.

def ScottishBookProblem55Claim : Prop :=
  ∀ (E : Type) [NormedAddCommGroup E] [NormedSpace ℝ E] [CompleteSpace E]
    (R : Set E) (B : E →L[ℝ] E →L[ℝ] ℝ),
    (∀ M : ℝ, ∃ r ∈ R, M < ‖r‖) →
    (∃ ε : ℝ, 0 < ε ∧ ∃ C : ℝ,
      ∀ x : E, (∃ r ∈ R, dist x r < ε) → |B x x| ≤ C) →
    ∃ (L : E →L[ℝ] ℝ) (c : ℝ) (V : Polynomial ℝ),
      (∀ x : E, B x x = V.eval (L x + c)) ∧
      ∃ M : ℝ, ∀ r ∈ R, |L r + c| ≤ M

Used by 1

Used byMazur's Problem 55 has a negative answer

TheoremMazur's Problem 55 has a negative answer#●ready to prove

The factorisation claim of Scottish Book Problem 55 is false: there are a real Banach space \(E\), an unbounded set \(R \subseteq E\) and a continuous bilinear form \(B : E \times E \to \mathbb{R}\) such that for some \(\varepsilon > 0\) the quadratic form \(W(x) = B(x, x)\) is bounded on the set of points at distance less than \(\varepsilon\) from \(R\), yet there are no continuous linear functional \(L : E \to \mathbb{R}\), constant \(c \in \mathbb{R}\) and real polynomial \(V\) with \(W(x) = V(L(x) + c)\) for all \(x \in E\) and \(|L(r) + c|\) bounded for \(r \in R\).

theorem_wanted mazur55_refuted : ¬ ScottishBookProblem55Claim

Sources

• S. Mazur, Problem 55 of the Scottish Book; refuted by V. Pestov, who constructs a continuous scalar-valued 2-polynomial W on the separable Hilbert space ℓ² and an unbounded set R such that W is bounded on an ε-neighbourhood of R but does not factor through any bounded first-degree polynomial sending R to a bounded set ("Two 1935 questions of Mazur about polynomials in Banach spaces: a counter-example", 1997, arXiv:funct-an/9705002). The same counterexample refutes Problem 75. In finite dimensions the answer is positive (Auerbach), so the infinite-dimensional Banach setting is essential.

Uses 1

Statement usesMazur's factorisation claim (Scottish Book Problem 55)

DefinitionScottish Book Problem 75 (Mazur), general Banach-space form#✓fully proved

The claim that for every real Banach space \(E\), every set \(R \subseteq E\), every continuous bilinear form \(B\) on \(E\) and every real \(\alpha \ne 0\) the following holds: if for some \(\varepsilon > 0\) the values \(|B(x, x)|\) are bounded on the open \(\varepsilon\)-neighbourhood of \(R\), that is, for the \(x \in E\) with \(\operatorname{dist}(x, r) < \varepsilon\) for some \(r \in R\), then for some \(\delta > 0\) the values \(|B(x, x)|\) are bounded on the open \(\delta\)-neighbourhood of the scaled set \(\alpha R = \{\alpha r : r \in R\}\).

def ScottishBookProblem75Claim : Prop :=
  ∀ (E : Type) [NormedAddCommGroup E] [NormedSpace ℝ E] [CompleteSpace E]
    (R : Set E) (B : E →L[ℝ] E →L[ℝ] ℝ) (α : ℝ), α ≠ 0 →
    (∃ ε : ℝ, 0 < ε ∧ ∃ C : ℝ,
      ∀ x : E, (∃ r ∈ R, dist x r < ε) → |B x x| ≤ C) →
    ∃ δ : ℝ, 0 < δ ∧ ∃ C : ℝ,
      ∀ x : E, (∃ r ∈ R, dist x (α • r) < δ) → |B x x| ≤ C

Used by 1

Used byMazur's Scottish Book Problem 75 fails in general Banach spaces

TheoremMazur's Scottish Book Problem 75 fails in general Banach spaces#●ready to prove

The Scottish Book Problem 75 claim is false: there are a real Banach space \(E\), a set \(R \subseteq E\), a continuous bilinear form \(B\) on \(E\) and a real \(\alpha \ne 0\) such that \(|B(x, x)|\) is bounded on some open \(\varepsilon\)-neighbourhood of \(R\) with \(\varepsilon > 0\), but is unbounded on every open \(\delta\)-neighbourhood of \(\alpha R = \{\alpha r : r \in R\}\) with \(\delta > 0\).

theorem_wanted mazur75_refuted : ¬ ScottishBookProblem75Claim

Sources

• S. Mazur, Problem 75 of the Scottish Book; refuted by V. Pestov, whose counterexample to Problem 55 is a continuous scalar-valued 2-polynomial W on ℓ², bounded on an ε-neighbourhood of an unbounded set R, but unbounded on (1 / 2) R, hence unbounded on every neighbourhood of (1 / 2) R ("Two 1935 questions of Mazur about polynomials in Banach spaces: a counter-example", 1997, arXiv:funct-an/9705002). Plichko and Zagorodnyuk gave an independent negative solution ("On automatic continuity and three problems of 'The Scottish Book' concerning the boundedness of polynomial functionals", J. Math. Anal. Appl. 220 (1998), 477–494). In the Euclidean case the statement is true, per typescript Ad 75.

Uses 1

Statement usesScottish Book Problem 75 (Mazur), general Banach-space form

3 fully proved3 ready to prove

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