Fréchet Differentiability of the Squaring Map #
This file establishes that the squaring map a ↦ a * a in the Banach algebra ℓ¹_ν
is Fréchet differentiable, with derivative D(a * a)h = 2(a * h).
Mathematical Background #
Theorem 7.4.7 (Differentiability in Banach Algebras) #
In any Banach algebra X with product ·, the squaring map G(x) = x · x is
Fréchet differentiable with derivative:
DG(x)h = x · h + h · x
Remark 7.4.8 (Commutative Case) #
When the algebra is commutative (as ℓ¹_ν is by Corollary 7.4.5), this simplifies to:
DG(x)h = 2(x · h)
Implementation Notes #
Why the Proofs Are So Simple #
Since l1Weighted ν has CommRing and NormedRing instances (from lpWeighted.lean),
we use ring operations *, + directly. The key algebraic identity:
(a + h)² - a² - 2ah = h²
reduces to a single ring tactic call, because the CommRing instance provides
all the necessary axioms (associativity, distributivity, commutativity).
Comparison: Before vs After Refactoring #
Before (manual Cauchy product proofs):
lemma sq_expansion (a h : l1Weighted ν) : ... := by
apply lpWeighted.ext; intro n
simp only [lpWeighted.sub_toSeq, lpWeighted.add_toSeq, sq_toSeq, leftMul_toSeq]
simp only [CauchyProduct.apply]
simp only [lpWeighted.add_toSeq]
rw [← Finset.sum_add_distrib, ← Finset.sum_sub_distrib, ← Finset.sum_sub_distrib]
apply Finset.sum_congr rfl; intro kl _; ring
After (using Ring instance):
lemma sq_expansion (a h : l1Weighted ν) : ... := by
simp only [sq, leftMul_apply]; ring
The refactoring philosophy: algebraic properties are inherited from typeclass instances, not reproven at each use site.
Role of SMulCommClass and IsScalarTower #
The instances SMulCommClass ℝ (l1Weighted ν) (l1Weighted ν) and
IsScalarTower ℝ (l1Weighted ν) (l1Weighted ν) (defined in lpWeighted.lean)
enable lemmas like smul_mul_assoc and smul_comm, which are used in:
leftMul.map_smul': Provinga * (c • h) = c • (a * h)leftMul_smul: ProvingleftMul (c • a) = c • leftMul a
These instances are proven using CauchyProduct.smul_mul and CauchyProduct.mul_smul.
Main Results #
leftMul: Left multiplicationh ↦ a * has a continuous linear maphasFDerivAt_sq: The squaring map has Fréchet derivative2 • leftMul adifferentiable_sq: The squaring map is differentiable everywherehasFDerivAt_F_sub_const: The mapF(a) = a * a - chas derivative2 • leftMul a
References #
- Theorem 7.4.7: Fréchet differentiability in Banach algebras
- Remark 7.4.8: Derivative formula for commutative case
- Corollary 7.4.5: ℓ¹_ν is a commutative Banach algebra
Part 1: Left Multiplication as Continuous Linear Map #
For fixed a ∈ ℓ¹_ν, the map h ↦ a * h is a bounded linear operator.
- Linearity: From Ring distributivity
a * (h₁ + h₂) = a * h₁ + a * h₂ - Boundedness: From submultiplicativity
‖a * h‖ ≤ ‖a‖ · ‖h‖
Left multiplication by a fixed element: h ↦ a * h
This is the key map appearing in the derivative formula DG(a)h = 2(a * h).
Construction: We use LinearMap.mkContinuous with:
- Linearity from Ring distributivity
- Scalar compatibility from
smul_comm - Operator norm bound from submultiplicativity
Equations
- l1Weighted.leftMul a = { toFun := fun (h : ↥(l1Weighted ν)) => a * h, map_add' := ⋯, map_smul' := ⋯ }.mkContinuous ‖a‖ ⋯
Instances For
Operator norm bound: ‖leftMul a‖ ≤ ‖a‖
Definitional equality: (leftMul a) h = a * h
The underlying sequence of (leftMul a) h equals the Cauchy product.
Linearity of leftMul in the First Argument #
These lemmas show that the map a ↦ leftMul a is itself linear, making
leftMul a bilinear operation. This is used in showing that the derivative
is a continuous linear map.
Scalar multiplication distributes: leftMul (c • a) = c • leftMul a
Uses smul_mul_assoc from IsScalarTower instance.
The Squaring Map #
The squaring map: a ↦ a * a
This is the main object of study. We prove it has Fréchet derivative 2(a * ·).
Equations
- l1Weighted.sq a = a * a
Instances For
Part 2: Fréchet Differentiability of the Squaring Map #
We prove that sq : ℓ¹_ν → ℓ¹_ν is Fréchet differentiable at every point a,
with derivative D(sq)(a) = 2 · leftMul a.
The key steps are:
- Algebraic identity:
(a+h)² - a² - 2ah = h² - Norm estimate:
‖h²‖ ≤ ‖h‖²(submultiplicativity) - Little-o bound:
‖h‖² = o(‖h‖)as h → 0
Together these show the remainder is o(‖h‖), which is the Fréchet condition.
Key algebraic identity for the Taylor expansion of squaring.
(a + h)² - a² - (ah + ha) = h²
In a commutative ring, ah = ha, so this becomes (a + h)² - a² - 2ah = h².
Proof: A single ring tactic, thanks to the CommRing instance!
Commutativity of leftMul: leftMul a h = leftMul h a
Direct consequence of mul_comm from CommRing instance.
Norm estimate for the remainder: ‖sq(a+h) - sq(a) - 2(a*h)‖ ≤ ‖h‖²
This is the key analytic estimate. By sq_expansion_comm, the LHS is ‖h * h‖,
and by submultiplicativity ‖h * h‖ ≤ ‖h‖ · ‖h‖ = ‖h‖².
This shows the remainder is O(‖h‖²), hence o(‖h‖).
Theorem 7.4.7 (for ℓ¹_ν): The squaring map has Fréchet derivative 2(a * ·)
For G(a) = a * a, we have DG(a)h = 2(a * h).
Proof outline:
- By
sq_remainder_norm: ‖G(a+h) - G(a) - 2(a*h)‖ ≤ ‖h‖² - For ‖h‖ < ε, we have ‖h‖² ≤ ε · ‖h‖
- This gives the little-o condition: remainder = o(‖h‖)
The squaring map is differentiable everywhere.
The Fréchet derivative of the squaring map.
Part 3: Application to F(a) = a * a - c #
For the ODE example (Section 7.7), we consider F(a) = a * a - c where c is a fixed sequence encoding the parameter. This is a translation of the squaring map.
The key point is that the derivative is unchanged: DF(a) = 2(a * ·), since differentiating a constant gives zero.
The map F(a) = a * a - c appearing in Section 7.7.
For the parameterized equilibrium problem x² - λ = 0, the sequence formulation becomes a * a - c = 0 where c encodes the parameter λ.
Equations
- l1Weighted.F_sub_const c a = l1Weighted.sq a - c
Instances For
F(a) = a * a - c has Fréchet derivative 2(a * ·) at a.
The constant c disappears upon differentiation.
F is differentiable everywhere.
The Fréchet derivative of F.