Sheafification Functor and Arbitrary Monetary Value Measure. Part 3

Here are Part 1 and Part 2

Let A be a fixed set of axioms. Then for a given arbitrary monetary value measure Ψ can we make a good alternative for it? In other words, can we find a monetary value measure that satisfies A and is the best approximation of the original Ψ? For a Grothendieck topology J on χ, define Sh(χ, J) ⊂ Setχop to be a full sub-category whose objects are all sheaves for J. Then, it is well known that  a left adjoint πJ in the following diagram.

Sh(χ, J) → Setχop

Sh(χ, J) ←πJ Setχop

πJ (Ψ) ← Ψ

The functor πJ is known as Sheafification functor, which has the following limit cone:

img_20170117_125305

for sieves I, K and U. This also satisfies the following theorem.

1.0 If πJ (Ψ) is a sheaf for J

1.1 If Ψ is a sheaf for J, then for any U ∈ χ,  πJ (Ψ)(U) ≅ L(U)

The theorem suggests that for an arbitrary monetary value measure, the sheafification functor provides one of its closest monetary value measures that may satisfy the given set of axioms. To make this certain, we need a following definition.

2.0 Let A be a set of axioms of monetary value measures

2.1 M(A) := the collection of all monetary value measures satisfying A

2.2 MO := collection of all monetary value measures

2.3 A is called complete if

πJM(A) (MO) ⊂ M(A)

3.0 Let A be a complete set of axioms. Then, for a monetary value measure Ψ ∈ MO, πJM(A(Ψ) is the monetary value measure that is the best approximation satisfying A.

Let us investigate if the set of axioms of concave monetary value measures is complete in the case of Ω = {1, 2, 3} with a σ-field F := 2Ω

We enumerate all possible sub-σ-fields of Ω, that is, the shape of the category χ = χ(Ω),

img_20170117_132606

where,

U := F := 2Ω

U1 := {Φ, {1}, {2, 3}, Ω}

U2 := {Φ, {2}, {1, 3}, Ω}

U3 := {Φ, {3}, {1, 2}, Ω}

U4 := {Φ, Ω}

The Banach spaces defined by the elements of χ are

L := L := L(U) := {a, b, c | a, b, c ∈ ℜ}

L1 := L(U1) := {a, b, b | a, b ∈ ℜ}

L2 := L(U2) := {a, b, a | a, b ∈ ℜ}

L3 := L(U3) := {a, a, c | a, c ∈ ℜ}

L0 := L(U0) := {a, a, a, | a ∈ ℜ}

Then a monetary value measure Ψ : χop → Set on χ is determined by the following six functions

img_20170117_142341

We will investigate its concrete shape one by one by considering axioms it satisfies.

For Ψ1 : L → L1, we have by the cash invariance axiom,

Ψ1 (a, b, c) = Ψ1 ((0, b – c, 0) + (a, c, c))

= Ψ1 ((0, b – c, 0)) + (a, c, c)

= (f12 (b – c), f11 (b – c), f11 (b – c)) + (a, c, c)

= (f12 (b – c) + a, f11 (b – c) + c, f11 (b – c)+ c)

where f11, f12 : ℜ → ℜ are defined by (f12(x), f11(x), f11(x)) = Ψ1 (0, x, 0).

Similarly, if we define nine functions

f11, f12, f21, f22, f31, f32, g1, g2, g3ℜ → ℜ by

(f12(x), f11(x), f11(x)) = Ψ1(0, x, 0)

(f21(x), f22(x), f21(x)) = Ψ2(0, 0, x)

(f31(x), f31(x), f32(x)) = Ψ3(x, 0, 0)

(g1(x), g1(x), g1(x)) = Ψ01(x, 0, 0)

(g2(x), g2(x), g2(x)) = Ψ02(0, x, 0)

(g3(x), g3(x), g3(x)) = Ψ03(0, 0, x)

We can represent the original six functions by nine functions

Ψ1(a, b, c) = (f12(b – c) + a, f11(b – c) + c, f11(b – c) + c),

Ψ2(a, b, c) = (f21(c – a) + a, f22(c – a) + a, f21(c – a) + a),

Ψ3(a, b, c) = (f31(a – b) + b, f31(a – b) + b, f32(a – b) + c),

Ψ01(a, b, b) = (g1(a – b) + b, g1(a – b) + b, g1(a – b) + b),

Ψ02(a, b, a) = (g2(b – a) + a, g2(b – a) + a, g2(b – a) + a),

Ψ03(a, a, c) = (g3(c – a) + a, g3(c – a) + a, g3(c – a) + a)

Next by the normahzation axiom, we have

f11(0) = f12(0) = f21(0) = f22(0) = f31(0) = f32(0) = g1(0) = g2(0) = g3(0) = 0

partially differentiating the function in Ψ1(a, b, c), we have

Ψ1(a, b, c)/∂a = (1, 0, 0)

∂Ψ1(a, b, c)/∂b = (f’12(b – c), f’11(b – c), f’11(b – c))

Ψ1(a, b, c)/∂c = (- f’12(b – c), 1 –  f’11(b – c), 1 –  f’11(b – c))

Therefore, by the monotonicity, we have f’12(x) = 0 and 0 ≤ f’11 ≤ 1. Then by the result of the normalization axiom, we have

x ∈ ℜ, f12(x) = 0. Hence, ∀ x ∈ ℜ,

f12(x) = f22(x) = f32(x) = 0

With this knowledge, let us redefine the three functions f1, f2, f3 : ℜ → ℜ by

(0, f1(x), f1(x)) = Ψ1(0, x, 0)

(f2(x), 0, f2(x)) = Ψ2(0, 0, x)

(f3(x), f3(x), 0) = Ψ3(x, 0, 0)

Then, we have a new representation of the original six functions

Ψ1(a, b, c) = (a, f1(b – c) + c, f1(b – c) + c)

Ψ2(a, b, c) = (f2(c – a) + a, b, f2(c – a) + a)

Ψ3(a, b, c) = (f3(a – b) + b, f3(a – b) + b, c)

Ψ01(a, b, b) = (g1(a – b) + b, g1(a – b) + b, g1(a – b) + b)

Ψ02(a, b, a) = (g2(b – a) + a, g2(b – a) + a, g2(b – a) + a)

Ψ03(a, a, c) = (g3(c – a) + a, g3(c – a) + a, g3(c – a) + a)

Thinking about the composition rule, we have

Ψ0 = Ψ01 o Ψ1 = Ψ02 o Ψ2 = Ψ03 o Ψ3

g1(a – f1(b – c) – c) + f1(b – c) + c

= g2(b – f2(c – a) – a) + f2(c – a) + a

=g3(c – f3(a – b) – b) + f3(a – b) + b

………..

 

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