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Convex Analysis And Optimization

Convex Analysis And OptimizationChap.1 Basic Convexity Concepts

Section1.4 Relative Interior, Closure, and Continuity

Naoki Ito

July 8, 2012

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Convex Analysis And Optimization

Basic Convexity Concepts

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The proof of Proposition 1.4.1

Proposition 1.4.1

C : Nonempty convex set

a (Line Segment Principle)x ∈ ri(C), x ∈ cl(C) ⇒ αx+ (1− α)x ∈ ri(C) for ∀α ∈ (0, 1]

Proof)(i)The case in x ∈ CLet x = αx+ (1− α)x (α ∈ (0, 1])

Since x ∈ ri(C),∃δ > 0 s.t. B(x, δ) ∩ aff(C) ⊂ C

The each point of B(xα, αδ)∩ aff(C) is a convex combinationof x and some point of B(x, δ) ∩ aff(C)

Therefore, by the covexity of C, B(xα, αδ) ∩ aff(C) ⊂ C,implying that xα ∈ ri(C)

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The proof of Proposition 1.4.1(a)(cont)

(ii)The case in x /∈ C

Let xk ⊂ C −→ x

Let xk,α = αx+ (1− α)xk (α ∈ (0, 1])xk,α → xα

As (i), we see that

∃δ > 0 s.t. B(x, δ) ∩ aff(C) ⊂ C

and B(xk,α, αδ) ∩ aff(C) ⊂ C for allk ≥ 0

Since xk,α → xα

∃N ≥ 0, ∀k ≥ N s.t. B

(xα,

αδ

2

)⊂ B(xk,α, αδ)

It follows B(xα,αδ2 ) ∩ aff(C) ⊂ C

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The proof of Proposition 1.4.1(a)(cont)

(ii)The case in x /∈ C

Let xk ⊂ C −→ x

Let xk,α = αx+ (1− α)xk (α ∈ (0, 1])xk,α → xα

As (i), we see that

∃δ > 0 s.t. B(x, δ) ∩ aff(C) ⊂ C

and B(xk,α, αδ) ∩ aff(C) ⊂ C for allk ≥ 0

Since xk,α → xα

∃N ≥ 0, ∀k ≥ N s.t. B

(xα,

αδ

2

)⊂ B(xk,α, αδ)

It follows B(xα,αδ2 ) ∩ aff(C) ⊂ C

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The proof of Proposition 1.4.1(a)(cont)

(ii)The case in x /∈ C

Let xk ⊂ C −→ x

Let xk,α = αx+ (1− α)xk (α ∈ (0, 1])xk,α → xα

As (i), we see that

∃δ > 0 s.t. B(x, δ) ∩ aff(C) ⊂ C

and B(xk,α, αδ) ∩ aff(C) ⊂ C for allk ≥ 0

Since xk,α → xα

∃N ≥ 0, ∀k ≥ N s.t. B

(xα,

αδ

2

)⊂ B(xk,α, αδ)

It follows B(xα,αδ2 ) ∩ aff(C) ⊂ C

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The proof of Proposition 1.4.1(a)(cont)

(ii)The case in x /∈ C

Let xk ⊂ C −→ x

Let xk,α = αx+ (1− α)xk (α ∈ (0, 1])xk,α → xα

As (i), we see that

∃δ > 0 s.t. B(x, δ) ∩ aff(C) ⊂ C

and B(xk,α, αδ) ∩ aff(C) ⊂ C for allk ≥ 0

Since xk,α → xα

∃N ≥ 0, ∀k ≥ N s.t. B

(xα,

αδ

2

)⊂ B(xk,α, αδ)

It follows B(xα,αδ2 ) ∩ aff(C) ⊂ C

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Proposition 1.4.1(b)

Proposition 1.4.1

C : Nonempty convex set

b (Nonemptiness of Relative interior)ri(C) is a nonempty convex setaff(C)=aff(ri(C))

Furthermore...dim(aff(C)):= m

m > 0 ⇒ ∃x0, x1, ..., xm ∈ ri(C)

s.t. span{x1 − x0, ..., xm − x0}= {subspace parallel to aff(C)}

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The proof of proposition 1.4.1(b)

Proof)(Convexity of ri(C))Convexity of ri(C) follows from the Line Segment Principle.

ri(C) ⊂ C ⊂ cl(C)

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The proof of proposition 1.4.1(b)(cont)

( Nonemptiness of ri(C) and aff(C)=aff(ri(C)) )We assume without loss of generality that 0 ∈ Cdim(aff(C)) := m

(i)The case in m = 0

C 6= φ → C = {0}aff(C) = {0}ri(C) = {0}, thus, ri(C) 6= φ

aff(ri(C)) = {0} thus, aff(C)=aff(ri(C))

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The proof of proposition 1.4.1(b)(cont)

( Nonemptiness of ri(C) and aff(C)=aff(ri(C)) )We assume without loss of generality that 0 ∈ Cdim(aff(C)) := m

(i)The case in m = 0

C 6= φ → C = {0}

aff(C) = {0}ri(C) = {0}, thus, ri(C) 6= φ

aff(ri(C)) = {0} thus, aff(C)=aff(ri(C))

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The proof of proposition 1.4.1(b)(cont)

( Nonemptiness of ri(C) and aff(C)=aff(ri(C)) )We assume without loss of generality that 0 ∈ Cdim(aff(C)) := m

(i)The case in m = 0

C 6= φ → C = {0}aff(C) = {0}

ri(C) = {0}, thus, ri(C) 6= φ

aff(ri(C)) = {0} thus, aff(C)=aff(ri(C))

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The proof of proposition 1.4.1(b)(cont)

( Nonemptiness of ri(C) and aff(C)=aff(ri(C)) )We assume without loss of generality that 0 ∈ Cdim(aff(C)) := m

(i)The case in m = 0

C 6= φ → C = {0}aff(C) = {0}ri(C) = {0}, thus, ri(C) 6= φ

aff(ri(C)) = {0} thus, aff(C)=aff(ri(C))

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The proof of proposition 1.4.1(b)(cont)

( Nonemptiness of ri(C) and aff(C)=aff(ri(C)) )We assume without loss of generality that 0 ∈ Cdim(aff(C)) := m

(i)The case in m = 0

C 6= φ → C = {0}aff(C) = {0}ri(C) = {0}, thus, ri(C) 6= φ

aff(ri(C)) = {0} thus, aff(C)=aff(ri(C))

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The proof of proposition 1.4.1(b)(cont)

(ii)The case in m > 0

We can take m linearly independent vectorsz1, ..., zm ∈ C that span aff(C)

Consider the setX = {x|x =

∑mi=1 αizi,

∑mi=1 αi < 1, αi > 0 (i = 1, ...,m)}

( X is open. X ∈ C)

We will show

X is open relative to aff(C)

∀x ∈ X, ∃δ > 0 s.t. B(x, δ) ∩ aff(C) ⊂ X

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The proof of proposition 1.4.1(b)(cont)

(ii)The case in m > 0

We can take m linearly independent vectorsz1, ..., zm ∈ C that span aff(C)

Consider the setX = {x|x =

∑mi=1 αizi,

∑mi=1 αi < 1, αi > 0 (i = 1, ...,m)}

( X is open. X ∈ C)

We will show

X is open relative to aff(C)

∀x ∈ X, ∃δ > 0 s.t. B(x, δ) ∩ aff(C) ⊂ X

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The proof of proposition 1.4.1(b)(cont)

(ii)The case in m > 0

We can take m linearly independent vectorsz1, ..., zm ∈ C that span aff(C)

Consider the setX = {x|x =

∑mi=1 αizi,

∑mi=1 αi < 1, αi > 0 (i = 1, ...,m)}

( X is open. X ∈ C)

We will show

X is open relative to aff(C)

∀x ∈ X, ∃δ > 0 s.t. B(x, δ) ∩ aff(C) ⊂ X

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The proof of proposition 1.4.1(b)(cont)

Fix x ∈ X and let x ∈aff(C)We have unique m-dimensional vectors α, αsuch that x = Zα, x = Zα (where Z := [z1, ..., zm])

Since Z ′Z is symmetric and positive definite,we have for some γ > 0, which is independent of x and x,

‖x− x‖2 = (α− α)′Z ′Z(α− α) ≤ γ‖α− α‖2

Since x ∈ X, α lies in the open set

A =

{(α1, ..., αm)|

m∑i=1

αi < 1, αi > 0 (i = 1, ...,m)

}

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The proof of proposition 1.4.1(b)(cont)

Fix x ∈ X and let x ∈aff(C)We have unique m-dimensional vectors α, αsuch that x = Zα, x = Zα (where Z := [z1, ..., zm])

Since Z ′Z is symmetric and positive definite,we have for some γ > 0, which is independent of x and x,

‖x− x‖2 = (α− α)′Z ′Z(α− α) ≤ γ‖α− α‖2

Since x ∈ X, α lies in the open set

A =

{(α1, ..., αm)|

m∑i=1

αi < 1, αi > 0 (i = 1, ...,m)

}

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The proof of proposition 1.4.1(b)(cont)

Fix x ∈ X and let x ∈aff(C)We have unique m-dimensional vectors α, αsuch that x = Zα, x = Zα (where Z := [z1, ..., zm])

Since Z ′Z is symmetric and positive definite,we have for some γ > 0, which is independent of x and x,

‖x− x‖2 = (α− α)′Z ′Z(α− α) ≤ γ‖α− α‖2

Since x ∈ X, α lies in the open set

A =

{(α1, ..., αm)|

m∑i=1

αi < 1, αi > 0 (i = 1, ...,m)

}

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The proof of proposition 1.4.1(b)(cont)

Take the δ such that the α where γ‖α− α‖2 < δ belongs toA. (i.e. x = Zα belongs to X)

Such the δ can be constructed as below.Since A is open set, ∃ε > 0 s.t. B(α, ε) ⊂ Aδ := γε2 then, γ‖α− α‖2 < δ = γε2 ⇒ ‖α− α‖2 < ε2

...implying that x ∈ X

Hence X contains B(x)∩aff(C), so X is open relative toaff(C).

It follows that ∀x ∈ X, x ∈ ri(C), so that ri(C) 6= φ.

Also, aff(X) = aff(C) (since by construction),X ⊂ri(C)(⊂ C),we see that aff(ri(C))=ri(C).

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The proof of proposition 1.4.1(b)(cont)

Take the δ such that the α where γ‖α− α‖2 < δ belongs toA. (i.e. x = Zα belongs to X)

Such the δ can be constructed as below.Since A is open set, ∃ε > 0 s.t. B(α, ε) ⊂ Aδ := γε2 then, γ‖α− α‖2 < δ = γε2 ⇒ ‖α− α‖2 < ε2

...implying that x ∈ X

Hence X contains B(x)∩aff(C), so X is open relative toaff(C).

It follows that ∀x ∈ X, x ∈ ri(C), so that ri(C) 6= φ.

Also, aff(X) = aff(C) (since by construction),X ⊂ri(C)(⊂ C),we see that aff(ri(C))=ri(C).

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The proof of proposition 1.4.1(b)(cont)

Take the δ such that the α where γ‖α− α‖2 < δ belongs toA. (i.e. x = Zα belongs to X)

Such the δ can be constructed as below.Since A is open set, ∃ε > 0 s.t. B(α, ε) ⊂ Aδ := γε2 then, γ‖α− α‖2 < δ = γε2 ⇒ ‖α− α‖2 < ε2

...implying that x ∈ X

Hence X contains B(x)∩aff(C), so X is open relative toaff(C).

It follows that ∀x ∈ X, x ∈ ri(C), so that ri(C) 6= φ.

Also, aff(X) = aff(C) (since by construction),X ⊂ri(C)(⊂ C),we see that aff(ri(C))=ri(C).

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The proof of proposition 1.4.1(b)(cont)

Take the δ such that the α where γ‖α− α‖2 < δ belongs toA. (i.e. x = Zα belongs to X)

Such the δ can be constructed as below.Since A is open set, ∃ε > 0 s.t. B(α, ε) ⊂ Aδ := γε2 then, γ‖α− α‖2 < δ = γε2 ⇒ ‖α− α‖2 < ε2

...implying that x ∈ X

Hence X contains B(x)∩aff(C), so X is open relative toaff(C).

It follows that ∀x ∈ X, x ∈ ri(C), so that ri(C) 6= φ.

Also, aff(X) = aff(C) (since by construction),X ⊂ri(C)(⊂ C),we see that aff(ri(C))=ri(C).

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The proof of proposition 1.4.1(b)(cont)

Take the δ such that the α where γ‖α− α‖2 < δ belongs toA. (i.e. x = Zα belongs to X)

Such the δ can be constructed as below.Since A is open set, ∃ε > 0 s.t. B(α, ε) ⊂ Aδ := γε2 then, γ‖α− α‖2 < δ = γε2 ⇒ ‖α− α‖2 < ε2

...implying that x ∈ X

Hence X contains B(x)∩aff(C), so X is open relative toaff(C).

It follows that ∀x ∈ X, x ∈ ri(C), so that ri(C) 6= φ.

Also, aff(X) = aff(C) (since by construction),X ⊂ri(C)(⊂ C),we see that aff(ri(C))=ri(C).

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The proof of proposition 1.4.1(b)(cont)

Take the δ such that the α where γ‖α− α‖2 < δ belongs toA. (i.e. x = Zα belongs to X)

Such the δ can be constructed as below.Since A is open set, ∃ε > 0 s.t. B(α, ε) ⊂ Aδ := γε2 then, γ‖α− α‖2 < δ = γε2 ⇒ ‖α− α‖2 < ε2

...implying that x ∈ X

Hence X contains B(x)∩aff(C), so X is open relative toaff(C).

It follows that ∀x ∈ X, x ∈ ri(C), so that ri(C) 6= φ.

Also, aff(X) = aff(C) (since by construction),X ⊂ri(C)(⊂ C),we see that aff(ri(C))=ri(C).

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Proposition 1.4.1(b) (written again)

m > 0 ⇒ ∃x0, x1, ..., xm ∈ ri(C)

s.t. span{x1 − x0, ..., xm − x0}= {subspace parallel to aff(C)}

Proof)

consider vectors

x0 = α

m∑i=1

zi, xi = x0 + αzi, i = 1, ...,m

where α is positive scalar that α(m+ 1) < 1

xi ∈ X ⊂ ri(C) (i = 0, 1, ...,m)

xi − x0 = αzi (i = 0, 1, ...,m) span aff(C).

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Proposition 1.4.1(b) (written again)

m > 0 ⇒ ∃x0, x1, ..., xm ∈ ri(C)

s.t. span{x1 − x0, ..., xm − x0}= {subspace parallel to aff(C)}

Proof)

consider vectors

x0 = α

m∑i=1

zi, xi = x0 + αzi, i = 1, ...,m

where α is positive scalar that α(m+ 1) < 1

xi ∈ X ⊂ ri(C) (i = 0, 1, ...,m)

xi − x0 = αzi (i = 0, 1, ...,m) span aff(C).

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Proposition 1.4.1(b) (written again)

m > 0 ⇒ ∃x0, x1, ..., xm ∈ ri(C)

s.t. span{x1 − x0, ..., xm − x0}= {subspace parallel to aff(C)}

Proof)

consider vectors

x0 = α

m∑i=1

zi, xi = x0 + αzi, i = 1, ...,m

where α is positive scalar that α(m+ 1) < 1

xi ∈ X ⊂ ri(C) (i = 0, 1, ...,m)

xi − x0 = αzi (i = 0, 1, ...,m) span aff(C).

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Proposition 1.4.1 (c)

Proposition 1.4.1

c x ∈ ri(C) ⇔ ∀x ∈ C, ∃γ > 1 s.t. x+(γ−1)(x− x) ∈ C

Proof)⇒) (trivial)⇐)

Since ri(C) 6= φ, there exists some x ∈ ri(C)

If x = x, we are done. So, we assume x 6= x.

By the given condition, since x ∈ C,there exists a γ > 1 s.t. y = x+ (γ − 1)(x− x) ∈ C

Thus, we have x = γ−1γ x+ 1

γyBy using Line Segment Principle, we obtain x ∈ ri(C)

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Proposition 1.4.1 (c)

Proposition 1.4.1

c x ∈ ri(C) ⇔ ∀x ∈ C, ∃γ > 1 s.t. x+(γ−1)(x− x) ∈ C

Proof)⇒) (trivial)⇐)

Since ri(C) 6= φ, there exists some x ∈ ri(C)

If x = x, we are done. So, we assume x 6= x.

By the given condition, since x ∈ C,there exists a γ > 1 s.t. y = x+ (γ − 1)(x− x) ∈ C

Thus, we have x = γ−1γ x+ 1

γyBy using Line Segment Principle, we obtain x ∈ ri(C)

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Proposition 1.4.1 (c)

Proposition 1.4.1

c x ∈ ri(C) ⇔ ∀x ∈ C, ∃γ > 1 s.t. x+(γ−1)(x− x) ∈ C

Proof)⇒) (trivial)⇐)

Since ri(C) 6= φ, there exists some x ∈ ri(C)

If x = x, we are done. So, we assume x 6= x.

By the given condition, since x ∈ C,there exists a γ > 1 s.t. y = x+ (γ − 1)(x− x) ∈ C

Thus, we have x = γ−1γ x+ 1

γyBy using Line Segment Principle, we obtain x ∈ ri(C)

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Proposition 1.4.1 (c)

Proposition 1.4.1

c x ∈ ri(C) ⇔ ∀x ∈ C, ∃γ > 1 s.t. x+(γ−1)(x− x) ∈ C

Proof)⇒) (trivial)⇐)

Since ri(C) 6= φ, there exists some x ∈ ri(C)

If x = x, we are done. So, we assume x 6= x.

By the given condition, since x ∈ C,there exists a γ > 1 s.t. y = x+ (γ − 1)(x− x) ∈ C

Thus, we have x = γ−1γ x+ 1

γyBy using Line Segment Principle, we obtain x ∈ ri(C)

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Proposition 1.4.2

X : Nonempty convex subset of Rn

f : X 7→ R , concaveX∗ := {x∗ ∈ X|f(x∗) = infx∈X f(x)}

X∗ ∩ ri(X) 6= φ ⇒ f must be constant over X, i.e., X∗ = X

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Proposition 1.4.3

C, C: Nonempty convex sets.

a cl(C) = cl(ri(C))

b ri(C) = ri(cl(C))

c Following three conditions are equivalent.

i ri(C) = ri(C)ii cl(C) = cl(C)iii ri(C) ⊂ C ⊂cl(C)

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Proposition 1.4.4

C : Nonempty convex subset of Rn

A : m× n matrix

a A · ri(C) = ri(A · C)

b A · cl(C) ⊂ cl(A · C)Furthermore, if C is boundedA · cl(C) = cl(A · C)

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Proposition 1.4.5

C1, C2 : Nonempty convex sets

a ri(C1) ∩ ri(C2) ⊂ ri(C1 ∩ C2), cl(C1 ∩ C2) ⊂ cl(C1) ∩ ri(C2)Furthermore, if ri(C1) ∩ ri(C2) 6= φ, thenri(C1) ∩ ri(C2) = ri(C1 ∩ C2), cl(C1 ∩ C2) = cl(C1) ∩ ri(C2)

b ri(C1+C2) ⊂ ri(C1)+ ri(C2), cl(C1)+ cl(C2) ⊂ cl(C1+C2)Furthermore, if at least one of the sets C1 and C2 is bounded,then cl(C1) + cl(C2) = cl(C1 + C2)

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Proposition 1.4.6

f : Rn 7→ R is convex ⇒ f is continuous.

f : Rn 7→ (−∞,∞] is a proper convex function⇒ f is continuous over the ri(dom(f)).