M. Riesz extension theorem - Alchetron, the free social encyclopedia

The M. Riesz extension theorem is a theorem in mathematics, proved by Marcel Riesz during his study of the problem of moments.

Formulation

Let E be a real vector space, F ⊂ E a vector subspace, and let K ⊂ E be a convex cone.

A linear functional φ: F → R is called K-positive, if it takes only non-negative values on the cone K:

ϕ ( x ) ≥ 0 for x ∈ F ∩ K .

A linear functional ψ: E → R is called a K-positive extension of φ, if it is identical to φ in the domain of φ, and also returns a value of at least 0 for all points in the cone K:

ψ | F = ϕ and ψ ( x ) ≥ 0 for x ∈ K .

In general, a K-positive linear functional on F cannot be extended to a K -positive linear functional on E. Already in two dimensions one obtains a counterexample taking K to be the upper halfplane with the open negative x-axis removed. If F is the x-axis, then the positive functional φ(x, 0) = x can not be extended to a positive functional on the plane.

However, the extension exists under the additional assumption that for every y ∈ E there exists x∈F such that y − x ∈K; in other words, if E = K + F.

Proof

By transfinite induction it is sufficient to consider the case dim E/F = 1.

Choose y ∈ EF. Set

ψ | F = ϕ , ψ ( y ) = sup { ϕ ( x ) ∣ x ∈ F , y − x ∈ K } ,

and extend ψ to E by linearity. Let us show that ψ is K-positive.

Every point z in K is a positive linear multiple of either x + y or x − y for some x ∈ F. In the first case, z = a(y + x), therefore y− (−x) = z/a is in K with −x in F . Hence

ψ ( y ) ≥ ψ ( − x ) = − ψ ( x ) ,

therefore ψ(z) ≥ 0. In the second case, z = a(x − y), therefore y = x − z/a. Let x₁ ∈ F be such that z₁ = y − x₁ ∈ K and ψ(x₁) ≥ ψ(y) − ε. Then

ψ ( x ) − ψ ( x 1 ) = ψ ( x − x 1 ) = ψ ( z 1 + z / a ) = ϕ ( z 1 + z / a ) ≥ 0 ,

therefore ψ(z) ≥ −a ε. Since this is true for arbitrary ε > 0, we obtain ψ(z) ≥ 0.

Corollary: Krein's extension theorem

Let E be a real linear space, and let K ⊂ E be a convex cone. Let x ∈ E(−K) be such that R x + K = E. Then there exists a K-positive linear functional φ: E → R such that φ(x) > 0.

Connection to the Hahn–Banach theorem

The Hahn–Banach theorem can be deduced from the M. Riesz extension theorem.

Let V be a linear space, and let N be a sublinear function on V. Let φ be a functional on a subspace U ⊂ V that is dominated by N:

ϕ ( x ) ≤ N ( x ) , x ∈ U .

The Hahn–Banach theorem asserts that φ can be extended to a linear functional on V that is dominated by N.

To derive this from the M. Riesz extension theorem, define a convex cone K ⊂ R×V by

K = { ( a , x ) ∣ N ( x ) ≤ a } .

Define a functional φ₁ on R×U by

ϕ 1 ( a , x ) = a − ϕ ( x ) .

One can see that φ₁ is K-positive, and that K + (R × U) = R × V. Therefore φ₁ can be extended to a K-positive functional ψ₁ on R×V. Then

ψ ( x ) = − ψ 1 ( 0 , x )

is the desired extension of φ. Indeed, if ψ(x) > N(x), we have: (N(x), x) ∈ K, whereas

ψ 1 ( N ( x ) , x ) = N ( x ) − ψ ( x ) < 0 ,

leading to a contradiction.

References

M. Riesz extension theorem Wikipedia

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Contents

Formulation

Proof

Corollary: Krein's extension theorem

Connection to the Hahn–Banach theorem

References