M. Riesz extension theorem

The M. Riesz extension theorem is a theorem in mathematics, proved by Marcel Riesz [1] during his study of the problem of moments.[2]

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:

\phi(x) \geq 0 \quad \text{for} \quad x \in F \cap 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:

\psi|_F = \phi \quad \text{and} \quad \psi(x) \geq 0\quad \text{for} \quad x \in K.

In general, a K-positive linear functional on F can not 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 xF 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  E\F. Set

 
\psi|_F = \phi, \quad 
\psi(y) = \sup \left\{ \phi(x) \, \mid \, x \in F, \, y - x \in K \right\},

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

 \psi(y) \geq \psi(-x) = - \psi(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

 \psi(x) - \psi(x_1) = \psi(x-x_1) = \psi(z_1 + z/a) = \phi(z_1 + z/a) \geq 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 HahnBanach theorem

Main article: 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:

 \phi(x) \leq N(x), \quad x \in U.

The HahnBanach 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 = \left\{ (a, x) \, \mid \, N(x) \leq a \right\}.

Define a functional φ1 on R×U by

 \phi_1(a, x) = a - \phi(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

 \psi(x) = - \psi_1(0, x)

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

 \psi_1(N(x), x) = N(x) - \psi(x) < 0,

leading to a contradiction.

Notes

References

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