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
be a real vector space,
be a vector subspace, and
be a convex cone.
A linear functional
is called
-positive, if it takes only non-negative values on the cone
:
![{\displaystyle \phi (x)\geq 0\quad {\text{for}}\quad x\in F\cap K.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c8b698cce02077afd4d493421f7cbc3b9be078cc)
A linear functional
is called a
-positive extension of
, if it is identical to
in the domain of
, and also returns a value of at least 0 for all points in the cone
:
![{\displaystyle \psi |_{F}=\phi \quad {\text{and}}\quad \psi (x)\geq 0\quad {\text{for}}\quad x\in K.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9cf447523281df7a2ea071154a0b5d1d077f01f0)
In general, a
-positive linear functional on
cannot be extended to a
-positive linear functional on
. Already in two dimensions one obtains a counterexample. Let
and
be the
-axis. The positive functional
can not be extended to a positive functional on
.
However, the extension exists under the additional assumption that
namely for every
there exists an
such that
Proof
The proof is similar to the proof of the Hahn–Banach theorem (see also below).
By transfinite induction or Zorn's lemma it is sufficient to consider the case dim
.
Choose any
. Set
![{\displaystyle a=\sup\{\,\phi (x)\mid x\in F,\ y-x\in K\,\},\ b=\inf\{\,\phi (x)\mid x\in F,x-y\in K\,\}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/32985356dae32949dd695bee7ef709d27b369637)
We will prove below that
. For now, choose any
satisfying
, and set
,
, and then extend
to all of
by linearity. We need to show that
is
-positive. Suppose
. Then either
, or
or
for some
and
. If
, then
. In the first remaining case
, and so
![{\displaystyle \psi (y)=c\geq a\geq \phi (-x)=\psi (-x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/82a9d82767c3a90ccd4d93d59624a88c1cc3eded)
by definition. Thus
![{\displaystyle \psi (z)=p\psi (x+y)=p(\psi (x)+\psi (y))\geq 0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1140747059e32b2aa2693933c4fa2a579a11ea25)
In the second case,
, and so similarly
![{\displaystyle \psi (y)=c\leq b\leq \phi (x)=\psi (x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/533085cd9b609ec4447198d15b8748838d098d07)
by definition and so
![{\displaystyle \psi (z)=p\psi (x-y)=p(\psi (x)-\psi (y))\geq 0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/83b2f0b0a753c3d20e47c4c87fd8270516e84992)
In all cases,
, and so
is
-positive.
We now prove that
. Notice by assumption there exists at least one
for which
, and so
. However, it may be the case that there are no
for which
, in which case
and the inequality is trivial (in this case notice that the third case above cannot happen). Therefore, we may assume that
and there is at least one
for which
. To prove the inequality, it suffices to show that whenever
and
, and
and
, then
. Indeed,
![{\displaystyle x'-x=(x'-y)+(y-x)\in K}](https://wikimedia.org/api/rest_v1/media/math/render/svg/72075314839ba9b99fbb3e1a5091bdfac0701d19)
since
is a convex cone, and so
![{\displaystyle 0\leq \phi (x'-x)=\phi (x')-\phi (x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4818ea0506451635d3abc4dab7916a073434ef50)
since
is
-positive.
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:
![{\displaystyle \phi (x)\leq N(x),\quad x\in U.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/94532e9a44014d18f5d318816362f66e034c4b5f)
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
![{\displaystyle K=\left\{(a,x)\,\mid \,N(x)\leq a\right\}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a28bc07ab1577c173164f84813933a4a37f316e9)
Define a functional φ1 on R×U by
![{\displaystyle \phi _{1}(a,x)=a-\phi (x).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0e2c4036078d8cde96ce5abe3b07158ac5884461)
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
![{\displaystyle \psi (x)=-\psi _{1}(0,x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4fb12bd06904a5541900aee3b581cd3bf83d034a)
is the desired extension of φ. Indeed, if ψ(x) > N(x), we have: (N(x), x) ∈ K, whereas
![{\displaystyle \psi _{1}(N(x),x)=N(x)-\psi (x)<0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6a5bb9e83f2f285eace89b19409c9793f9a652dc)
leading to a contradiction.
References
Sources
- Castillo, Reńe E. (2005), "A note on Krein's theorem" (PDF), Lecturas Matematicas, 26, archived from the original (PDF) on 2014-02-01, retrieved 2014-01-18
- Riesz, M. (1923), "Sur le problème des moments. III.", Arkiv för Matematik, Astronomi och Fysik (in French), 17 (16), JFM 49.0195.01
- Akhiezer, N.I. (1965), The classical moment problem and some related questions in analysis, New York: Hafner Publishing Co., MR 0184042
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