Int. J. Contemp. Math. Sciences, Vol. 4, 2009, no. 16, 799 - 802
The Riesz Interpolation Property for the Space of Continuously Differentiable Functions Vasilios N. Katsikis Department of Mathematics, National Technical University of Athens Zographou 15780, Athens, Greece
[email protected],
[email protected] Abstract It is well known that the real vector space C (1) [a, b], of continuously differentiable functions defined on a closed interval [a, b] has the Riesz interpolation property. The purpose of this article is to provide a new, direct proof of this result without the use of additional theorems or other known results.
Mathematics Subject Classification: 46A40, 06F20 Keywords: Riesz interpolation property, Riesz decomposition property
1
Introduction
Let X = C (1) [a, b] be the real vector space of continuously differentiable functions defined on a closed interval [a, b]. The ordering in X is the usual functionspace order i.e., f ≤ g iff f (x) ≤ g(x), for all x ∈ [a, b]. We shall start with the basic definitions of our study, Definition 1.1. A partially ordered vector space E is said to have the Riesz interpolation property if for every a, b, c, d ∈ E with a, b ≤ c, d there exists an element v ∈ E such that a, b ≤ v ≤ c, d. Definition 1.2. A partially ordered vector space E is said to have the Riesz decomposition property if for every v, w1 , w2 ∈ E, v, w1, w2 ≥ 0 with v ≤ w1 + w2 then there exist v1 , v2 ∈ E such that v = v1 + v2 and 0 ≤ v1 ≤ w1 , 0 ≤ v2 ≤ w2 . Recall that in partially ordered vector spaces the Riesz interpolation property is equivalent with the Riesz decomposition property. For notions on ordered spaces and ordered subspaces with the Riesz decomposition property or
V. N. Katsikis
800
the Riesz interpolation property, that are not defined here, the reader may refer to [1]. Every Riesz space1 has the Riesz interpolation property but the converse is not always true, such an example is the space X. Note also that a partially ordered vector space with the Riesz interpolation property is a reasonable generalization of a Riesz space. It is easy to see that the space X fails to be a Riesz space. On the other hand, the proof that X has the Riesz interpolation property is not obvious. Next, we are addressing two different proofs, that X has the Riesz decomposition property: The first proof appears in [2, Example 6.3], where it is proved that X has the Riesz decomposition property by using the following theorem due to Ando: Theorem 1.3. [3, Theorem 2] Let L be an ordered Banach space with a closed and generating cone whose intervals are norm bounded. Then L has the Riesz decomposition property if and only if its norm dual has likewise the Riesz decomposition property. In particular, this proof(see [2, Example 6.3]) uses the fact that X is norm dense in C[a, b], therefore the norm dual of X is Riesz isomorphic to the norm dual of C[a, b]. Then, it follows that the norm dual of X is a Riesz space. Therefore, the norm dual of X has the Riesz decomposition property and by Ando’s Theorem 1.3, X has the Riesz decomposition property. The second proof can be found in [1]. Specifically, in [4] Theorem 1, it is proved that the space C (1) (R) has the Riesz decomposition property. By using a similar technique, in [1] Example 1.58 a direct proof(without the use of additional theorems) that the space X has the Riesz decomposition property is provided. In this article we give an alternative proof of this result by showing the equivalent property i.e., X has the Riesz interpolation property. The proof is direct, as in [1], and uses only properties of the elements of X.
2
Main Results
Suppose that f1 , f2 , g1 , g2 ∈ X with f1 , f2 ≤ g1 , g2 . Then, we define the function r = g1 + g2 − f1 − f2 . We shall start with an easy lemma: Lemma 2.1. (i) For each x0 such that r(x0 ) = 0 it holds g1 (x0 ) = g2 (x0 ) = f1 (x0 ) = f2 (x0 ). 1
A vector space E is a Riesz space(or a vector lattice) if for any two vectors x, y ∈ E the supremum and the infimum of {x, y} both exist in E.
Riesz interpolation property
801
(ii) For each x0 such that r(x0 ) = 0 it holds g 1 (x0 ) = g 2 (x0 ) = f 1 (x0 ) = f 2 (x0 ). Proof. (i) It is clear that r ≥ 0, g1 − f1 ≥ 0, g2 − f2 ≥ 0 hence the relation g1 (x0 ) + g2 (x0 ) − f1 (x0 ) − f2 (x0 ) = 0 implies g1 (x0 ) = f1 (x0 ) and g2 (x0 ) = f2 (x0 ). Also, g2 (x0 ) = f2 (x0 ) ≤ g1 (x0 ) and g1 (x0 ) = f1 (x0 ) ≤ g2 (x0 ) hence g1 (x0 ) = g2 (x0 ) = f1 (x0 ) = f2 (x0 ). (ii) For each x > x0 since f1 (x) ≤ g1 (x), f1 (x0 ) = g1 (x0 ) we have that f1 (x) − f1 (x0 ) g1 (x) − g1 (x0 ) ≤ . x − x0 x − x0 Similarly, for each x < x0 we have g1 (x) − g1 (x0 ) f1 (x) − f1 (x0 ) ≥ , x − x0 x − x0 and since f1 , g1 are differentiable at the point x0 it follows that f 1 (x0 ) = g 1(x0 ). Using similar arguments one can prove the rest of the equalities in (ii). 1 f2 Let p be the function p = g1 g2 −f , r = 0 and p(x0 ) = g1 (x0 ) = g2 (x0 ) = r f1 (x0 ) = f2 (x0 ), for each point x0 such that r(x0 ) = 0. We shall prove that p is the required interpolating function. First, observe that p is continuously differentiable at all points x0 where r(x0 ) = 0. Then,
Lemma 2.2. It holds, f1 , f2 ≤ p ≤ g1 , g2 . Proof. We can write p as p=
g2 − f1 g1 − f2 g1 + f1 . r r
Then, it follows that p is a convex combination of f1 , g1 and a similar argument shows that p is a convex combination of f2 , g2 . Therefore, f1 , f2 ≤ p ≤ g1 , g2 . Lemma 2.3. For each x0 such that r(x0 ) = 0 it holds p (x0 ) = g1 (x0 ) = = f1 (x0 ) = f2 (x0 ).
g2 (x0 )
Proof. Trivial, from lemma 2.1 and lemma 2.2. Lemma 2.4. p is continuous at x0 where r(x0 ) = 0. Proof. At points x where r(x) > 0 we have the following: Since g1 − f2 = r − g2 + f1 we have p = (
g2 − f1 g1 − f2 g2 − f1 g1 − f2 ) g1 + ( ) f1 + g1 + f1 = r r r r
V. N. Katsikis
802
g2 − f1 r − (g2 − f1 ) g2 − f1 g1 − f2 ) g1 + ( ) f1 + g1 + f1 = r r r r g2 − f1 g1 − f2 g2 − f1 g1 + f1 + ( ) (g1 − f1 ) r r r 1 2 g1 + g1 −f f1 then Let s = g2 −f r r (
g2 − f1 r − (g2 − f1 ) g2 − f1 g1 + f1 = (g1 − f1 ) + f1 r r r g2 −f1 and since 0 ≤ r ≤ 1, limx→x0 (g1 − f1 ) = 0 we have s=
lim s(x) = lim f1 (x) = f1 (x0 ) = p (x0 ).
x→x0
Also, let t =
1 ( g2 −f ) (g1 r
x→x0
− f1 ) then
(g2 − f1 ) (g2 − f1 ) + (g2 − f1 ) (g1 − f2 ) − (g2 − f1 )(g2 − f1 ) − (g2 − f1 )(g1 − f2 ) · r (g1 − f1 ) (g2 − f1 )(g1 − f2 ) (g2 − f1 )(f2 − g1 ) (g1 − f1 ) = + · , r r r r as before we have limx→x0 t(x) = 0. Therefore, since limx→x0 p (x) = limx→x0 (s(x) + t(x)) = p (x0 ) we have that p is continuous at x0 where r(x0 ) = 0. t=
As a consequence of the previous lemmas, we can state our main result: Theorem 2.5. The space X = C (1) [a, b] has the Riesz interpolation property. ACKNOWLEDGEMENTS. The research of the author was financially supported by the State Scholarship Foundation (IKY) in his postdoctoral studies.
References [1] C.D. Aliprantis , R. Tourky. Cones and duality, American Mathematical Society, Providence, Rhode Island, 2007. [2] C.D. Aliprantis , M. Florenzano , R. Tourky. General equilibrium analysis in ordered topological vector spaces. Journal of mathematical economics 40(2004),247-269. [3] T. Ando, On fundamental properties of a Banach space with a cone, Pacific J. Math. ,12(1962),1163-1169. [4] A. Nagel , W. Rudin, Riesz Decompositions, Can. J. Math. , Vol. XXVI, No 3(1974),753-760. Received: November, 2008
