Vector valued Banach limits and generalizations applied to the inhomogeneous Cauchy equation

In Prager and Schwaiger (Grazer Math Ber 363:171–178, 2015) the classical notion of Banach limits was used to solve the inhomogeneous Cauchy equation f(x+y)-f(x)-f(y)=ϕ(x,y)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$f(x+y)-f(x)-f(y)=\phi (x,y)$$\end{document} for real functions of one real variable. Here these methods are generalized to more general target spaces, namely Banach spaces which admit vector valued Banach limits.


Introduction
The main topic of this paper is the investigation of the inhomogeneous Cauchy equation where φ : V × V → W is given and f : V → W is to be determined. The emphasis lies on giving explicit formulas for solutions when certain hypotheses on the inhomogeneity are fulfilled. It is well known (see [7]) that the solvability of this equation is equivalent to the properties of the inhomogeneity φ. Solving the equation in general is based on the application of Zorn's Lemma. In our case, if the domain and co-domain are rational vector spaces, assuming the existence of a basis for such vector spaces is enough to give explicit solutions. In the generic case the existence of such a basis is granted only by applying also Zorn's Lemma. Nevertheless situations may In what follows, given a (Hamel) basis of V over Q, all solutions of (1) will be given. The ideas for the definition of the solution f arise from some necessary conditions on f in (1).
Rewriting this equation as f (x + y) = f (x) + f (y) + φ(x, y) we get by induction that which for x 1 = x 2 = · · · = x n = x means Now we gather some auxiliary results needed for the construction. (2) and (3). Then φ satisfies
Let B ⊂ V be a Hamel basis of V over Q. Then for each x ∈ V there is a unique finite subset B = B x ⊆ B and a unique family (For x = 0 the set B x is the empty set and the corresponding family of λ b is the empty family.) This implies that for any C ⊆ B and any family (μ c ) c∈C ∈ Q C such that x = c∈C μ c c it follows that Proof. This follows immediately from Lemma 3.
According to this lemma the function F , Proof. Let n := #C, m := #B and let, which is always possible, α : {1, 2, . . . , n} → C be such that α ({1, 2, . . . , m}) = B. Then Proof. If p = 0 then r = 0, too. Thus G * (b, p, q) = G * (b, r, s) in this case. If p = 0 also r = 0 and we may assume that, say, p and q are coprime. Then, since p q = r s there is some l ∈ N such that r = ln and s = lq. We have to show that Vol. 93 (2019)

Vector valued Banach limits and generalizations 263
After division by sgn(λ), observing Lemma 3 and (1), we get for the right hand side of (9) that hence (9) is true.
Then, using the considerations above, we may define functions Taking for granted that f is a solution of (1), the second part obviously becomes true.
It remains to show that f indeed is a solution. Let Now we will show that for all b ∈ B we have We set Vol. 93 (2019) Vector valued Banach limits and generalizations 265 and with an analogous application for H Corresponding to the possible distributions of signs we consider the cases listed in the following table. A column without a numeral contains a case which is obtained from the immediately preceding one, considering (2) by renaming λ b by μ b and vice versa. Therefore these cases need not be treated separately.
Adding together we thus obtain which with (10) yields the result.

Different types of Banach limits
In [1] one may find the following definition of a vector valued Banach limit. It is also shown there that this includes the usual definition of a Banach limit on bounded sequences of reals (see [3]).

Definition 1.
Let X be a normed space and let ∞ (X) be the space of bounded sequences on X equipped with the supremum norm. Then L : ∞ (X) → X is a Banach limit if (i) L is linear and continuous (ii) L(x) = lim n→∞ x n for any convergent sequence x = (x n ) n∈N in X, ) n∈N , and (iv) the operator norm of L equals 1: L = 1.
In this paper it was proved that Banach limits exist on the dual X * of any normed space X. The proof used an ultrafilter U on N containing {A ⊆ N | N\S is finite} and the definition which is meaningful with respect to the w * -topology since then any ball is compact.
the closure with respect to the w * -topology of the set 1 Theorem 2. Let X be a normed space and X * its dual. Then the Banach limit L defined by (16) has the property that for any sequence ξ = (x * n ) n∈N ∈ ∞ (X * ) the value L(ξ) is contained in cl w * conv{x * n | n ∈ N}, the closure with respect to the w * -topology of the convex hull of the elements in the sequence. But there are Banach limits which do not fulfill this property.
Proof. The first part is clear by the lines above if one takes into account that The second part follows from the last item of the following remark, since by [2] there are different Banach limits on the reals.
Proof of the remark above. The first part follows immediately from the definition. This implies the second item. A proof of the third one is contained in [11].
Regarding the fourth item one may proceed in the following manner. Let (f n ) n∈N be any sequence in ∞ (X * ). Then the sequences f n (e 1 ) and f n (e 2 ) are bounded real sequences. Thus their Banach limits exist for all Banach limits on R. Accordingly L((f n ) n∈N ) ∈ X * . Obviously L is shift invariant since L 1 , L 2 are. If the sequence f n converges to f , the sequences f n (e i ) will converge to f (e i ), i = 1, 2. Thus L i ((f n (e i ))) = f (e i ) for i = 1, 2. Since obviously L is also linear, it remains to show that L = 1. Note that L((f n ) n∈N ) = max{|L 1 ((f n (e 1 )) n∈N )| , |L 2 ((f n (e 2 )) n∈N )|}. Note also that L 1 = L 2 = 1. This implies |L 1 ((f n (e 1 )) n∈N | ≤ 1 · (f n (e 1 )) n∈N ∞ and |L 2 ((f n (e 2 )) n∈N | ≤ 1 · (f n (e 2 )) n∈N ∞ . Using 2. this also gives L( This implies L ≥ 1 and altogether L = 1. Now we deal with the final part. Let c = (c n ) n∈N ∈ ∞ (R) be such that L 1 (c) = L 2 (c) and define f n := c n (π 1 − π 2 ). Then f n = |c n | for all n.
Let A : A x for some x = 0. By 2. this set is also closed with respect to the norm topology. It is also convex since it is a linear subspace of X * . Assume now that Using Banach limits L 1 , L 2 , . . . , L m it is easy to construct Banach limits on R n . The following theorem answers the question whether a result similar to that of Theorem 2 holds true.

This Banach limit has the property that
Proof. The properties (i)-(iii) for L immediately follow from the corresponding properties of L i . Since L i = 1 and since we use the supremum norm on R m it follows that L ≤ 1. Now we consider the constant sequence determined by 1 m := (1, 1, . . . , 1). L i are Banach limits. Thus L i ((1, 1, . . .)) = 1. Therefore L ((1 m , 1 m , . . .)) = 1 m . Thus L ≥ 1 m ∞ = 1 and also L = 1, as desired.
Suppose that L 1 = L 2 = · · · = L m and assume that Then by the separation theorem for closed convex sets (see [4,Théoréme 1.7]) there is some a = (a 1 , a 2 , . . . , a m ) ∈ R m and some α ∈ R such that n ) n∈N and by the property of a Banach limit this value must be ≤ α since n < α for all n. Thus L((x n ) n∈N ) ∈ K, as desired.
Finally assume (without loss of generality) that L 1 = L 2 and that L 1 ((z n ) n∈N ) = L 2 ((z n ) n∈N ) for some (z n ) n∈N ∈ ∞ (R m ). Then the sequence x n , x n := (z n , z n , 0, . . . , 0) is contained in ∞ (R m ). Moreover all x n are contained in the (closed) linear subspace E : and L 1 ((z n ) n∈N ) = L 2 ((z n ) n∈N ), a contradiction.

Solutions of the inhomogeneous Cauchy equation expressed in terms of Banach limits
In [10] it was shown that for V = W = R the relations (2) and (3) imply provided that (1) has a solution at all. The proof for arbitrary rational vector spaces V, W is the same. Thus (17) holds true in view of Theorem 1. The following generalizes Theorem 2 of [10].

Theorem 4.
Let V be a rational vector space and X a normed space which admits a Banach limit L : ∞ (X) → X. Assume that φ : V × V → X satisfies (2) and (3) and that the sequences φ(nx, x) and φ(nx, ny) are contained in Proof. Note that L may be applied to all sequences generated by (17) separately. Then the result follows from the fact that L is shift invariant and that L maps constant functions to the corresponding constant.
Example 1 (Unbounded φ). Let the rational vector space V as above be of infinite dimension and let X admit a Banach limit L : where Q >0 denotes the set of all positive rational numbers. Then the set of all C x gives a partition of V . Let R ⊆ V be a set of representatives for this partition, i.e., V = r∈R C r and C r ∩ C s = ∅ for r, s ∈ R, r = s. AEM Take g : R → X arbitrary and h : V → X bounded and define f : Then Theorem 4 works for φ: If x ∈ C r , y ∈ C s , x + y ∈ C t the Cauchy difference for f is given by Since nx, (n + 1)x ∈ C r , we get φ(nx, x) = h((n + 1)x) − h(nx) − h(x). A similar calculation shows that φ(nx, ny) = g(t) − g(r) − g(s) + h(n(x + y)) − h(nx) − h(ny). Thus the hypotheses of the theorem are satisfied.
g may be chosen unbounded since by the assumption on V the set R of representatives has to be infinite. This is well defined, since sin •f is measurable and even square integrable since sin is bounded. Moreover φ itself is bounded because 1 0 sin 2 (f (x)) dx ≤ 1. Thus a fortiori φ satisfies the hypotheses of the theorem above.

Abstract Banach limits and solutions of the inhomogeneous Cauchy equation
Analyzing the considerations of the previous section it turns out that condition (iv) of a Banach limit is not used when proving Theorem 4. Here we want to discuss a more general situation. Throughout we consider an Abelian semigroup S, an Abelian group G = {0} and an inhomogeneity φ : S × S → G, which satisfies (2) and (3)