Abstract
Let (X,d) be a metric linear space and a∈X. The point a divides the space into three sets: Ha = {x ∈ X: d(0,x) < d(x,a)}, Ma = {x ∈ X: d(0,x) = d(x,a)} and La = {x ∈ X: d(0,x) > d(x,a)}. If the distance is generated by a norm, Ha is called the Leibnizian halfspace of a, Ma is the perpendicular bisector of the segment 0,a and La is the remaining set La = X\(Ha∪ Ma). It is known that the perpendicular bisector of the segment [0,a] is an affine subspace of X for all a ∈ X if, and only if, X is an inner product space, that is, if and only if the norm is generated by an inner product. In this case, it is also true that if x,y ∈ La ∪ Ma, then x + y ∈ La∪ Ma. Otherwise written, the set La∪ Ma is a semi-group with respect to addition. We investigate the problem: for what kind of norms in X the pair (La ∪ Ma,+) is a semi-group for all a ∈ X? In that case, we say that “(X,‖.‖)has the semi-group property” or that “the norm ‖.‖ has the semi-group property”. This is a threedimensional property, meaning that if all the three-dimensional subspaces of X have it, then X also has it. We prove that for two-dimensional spaces, (La,+) is a semi-group for any norm, that (X,‖.‖) has the semi-group property if, and only if, the norm is strictly convex, and, in higher dimensions, the property fails to be true even if the norm is strictly convex. Moreover, studying the Lp norms in higher dimensions, we prove that the semi-group property holds if, and only if, p = 2. This fact leads us to the conjecture that in dimensions greater than three, the semi-group property holds if, and only if, X is an inner-product space.
Highlights
Let (X, k k) be a real normed space.In [1], the authors were interested in the following property of XFor every x,y,z ∈ X, the inequalities (A) For every x,y,z ∈ X, the inequalities kxk ≥ ky + zk, kyk ≥ ky + zk, kzk ≥ kx + yk, imply the equalities kxk = ky + zk, kyk = ky + zk, kzk = kx + ykThe typical examples of spaces with this property are the inner product spaces
We obtain even more than the simple equalities kxk = ky + zk, kyk = kz + xk, kzk = kx + yk, namely, x = −(y + z), y = −(x + z), z = −(y + x)
In an inner-product space, (La ∪ Ma,+) is a semigroup. This is obvious: squaring the inequalities kxk ≥ ka − xk, kyk ≥ ka − ykand adding them, we obtain kx + yk2 ≥ 2kak2 − 2 + kx + yk2 ≥ kx + y − ak2, meaning that x + y is in La
Summary
Let (X, k k) be a real normed space. For every x,y,z ∈ X, the inequalities (A) For every x,y,z ∈ X, the inequalities kxk ≥ ky + zk, kyk ≥ ky + zk, kzk ≥ kx + yk, imply the equalities kxk = ky + zk, kyk = ky + zk, kzk = kx + yk. We always assume that a 6= 0, since a = 0 is nonsensical Using these notations, the property (A) becomes (A) For any 0 6= a ∈ X, if x ∈ La ∪ Ma , y ∈ La ∪ Ma , x + y ∈ Ha ∪ Ma x,y,x + y ∈ Ma. For any 0 6= a ∈ X, if x ∈ La ∪ Ma , y ∈ La ∪ Ma , x + y ∈ Ha ∪ Ma x + y ∈ Ma. For any 0 6= a ∈ X, if x ∈ La ∪ Ma , y ∈ La ∪ Ma , x + y ∈ Ha ∪ Ma x + y ∈ Ma Is it possible that x,y,x + y ∈ Ma ? In an inner-product space, (La ∪ Ma ,+) is a semigroup This is obvious: squaring the inequalities kxk ≥ ka − xk, kyk ≥ ka − ykand adding them, we obtain kx + yk2 ≥ 2kak2 − 2 + kx + yk2 ≥ kx + y − ak , meaning that x + y is in La
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