Georgian Mathematical Journal Volume 8 (2001), Number 2, 231–236

ON A CHARACTERISATION OF INNER PRODUCT SPACES

G. CHELIDZE

Abstract. It is well known that for the Hilbert space H the minimum value of the functional F_{μ}(f) = ∫_{H} ∥f −g∥^{2}dμ(g), f ∈ H, is achived at the mean of μ for any probability measure μ with strong second moment on H. We show that the validity of this property for measures on a normed space having support at three points with norm 1 and arbitrarily fixed positive weights implies the existence of an inner product that generates the norm.

2000 Mathematics Subject Classification: 46C15.

Key words and phrases: Inner product space, Hilbert space, optimal location, probability measure.

Let X be a real normed space, dim X ≥ 2, and μ be a Borel probability measure on X with strong second moment. Denote by F_{μ} the following functional on X:

F_{μ}(f) = ∫_{X} ∥f − g∥^{2}dμ(g), f ∈ X.

It is easy to show that if X is an inner product space and there exists the mean m of μ (in the usual weak sense as the Pettis integral), then F_{μ}(f) ≥ F_{μ}(m) for all f ∈ X.

The problem which was brought to my attention by N. Vakhania was to find a class of probability measures as small as possible, for which this property of F_{μ} characterizes the inner product spaces. It is easy to see that the class of measures with supports containing two points is not a sufficient class since the minimum of F_{μ} is attained at the mean of μ for any such μ whatever the normed space X is. Indeed, for any normed space X let μ be a probability measure concentrated at two points f, g ∈ X and let μ(f) = α, μ(g) = β, α > 0, β > 0, α+β = 1. It is clear that m = αf +βg and F_{μ}(m) = αβ∥f −g∥^{2}. The condition F_{μ}(h) ≥ F_{μ}(m), h ∈ X, gives the inequality

α∥f − h∥^{2} + β∥g − h∥^{2} ≥ αβ∥f − g∥^{2}.

Denoting f − h and g − h by p and q respectively we obtain

α∥p∥^{2} + β∥q∥^{2} ≥ αβ∥p − q∥^{2}.  (1)

However, this inequality is true for any normed space since by the triangle inequality we have

αβ∥p − q∥^{2} ≤ αβ∥p∥^{2} + 2αβ∥p∥ · ∥q∥ + αβ∥q∥^{2}

ISSN 1072-947X / $8.00 / \begin{matrix}c \\ ◯ \end{matrix} Heldermann Verlag www.heldermann.de

and, using the obvious relation 2αβ∥p∥ · ∥q∥ ≤ α^{2}∥p∥^{2} + β^{2}∥q∥^{2}, we get (1).

As the referee of the present paper noticed^{1}, a sufficient class can be constructed using measures concentrated at three points. There are two types of results in this direction:

a) the sufficient class consists of measures μ = \frac{1}{3}δ_{f_{1}} + \frac{1}{3}δ_{f_{2}} + \frac{1}{3}δ_{f_{3}} for all triplets {f_{1}, f_{2}, f_{3}} from X (δ_{p} being the Dirac measure at p ∈ X) (see [1], Proposition (1.12), p. 10)

b) the sufficient class consists of measures μ = αδ_{f_{1}} +βδ_{f_{2}} +γδ_{f_{3}} for all triplets {f_{1}, f_{2}, f_{3}} with unit norms and all weights α, β, γ such that αf_{1} +βf_{2} +γf_{3} = 0 (Theorem 5.3 in [2], p. 236).

The aim of this paper is to show that in fact yet a smaller class of measures can be taken.

Theorem. Let X be a real normed space, dim X ≥ 2, S(X) be the set of points of norm one, α, β, γ be arbitrarily fixed positive numbers and δ_{p} be the Dirac measure at p ∈ X. The following propositions are equivalent:

(i) X is an inner-product space.

(ii) For any two points f, g from S(X) and the point h = 0, the mean of the measure μ = \frac{1}{α+β+γ} (αδ_{f} + βδ_{g} + γδ_{0}) is a point of a local minimum for the functional

F_{μ}(t) = α∥t − f∥^{2} + β∥t − g∥^{2} + γ∥t∥^{2}, t ∈ X.

(iii) For any three points f, g, h from S(X) the mean of μ = \frac{1}{α+β+γ} (αδ_{f} +βδ_{g}+ γδ_{h}) is a point of a local minimum for the functional

F_{μ}(t) = α∥t − f∥^{2} + β∥t − g∥^{2} + γ∥t − h∥^{2}, t ∈ X.

According to the well-known von Neumann–Jordan criterion it is enough to prove the Theorem for the case dim X = 2. Thus we should prove that the surface S(X) of the unit ball in (R^{2}, ∥ · ∥) is an ellipse. It is clear that we may assume α + β + γ = 1.

The proof of the Theorem is based on the following auxiliary results.

Lemma 1. Let α, β, γ be given positive reals with α + β + γ = 1. For any two noncollinear A and B from S(X) there exist:

(i) A_{1} and B_{1} on S(X) such that

\frac{A_{1} − M}{∥A_{1} − M∥} = A, \frac{B_{1} − M}{∥B_{1} − M∥} = B,

where M = αA_{1} + βB_{1}.

(ii) \overline{A}_{1}, \overline{B}_{1}, \overline{C}_{1} on S(X) such that

\frac{\overline{A}_{1} − \overline{M}}{∥\overline{A}_{1} − \overline{M}∥} = A, \frac{\overline{B}_{1} − \overline{M}}{∥\overline{B}_{1} − \overline{M}∥} = B,

where \overline{M} = α\overline{A}_{1} + β\overline{B}_{1} + γ\overline{C}_{1}.

1The author takes this opportunity to express his deep gratitude to the referee for his comments including this information.

Proof. (i). Denote by S′(X) the part of S(X) which is inside the smaller angle generated by the vectors A and B. Let C be any point of S′(X). It is clear that for all u, 0 < u < 1, there exist A_{u} and B_{u} on S′(X) such that for some u_{1} > 0, u_{2} > 0 we have A_{u} − uC = u_{1}A and B_{u} − uC = u_{2}B. Since M_{u} = uC is inside the triangle A_{u}OB_{u}, where O denotes the zero vector, we have

M_{u} = α_{u}A_{u} + β_{u}B_{u}

for some α_{u} > 0, β_{u} > 0, γ_{u} > 0, α_{u} + β_{u} + γ_{u} = 1. Therefore we have to prove that for some C ∈ S′(X) and u > 0 there exist α_{u}, β_{u} and γ_{u} such that α_{u} = α, β_{u} = β, γ_{u} = γ. It is clear that \frac{∥M_{u}∥}{∥O_{u}−M_{u}∥} = \frac{1−γ_{u}}{γ_{u}} where O_{u} is the intersection

of the lines (A_{u}B_{u}) and (OC). Consider the function φ(u) = \frac{∥M_{u}∥}{∥O_{u}−M_{u}∥} = \frac{1−γ_{u}}{γ_{u}} . Since S(X) is a continuous curve, the function φ defined on the interval (0, 1) is continuous and lim_{u→1} φ(u) = +∞, lim_{u→0} φ(u) = 0. Therefore there exists u_{C} such

that φ(u_{C}) = \frac{∥M_{u_{C}} ∥}{∥M_{u_{C}} −O_{u_{C}} ∥} = \frac{1−γ}{γ} . Now we consider the following two continuous functions on S′(X) :

ψ_{1}(C) = \frac{∥A_{u_{C}} − M_{u_{C}}∥}{∥A′_{u_{C}} − M_{u_{C}}∥} = \frac{1 − α_{u_{C}}}{α_{u_{C}}} , ψ_{2}(c) = \frac{∥B_{u_{C}} − M_{u_{C}}∥}{∥B′_{u_{C}} − M_{u_{C}}∥} = \frac{1 − β_{u_{C}}}{β_{u_{C}}} ,

where A′_{u_{C}} (B′_{u_{C}}) denotes the intersection of the lines (A_{u_{C}}M_{u_{C}}) and (OB_{u_{C}}) ((B_{u_{C}}M_{u_{C}}) and (OA_{u_{C}})). Obviously, lim_{C→B} ψ_{1}(C) = +∞, lim_{C→A} ψ_{2}(C) = +∞.

Since γ_{u_{C}} = γ and α_{u_{C}} + β_{u_{C}} + γ = 1, we get \frac{1}{1+ψ_{1}(C)} + \frac{1}{1+ψ_{2}(C)} + γ = 1. This equality gives lim_{C→A} ψ_{1}(C) = \frac{γ}{1−γ} . As ψ_{1} receives all values from the interval

(\frac{γ}{1−γ} , +∞), the inequality \frac{1−α}{α} > \frac{γ}{1−γ} shows the existence of a point C_{1} ∈ S′(X) such that ψ_{1}(C_{1}) = \frac{1−α}{α}, ψ_{2}(C_{1}) = \frac{1−β}{β} . For such C_{1} we have α_{u_{C_{1}}} = α, β_{u_{C_{1}}} = β, γ_{u_{C_{1}}} = γ which proves the statement (i).

(ii). Now we consider the same points A_{1}, B_{1} as in (i) and the other point A_{2} of the intersection S(X) ∩ (A_{1}M).

Let M_{1} be a point on the line (A_{1}A_{2}) which is inside the unit ball B(X). Let now B_{2} be the point on S(X) for which B_{2} −M_{1} = uB, u > 0. Denote by C_{2} the point \frac{1}{γ}(M_{1} − αA_{1} − βB_{2}). Since \frac{A_{1}−M_{1}}{∥A′_{1}−M_{1}∥} = \frac{1−α}{α}, where A′_{1} = (B_{2}C_{2}) ∩ (A_{1}A_{2}), A′_{1} is outside of B(X) if ∥M_{1} − A_{2}∥ is small enough and hence C_{2} is outside of B(X) as well. Therefore there exists a point M_{1} on (A_{1}A_{2}) such that the points B_{2} = M_{1} + uB, u > 0, and C_{2} = \frac{1}{γ}(M_{1} − αA_{1} − βB_{2}) are on S(X) and the proof is complete.

Lemma 2. There exists an ellipse which is inside the unit ball B(X) and touches S(X) at four points at least.

Proof. It is easy to show that an ellipse of maximum area inside B(X) touches S(X) at four points at least (this argument seems to be used frequently, see, e.g., [3], p. 322).

Lemma 3. Let φ and ψ be two functions defined on the interval I = (a − ε, a + ε), ε > 0, such that ψ(x) ≥ φ(x), ∀x ∈ I, ψ(a) = φ(a) and the derivatives φ′(a), ψ′_{−}(a), ψ′_{+}(a) exist. If ψ′_{−}(a) ≥ ψ′_{+}(a), then ψ′_{−}(a) = ψ′_{+}(a) = φ′(a).

Proof.

φ′(a) = lim_{u→0, u>0} \frac{φ(a) − φ(a − u)}{u} ≥ lim_{u→0, u>0} \frac{ψ(a) − ψ(a − u)}{u} = ψ′_{−}(a)

≥ ψ′_{+}(a) = lim_{u→0, u>0} \frac{ψ(a + u) − ψ(a)}{u} ≥ lim_{u→0, u>0} \frac{φ(a + u) − φ(a)}{u} = φ′(a)

which proves the lemma.

Proof of the Theorem. Let E be the ellipse from Lemma 2 and A′, B′ be the points of the intersection S(X) ∩ E, A′ ≠ B′, A′ ≠ −B′. Apply an affine transformation T that carries E into the unit circle of (R^{2}, ∥·∥_{2}), ∥·∥_{2} being the usual l_{2} norm. Let XOY be an orthogonal Cartesian system on R^{2} such that T (A′) = (−1, 0). Denote (−1, 0) by A, and T (B′) by B = (b_{1}, b_{2}). Obviously, b_{1}^{2} + b_{2}^{2} = 1 and b_{2} ≠ 0. By Lemma 1 there exist the points \overline{A}_{1}, \overline{B}_{1}, \overline{C}_{1} from T (S(X)) for which the following equalities hold:

\frac{\overline{A}_{1} − \overline{M}}{∥\overline{A}_{1} − \overline{M}∥} = A, \frac{\overline{B}_{1} − \overline{M}}{∥\overline{B}_{1} − \overline{M}∥} = B,  (2)

where

\overline{M} = α\overline{A}_{1} + β\overline{B}_{1} + γ\overline{C}_{1} .  (3)

Denote now

(x′, y′) = \frac{\overline{C}_{1} − \overline{M}}{∥\overline{C}_{1} − \overline{M}∥}.  (4)

Since β > 0 relation (2) and (3) show that y′ ≠ 0.

Let M_{ε} be the point M_{ε} = (aε, ε), a = \frac{x′}{y′} . Introduce the notation:

\overline{A}_{1} − \overline{M} − M_{ε} = (x_{1} − x_{0} − aε, y_{1} − y_{0} − ε) := (m_{1} − aε, n_{1} − ε),

\overline{B}_{1} − \overline{M} − M_{ε} = (x_{2} − x_{0} − aε, y_{2} − y_{0} − ε) := (m_{2} − aε, n_{2} − ε),

\overline{C}_{1} − \overline{M} − M_{ε} = (x_{3} − x_{0} − aε, y_{3} − y_{0} − ε) := (m_{3} − aε, n_{3} − ε).

It is clear that n_{1} = 0, m_{1} ≠ 0, n_{2} ≠ 0, n_{3} ≠ 0 and ∥\overline{A}_{1} − \overline{M}∥ = −m_{1}, ∥\overline{B}_{1} − \overline{M}∥ = \frac{n_{2}}{b_{2}} , ∥\overline{C}_{1} − \overline{M}∥ = \frac{n_{3}}{y′} . Since a = \frac{x′}{y′} = \frac{x_{3}−x_{0}}{y_{3}−y_{0}} = \frac{m_{3}}{n_{3}} we get \overline{C}_{1} − \overline{M} − M_{ε} = (m_{3} − aε, n_{3} − ε) = (m_{3} − \frac{m_{3}}{n_{3}} ε, n_{3} − ε) = \frac{n_{3}−ε}{y′}(x′, y′) and hence

∥\overline{C}_{1} − \overline{M} − M_{ε}∥ = \frac{n_{3} − ε}{y′}.  (5)

We are going to estimate the norms of the two other vectors. First we consider the case ε > 0. Without loss of generality we may assume that b_{2} < 0. Consider the two lines (L_{1}) : y = −ux−u and (L_{2}) : y = (−b−ω(u))(x−b_{1})+b_{2} where u > 0, b = b_{1}/b_{2} and ω is a positive continuous function defined on [0, ∞) such that lim_{u→∞} ω(u) = 0. By Lemma 3 there exist the tangents to T (S(X)) at the points A and B and they are expressed by the equations x = −1, y = −b(x − b_{1}) + b_{2}, respectively. The line (L_{1}) passes the point A and is different from the tangent at A. Therefore (L_{1}) intersects T (S(X)) at some other point A_{u} ≠ A. By the convexity of the unit ball the segment \overline{A_{u}A} = \{vA + (1 − v)A_{u}, 0 ≤ v ≤ 1\} is inside T (B(X)). Let (\overline{x}, \overline{y}) be the point of intersection of the lines \{v(\overline{A}_{1} − \overline{M} − M_{ε}), v ∈ R\} and (L_{1}), i.e., \overline{x} = \frac{−u(m_{1}−aε)}{(m_{1}−aε)u−ε}. If ε is small enough, then the point (\overline{x}, \overline{y}) is on the segment \overline{A_{u}A} and therefore we get the inequality

∥\overline{A}_{1} − \overline{M} − M_{ε}∥ ≤ \frac{∥\overline{A}_{1} − \overline{M} − M_{ε}∥_{2}}{∥(\overline{x}, \overline{y})∥_{2}} = \frac{m_{1} − aε}{\overline{x}} = −m_{1} + aε + ε/u (6)

for all ε, 0 < ε < ε′_{u}, ε′_{u} > 0.

Now we consider the intersection (\overline{\overline{x}}, \overline{\overline{y}}) of the lines \{v(\overline{B}_{1} −\overline{M} −M_{ε}), v ∈ R\} and (L_{2}). We get \overline{\overline{x}} = \frac{(m_{2}−aε)(b_{1}b+b_{2}+b_{1}ω(u))}{n_{2}−ε+(m_{2}−aε)(b+ω(u))}. The same arguments show that there exists ε′′_{u} > 0 such that

∥\overline{B}_{1} − \overline{M} − M_{ε}∥ ≤ \frac{m_{2} − aε}{\overline{\overline{x}}} = \frac{n_{2} − ε + (m_{2} − aε)(b + ω(u))}{b_{1}b + b_{2} + b_{1}ω(u)}

for all ε, 0 < ε < ε′′_{u}.

Since \frac{m_{2}}{n_{2}} = \frac{b_{1}}{b_{2}} = b, we get

∥\overline{B}_{1} − \overline{M} − M_{ε}∥ ≤ \frac{n_{2}}{b_{2}} − \frac{1 + ab + aω(u)}{(1 + b^{2} + bω(u))b_{2}} · ε.  (7)

By the property of the functional F_{μ}, there exists ε′ > 0 such that F (\overline{M}) ≤ F (\overline{M} + M_{ε}) for all ε, 0 < ε < ε′. If ε < min(ε′, ε′_{u}, ε′′_{u}) = ε_{u}, we obtain, using relations (5), (6) and (7),

F (\overline{M}) = αm_{1}^{2} + β \frac{n_{2}^{2}}{b_{2}^{2}} + γ \frac{n_{3}^{2}}{y′^{2}}

≤ α (m_{1} − (a + \frac{1}{u}) ε)^{2} + β (\frac{n_{2}}{b_{2}} − \frac{1 + ab + aω(u)}{(1 + b^{2} + bω(u))b_{2}} ε)^{2} + γ (\frac{n_{3} − ε}{y′})^{2} ,

i.e., 0 ≤ 2h_{u}ε + h′_{u}ε^{2}, where

h_{u} = −αm_{1} (a + \frac{1}{u}) − β \frac{n_{2}(1 + ab + aω(u))}{b_{2}^{2}(1 + b^{2} + bω(u))} − γ \frac{n_{3}}{y′^{2}} .

Since ε > 0, we have

h_{u} ≥ −\frac{εh′_{u}}{2}

for all ε, 0 < ε < ε_{u}, i.e. h_{u} ≥ 0 for all u > 0. Let \overline{h} = −αm_{1}a − \frac{βn_{2}(1+ab)}{b_{2}^{2}(1+b^{2})} − γ \frac{n_{3}}{y′^{2}} . We have \overline{h} = lim_{u→∞} h_{u} ≥ 0. Passing now to the case ε < 0, we consider the two lines y = ux + u and y = (−b + ω(u))(x − b_{1}) + b_{2}. Using the same arguments as for the case ε > 0, we can derive the inequality \overline{h} ≤ 0. Therefore \overline{h} = 0, which gives

y′^{2} = −\frac{γn_{3}}{αam_{1} + β(1 + ab)n_{2}} .

Using the relations x_{0} = x_{2}+b(y_{1}−y_{2}), y_{0} = y_{1}, x_{3} =−\frac{α}{γ}x_{1} + \frac{1−β}{γ} x_{2} + \frac{b}{γ} (y_{1} − y_{2}), y_{3} = \frac{1−α}{γ}y_{1} − \frac{β}{γ} y_{2} which follow from (2), (3), (4), we get αm_{1} = βan_{2} −βbn_{2} and n_{3} = −\frac{β}{γ} n_{2}, i.e.,

x′^{2} + y′^{2} = (1 + a^{2})y′^{2} = \frac{βn_{2}(1 + a^{2})}{βa^{2}n_{2} − βabn_{2} + βn_{2} + βabn_{2}} = 1.

Denote by arc(A, B) the part of the circle T (E) which is inside the smaller angle generated by the vectors A and B. As we have just proved, if T (S(X)) and T (E) coincide at two points A and B they coincide at one more point C ∈ arc(A, B). Continuing this process, we see that T (S(X)) and arc(A, B) coincide on a dense set of points. Hence arc(A, B) ⊂ T (S(X)) and by the symmetry argument arc(−A, −B) ⊂ T (S(X)). The same reasoning for the points A and −B shows that arc(A, −B) ⊂ T (S(X)) and therefore arc(−A, B) ⊂ T (S(X)) as well. The proof of statement (ii) is complete. Statement (i) can be proved similarily.

Remarks: 1. In the Theorem we can replace the unit sphere S(X) by any sphere with center at \overline{x} and radius R. Moreover, in the case dim X = 2, S(X) and its center can be replaced by any continuous convex closed curve S on R^{2} and any point from the area which is bounded by S.

2. The Theorem holds true for measures concentrated at n points of S(X), n ≥ 3, with any fixed positive weights α_{1}, α_{2}, …, α_{n}.

3. The complex and quaternion versions of the Theorem are easily derived from the real one.

References

1. D. Amir, Characterization of inner product spaces. Birkhäuser Verlag, Basel, 1986.

2. R. Durier, Optimal locations and inner products. J. Math. Anal. Appl. 207(1991), 220–230.

3. M. Day, Some characterization of inner-product spaces. Trans. Amer. Math. Soc. 62(1947), 320–327.

(Received 15.07.2000; revised 2.03.2001)

Author’s address:

N. Muskhelishvili Institute of Computational Mathematics

Georgian Academy of Sciences

8, Akuri St., Tbilisi 380093

Georgia

E-mail: vvk@gw.acnet.ge