Page 1

arXiv:math/9212203v1 [math.FA] 4 Dec 1992

Operators preserving orthogonality are isometries

Alexander Koldobsky

Department of Mathematics

University of Missouri-Columbia

Columbia, MO 65211

Abstract. Let E be a real Banach space. For x,y ∈ E, we follow R.James in saying that

x is orthogonal to y if ?x+αy? ≥ ?x? for every α ∈ R. We prove that every operator from

E into itself preserving orthogonality is an isometry multiplied by a constant.

Let E be a real Banach space. For x,y ∈ E, we follow R.James in saying that x is

orthogonal to y (x⊥y) if ?x + αy? ≥ ?x? for every α ∈ R.

It is clear that every isometry T : E → E preserves orthogonality, i.e. x⊥y implies

Tx⊥Ty. We prove here that the converse statement is valid, namely, every linear operator

preserving orthogonality is an isometry multiplied by a constant.

D.Koehler and P.Rosenthal [4] have proved that an operator is an isometry if and only

if it preserves any semi-inner product. It is easy to show (see [2]) that orthogonality of

vectors with respect to any semi-inner product implies James’ orthogonality. So the result

of this paper seems to refine that from [4].

We start with some auxiliary facts.

For x ∈ E,x ?= 0, denote by S(x) = {x∗∈ E∗: ?x∗? = 1,x∗(x) = ?x?} the set of

support functionals at the point x. It is well-known [1] that, for every x,y ∈ E,x ?= 0,

(1)lim

α→0(?x + okstate.αy? − ?x?)/α =sup

x∗∈S(x)

x∗(y)

The limit in the left-hand side as α → 0 is equal to inf{x∗(y) : x∗∈ S(x)}. Therefore,

the function φ(α) = ?x+αy? is differentiable at a point α ∈ R if and only if x∗

for every x∗

1,x∗

Fix linearly independent vectors x,y ∈ E. The function φ(α) = ?x+αy? is convex on

R and, hence, φ is differentiable almost everywhere on R with respect to Lebesgue measure

(see [5]).

Denote by D(x,y) the set of points α at which φ is differentiable.

1(y) = x∗

2(y)

2∈ S(x + αy).

1

Page 2

Lemma 1. Let α ∈ D(x,y),a,b ∈ R. Then

(i) the number x∗(ax + by) does not depend on the choice of x∗∈ S(x + αy),

(ii) x + αy⊥ax + by if and only if x∗(ax + by) = 0 for every x∗∈ S(x + αy).

Proof: (i) As shown above, x∗(y) does not depend on the choice of x∗∈ S(x + αy).

Besides,

x∗(x) = x∗(x + αy) − αx∗(y) = ?x + αy? − αx∗(y)

for every x∗∈ S(x + αy), so x∗(x) does not depend on the choice of a functional x∗also.

(ii) If x + αy⊥ax + by then, by the definition of orthogonality and (1), we have

sup{x∗(ax+ by) : x∗∈ S(x +αy)} ≥ 0 and inf{x∗(ax+ by) : x∗∈ S(x +αy)} ≤ 0. By (i),

sup = inf = 0.

On the other hand, if x∗(ax + by) = 0 for any x∗∈ S(x + αy) then

x∗((x + αy) + γ(ax + by)) = x∗(x + αy) = ?x + αy?

for every γ ∈ R. Since ?x∗? = 1 we have x + αy⊥ax + by.

The following fact is an easy consequence of the convexity of the function α → ?x+αy?.

Lemma 2. The set of numbers α for which x+αy⊥y is a closed segment [m,M] in R and

?x + αy? = ?x + my? for every α ∈ [m,M].

Lemma 3. Let α ∈ D(x,y). Then either x + αy⊥y or there exists a unique number

f(α) ∈ R such that x + αy⊥x − f(α)y.

Proof: By Lemma 1, the numbers x∗(x) and x∗(y) does not depend on the choice of

x∗∈ S(x + αy). Fix x∗∈ S(x + αy). If x∗(y) = 0 then, by Lemma 1, x + αy⊥y. If

x∗(y) ?= 0 then, again by Lemma 1, x + αy⊥x − βy if and only if x∗(x − βy) = 0. Thus,

f(α) = x∗(x)/x∗(y).

By Lemma 2, the function f is defined on R \ [m,M]. It appears that the norm can

be expressed in terms of the function f.

Lemma 4. For every α > M,

(2)

?x + αy? = ?x + My?exp(

?α

M

(t + f(t))−1dt)

and, for every α < m,

(3)

?x + αy? = ?x + my?exp(−

?m

α

(t + f(t))−1dt)

2

Page 3

Proof: Let α ∈ D(x,y),α > M. Fix x∗∈ S(x + αy). By Lemma 3, x∗(x) = f(α)x∗(y)

and, by (1), x∗(y) = ?x + αy?

α. Therefore, x∗(x) = x∗(x + αy) − αx∗(y) = ?x + αy? −

α?x + αy?

α. We have

?x + αy?

′

′

′

α/?x + αy? = (α + f(α))−1

Since α is an arbitrary number from D(x,y) ∩ [M,∞] and Lebesgue measure of the set

R \ D(x,y) is zero, we get

(4)

?α

M

(?x + ty?

′

t/?x + ty?)dt =

?α

M

(t + f(t))−1dt

for every α > M. It is easy to see that the function α → ln?x+αy? satisfies the Lipschitz

condition and, therefore, is absolutely continuous. Every absolutely continuous function

coincides with the indefinite integral of its derivative [5], so the integral in the left-hand

side of (4) is equal to ln(?x+αy?/?x+My?) and we get (2). The proof of (3) is similar.

Now we can prove the main result.

Theorem. Let E be a real Banach space and T : E → E be a linear operator preserving

orthogonality. Then T = kU where k ∈ R and U is an isometry.

Proof: Assume that T is not the zero operator and fix x ∈ E such that Tx ?= 0. Consider

an arbitrary y ∈ E such that x ?= αy for every α ∈ R. Denote by I1and I2the intervals

[m,M] corresponding to the pairs of vectors (x,y) and (Tx,Ty).

Since T preserves orthogonality we have I1⊂ I2. Let us prove that I1= I2. Assume

that I = I2\ I1?= ∅ and consider a number α ∈ I such that α ∈ D(x,y) ∩ D(Tx,Ty).

Since α ∈ I2we have Tx + αTy⊥Ty. By Lemma 3, there exists a number f(α) such that

x + αy⊥x − f(α)y and, consequently, Tx + αTy⊥Tx − f(α)Ty. By Lemma 1, for every

functional x∗∈ S(Tx + αTy), we have x∗(Ty) = 0 and x∗(Tx − f(α)Ty) = 0. But then

0 = x∗(x + αy) = ?x + αy? and we get a contradiction.

Thus, the numbers m,M and, obviously, the function f(α) are the same for both pairs

of vectors (x,y) and (Tx,Ty).

The functions ?x + αy? and ?Tx + αTy? are constant and non-zero on [m,M], so

there exist k1,k2∈ R such that ?x+αy? = k1and ?Tx+αTy? = k2for every α ∈ [m,M].

Using this fact and (2), (3) for both pairs of vectors (x,y) and (Tx,Ty) we get ?Tx+

αTy? = (k2/k1)?x + αy? for every α ∈ R. First put α = 0 and then divide the latter

equality by α and tend α to infinity. We get ?Tx?/?x? = ?Ty?/?y? for every non-zero

x,y ∈ E which completes the proof.

Acknowledgements.. Part of this work was done when I was visiting Memphis State

University. I am grateful to Professors J.Jamison and A.Kaminska for fruitful discussions

and hospitality. I express my gratitude to Professor J.Arazy for helpful remarks.

3

Page 4

References.

1. Dunford, N. and Schwartz, J. Linear operators. Vol.1, Interscience, New York 1958.

2. Giles, J.R. Classes of semi-inner product spaces, Trans. Amer. Math. Soc. 129

(1967), 436-446.

3. James, R.C. Orthogonality and linear functionals in normed linear spaces, Trans.

Amer. Math. Soc. 61 (1947), 265-292.

4. Koehler, D. and Rosenthal, P. On isometries of normed linear spaces, Studia Mathe-

matica 36 (1970), 213-216.

5. Riesz, F. and Sz-Nagy, B. Lecons d’analyse fonctionelle, Akademiai Kiado, Budapest

1972.

4