Operators preserving orthogonality are isometries

arXiv:math/9212203v1 [math.FA] 4 Dec 1992Operators preserving orthogonality are isometriesAlexander KoldobskyDepartment of MathematicsUniversity of Missouri-ColumbiaColumbia, MO 65211Abstract. Let E be a real Banach space.

For x, y ∈E, we follow R.James in saying thatx is orthogonal to y if ∥x+αy∥≥∥x∥for every α ∈R. We prove that every operator fromE 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 isorthogonal 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 impliesTx⊥Ty.

We prove here that the converse statement is valid, namely, every linear operatorpreserving orthogonality is an isometry multiplied by a constant.D.Koehler and P.Rosenthal [4] have proved that an operator is an isometry if and onlyif it preserves any semi-inner product. It is easy to show (see [2]) that orthogonality ofvectors with respect to any semi-inner product implies James’ orthogonality.

So the resultof 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 ofsupport functionals at the point x. It is well-known [1] that, for every x, y ∈E, x ̸= 0,(1)limα→0(∥x + okstate.αy∥−∥x∥)/α =supx∗∈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∗1(y) = x∗2(y)for every x∗1, x∗2 ∈S(x + αy).Fix linearly independent vectors x, y ∈E. The function φ(α) = ∥x + αy∥is convex onR 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

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 havesup{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) thenx∗((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 numberf(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 ofx∗∈S(x + αy). Fix x∗∈S(x + αy).

If x∗(y) = 0 then, by Lemma 1, x + αy⊥y. Ifx∗(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 canbe expressed in terms of the function f.Lemma 4.

For every α > M,(2)∥x + αy∥= ∥x + My∥exp(Z αM(t + f(t))−1dt)and, for every α < m,(3)∥x + αy∥= ∥x + my∥exp(−Z mα(t + f(t))−1dt)2

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))−1Since α is an arbitrary number from D(x, y) ∩[M, ∞] and Lebesgue measure of the setR \ D(x, y) is zero, we get(4)Z αM(∥x + ty∥′t/∥x + ty∥)dt =Z αM(t + f(t))−1dtfor every α > M. It is easy to see that the function α →ln∥x + αy∥satisfies the Lipschitzcondition and, therefore, is absolutely continuous.

Every absolutely continuous functioncoincides with the indefinite integral of its derivative [5], so the integral in the left-handside 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 preservingorthogonality. 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.

Consideran arbitrary y ∈E such that x ̸= αy for every α ∈R. Denote by I1 and I2 the 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. Assumethat I = I2 \ I1 ̸= ∅and consider a number α ∈I such that α ∈D(x, y) ∩D(Tx, Ty).Since α ∈I2 we have Tx + αTy⊥Ty.

By Lemma 3, there exists a number f(α) such thatx + αy⊥x −f(α)y and, consequently, Tx + αTy⊥Tx −f(α)Ty. By Lemma 1, for everyfunctional x∗∈S(Tx + αTy), we have x∗(Ty) = 0 and x∗(Tx −f(α)Ty) = 0.

But then0 = 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 pairsof vectors (x, y) and (Tx, Ty).The functions ∥x + αy∥and ∥Tx + αTy∥are constant and non-zero on [m, M], sothere exist k1, k2 ∈R such that ∥x+αy∥= k1 and ∥Tx+αTy∥= k2 for 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 latterequality by α and tend α to infinity.

We get ∥Tx∥/∥x∥= ∥Ty∥/∥y∥for every non-zerox, y ∈E which completes the proof.Acknowledgements.. Part of this work was done when I was visiting Memphis StateUniversity. I am grateful to Professors J.Jamison and A.Kaminska for fruitful discussionsand hospitality.

I express my gratitude to Professor J.Arazy for helpful remarks.3

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, Budapest1972.4

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