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Математичнi Студiї. Т.57, №2 Matematychni Studii. V.57, No.2
I. Karpenko, O. Zavarzina
LINEAR EXPAND-CONTRACT PLASTICITY OF ELLIPSOIDS
REVISITED
I. Karpenko, O. Zavarzina. Linear expand-contract plasticity of ellipsoids revisited , Mat. Stud.
57 (2022), 192–201.
This work is aimed to describe linearly expand-contract plastic ellipsoids given via quadratic
form of a bounded positively defined self-adjoint operator in terms of its spectrum.
Let Ybe a metric space and F:Y→Ybe a map. Fis called non-expansive if it does not
increase distance between points of the space Y. We say that a subset Mof a normed space X
is linearly expand-contract plastic (briefly an LEC-plastic) if every linear operator T:X→X
whose restriction on Mis a non-expansive bijection from Monto Mis an isometry on M.
In the paper, we consider a fixed separable infinite-dimensional Hilbert space H. We define
an ellipsoid in Has a set of the following form E={x∈H:⟨x, Ax⟩ ≤ 1}where Ais a self-
adjoint operator for which the following holds: inf∥x∥=1 ⟨Ax, x⟩>0and sup∥x∥=1 ⟨Ax, x⟩<∞.
We provide an example which demonstrates that if the spectrum of the generating operator
Ahas a non empty continuous part, then such ellipsoid is not linearly expand-contract plastic.
In this work, we also proof that an ellipsoid is linearly expand-contract plastic if and only
if the spectrum of the generating operator Ahas empty continuous part and every subset of
eigenvalues of the operator Athat consists of more than one element either has a maximum of
finite multiplicity or has a minimum of finite multiplicity.
1. Introduction. Let Mbe a metric space and F:M→Mbe a map. Fis called non-
expansive if it does not increase distance between points of the space M.Mis called expand-
contract plastic (or just plastic for short) if every non-expansive bijection F:M→Mis an
isometry.
There is a number of relatively recent publications devoted to plasticity of the unit balls
of Banach spaces (see [1, 3, 4, 6, 12]). Here we give only one theorem which is a simple
consequence of Theorem 1 in [7] or Theorem 3.8 in [12].
Theorem 1. Let Xbe a finite-dimensional Banach space. Then its unit ball is plastic.
However, the question about plasticity of the unit ball of an arbitrary infinite-dimensional
Banach space is open. At least, there are no counterexamples. On the other hand, an example
of non-plastic ellipsoid in separable Hilbert space was built in [3]. In [14], this example was
generalized and the following definition was introduced.
Definition 1. Let Mbe a subset of a normed space X. We say that Mis linearly expand-
contract plastic (briefly an LEC-plastic) if every linear operator T:X→Xwhose restriction
on Mis a non-expansive bijection from Monto Mis an isometry on M.
2010 Mathematics Subject Classification: 46B20, 54E15.
Keywords: non-expansive map; ellipsoid; linearly expand-contract plastic space.
doi:10.30970/ms.57.2.192-201
©I. Karpenko, O. Zavarzina, 2022
LINEAR PLASTICITY OF ELLIPSOIDS 193
In the mentioned article [14] the ellipsoids of the following form were considered
E=nx=X
n∈N
xnen∈H:X
n∈N
xn
a(n)
2
≤1o,
where His a separable Hilbert space with basis {en}∞
1and a(n)>0. There was given a
description of the LEC-plastic ellipsoids of such a form.
In what follows, we use the notations from [5]. The letter Hdenotes a fixed separable
infinite-dimensional Hilbert space (real or complex), the symbol ⟨x, y⟩stays for the scalar
product of elements x, y ∈H. We use the symbol Lin to denote the linear span, and the
symbol Lin to denote the closed linear span.
In the present paper, we will consider a more general definition of an ellipsoid.
Definition 2. An ellipsoid in His a set of the form
E={x∈H:⟨x, Ax⟩ ≤ 1},
where Ais a self-adjoint operator such that inf∥x∥=1 ⟨Ax, x⟩>0and sup∥x∥=1 ⟨Ax, x⟩<∞.
We will denote the boundary of Eby
S={x∈H:⟨x, Ax⟩= 1}.
In what follows, σ(A)will stand for the spectrum of A. Note that in this case σ(A)is bounded
from below and above by some positive constants.
In this paper we will show, that in fact the description of LEC-plastic ellipsoids in [14] was
complete. In other words, there is no other LEC-plastic ellipsoids, except for those already
described.
2. Basic facts. For our purpose, we will need some results related to measure theory (see,
e.g., [5], [9], [13]). Let us collect these results.
Recall that a distribution function of a given positive finite Borel measure on the real
numbers µis given by
Fµ(t) = µ([0, t]).
Notice that this function is non-decreasing, and hence its generalized inverse
F−1
µ(t) = sup{x:Fµ(x)≤t}
is well-defined and also non-decreasing.
Notice that any normalized atomless Borel measure µcan be mapped into Lebesgue
measure λon [0,1] (Indeed, performing a simple computation, we obtain
µ([0, F −1
µ([0, t])]) = t=λ([0, t]),t∈[0,1]).
Hence, we get the following theorem.
Theorem 2 ([13] or Theorem 9.2.2 [2]).All atomless standard probability spaces are mu-
tually almost isomorphic.
Corollary 1. Let µand νbe (finite and compactly supported) atomless Borel measures on
Rwith Mν=ν(R)and Mµ=µ(R). Then there exists a map Gµ,ν such that ν=Mν
Mµµ◦Gµ,ν .
Remark 1. Observe that:
[1.]Gµ,ν can be written explicitly in terms of corresponding distribution functions, namely,
Gµ,ν =F−1
µ◦Mµ
Mν
Fν.
[2.]Gµ,ν : supp (ν)−→ supp (µ).
194 I. KARPENKO, O. ZAVARZINA
The next theorem can be found in [9], but since this source was not published yet, we
will provide the proof.
Theorem 3 ([9], Theorem 2.16).For given measure spaces (X, Σ) and (Y, Ω), a measure µ
on Σand a function f:X−→ Ywhich is measurable w.r.t. µ, we define a measure f∗µon Ω
as f∗µ(B) = µ(f−1(B)) for B∈Ω. Let g:Y−→ Cbe a Borel function. Then the function
g◦f:X−→ Cis integrable w.r.t. µif and only if gis integrable w.r.t f∗µ. Moreover
ZY
gd(f∗µ) = ZY
g◦fdµ.
Proof. It suffices to check this formula for simple functions g, which follows since χB◦f=
χf−1(B), where χBis the characteristic function of the set B.
Furthermore, we will need some results related to operator theory (see, e.g., [8], [11]). Let
Abe a bounded self-adjoint operator on a separable Hilbert space H. Then we can introduce
continuous functions of A, as it is shown in the following theorem. Before formulating the
theorem, let us recall that C(σ(A)) denotes the set of continuous functions on σ(A)and
L(H)stands for the space of linear operators acting from Hto H.
Theorem 4 ([8], Theorem VII.1; [11], Theorem 3.1).Let Abe a bounded self-adjoint
operator on a Hilbert space H. Then there is a unique map ϕA:C(σ(A)) → L(H)wi-
th the following properties: 1. ϕA(fg) = ϕA(f)ϕA(g),ϕA(λf) = λϕA(f),ϕA(1) = I,
ϕA(¯
f) = ϕA(f)∗;2.||ϕA(f)||L(H)≤C||f||∞;3.if f(x) = x, then ϕA(f) = A.
Moreover, 4.if Aψ =λψ, than ϕA(f)ψ=f(λ)ψ;5. σ(ϕA(f)) = {f(λ)|λ∈σ(A)};
6.if f≥0, then ϕA(f)≥0;7.||ϕA(f)|| =||f||∞.
Then we define f(A) := ϕA(f).
For every ψ∈Hwe can define a corresponding linear functional on C(σ(A)) mapping
f→ ⟨ψ, f (A)ψ⟩. Then by Riesz theorem, there exist a unique measure µψon σ(A)such
that ⟨ψ, f (A)ψ⟩=Rσ(A)f(λ)dµψ. The measure µψis called the spectral measure associated
with the vector ψ.
The next important result (spectral theorem) states that every bounded self-adjoin
operator can be realized as multiplication operator on a suitable measure space. Let us
specify that in the following theorem and further in the text we use the notion L2(R, dµ)for
the space of measurable scalar-valued functions fon Rfor which the integral RR∥f(t)∥2dµ
exists and ∥f∥= (RR∥f(t)∥2dµ)1/2.
Theorem 5 ([8], Theorem VII.3; [11], Lemma 3.4 and Theorem 3.6).Let Abe a bounded
self-adjoint operator on a separable Hilbert space H. Then, there exist measures {µn}N
n=1(N∈
Nor N=∞)on σ(A)and a unitary operator
U:H→
N
M
n=1
L2(R, dµn)
so that (UAU −1ψ)n(λ) = λψn(λ),where we write an element ψ∈ ⊕N
n=1L2(R, dµn)as an
N-tuple (ψ1(λ), ..., ψN(λ)).This realization of Ais called a spectral representation.
Let us define Hpp ={ψ∈H|µψis pure point},Hac ={ψ∈H|µψis absolutely
continuous},Hsc ={ψ∈H|µψis singularly continuous}.
LINEAR PLASTICITY OF ELLIPSOIDS 195
Theorem 6 ([8], Theorem VII.4; [11], Lemma 3.19).H=Hpp ⊕Hac ⊕Hsc. Each of these
subspaces is invariant under A.A|Hpp has a complete set of eigenvectors, A|Hac has only
absolutely continuous spectral measures and A|Hsc has only singularly continuous spectral
measures.
We will use the following notations:
σpp =σ(A|Hpp ), σcont =σ(A|Hcont ),where Hcont =Hac ⊕Hsc,
σac =σ(A|Hac ), σsc =σ(A|Hsc ), σp={λ:λis an eigenvalue of A}.
Note that
σcont =σac ∪σsc, σ(A) = σp∪σcont.
The following useful results can be found in [11] and [10].
Theorem 7 ([11], Theorem 2.20).Let A be bounded self-adjoint. Then
inf{σ(A)}= inf
||x||=1 ⟨x, Ax⟩,sup{σ(A)}= sup
||x||=1
⟨x, Ax⟩
Note that
inf
||x||=1 ⟨x, Ax⟩= inf
x∈H, x=0
⟨x, Ax⟩
||x||2,sup
||x||=1
⟨x, Ax⟩= sup
x∈H, x=0
⟨x, Ax⟩
||x||2.
Moreover, one can show the following:
Theorem 8 (see Problem 13.1 in [10]).Let A be bounded self-adjoint (particularly, σ(A)⊂
[a, b]). Then λ0:= inf{σ(A)}is an eigenvalue iff inf||x||=1 ⟨x, Ax⟩is a minimum. In this case,
eigenvectors are precisely the minimizers.
Since [10] also was not published yet, we provide the proof.
Proof. Consider the functional F:H→Rgiven by F(x) = ⟨x, (A−λ0)x⟩.
By assumption we have F(x)≥0and F(x0)=0, where x0is the minimizer of ⟨x, Ax⟩.
Then one may calculate the Gateaux derivative δF (x0, x) = 2 Re (⟨(A−λ0)x0, x⟩) = 0 for
x∈H. Replacing xby ix we also get 2 Im (⟨(A−λ0)x0, x⟩)=0. Hence ⟨(A−λ0)x0, x⟩= 0,
for x∈H, which means (A−λ0)x0= 0.
3. Main result.
Proposition 1. Suppose the spectrum σ(A)of the self-adjoint operator Acontains a set of
eigenvalues Bpossessing the following properties:
1. Bhas at least two elements;
2. either Bdoesn’t have minimum or the multiplicity of the minimum is infinite;
3. either Bdoesn’t have maximum or the multiplicity of the maximum is infinite.
Then Eis not LEC-plastic.
Proof. Denote r= inf B,R= sup B; according to (1) r < R. The property (2) ensures the
existence of distinct nk∈N,k= 1,2, . . . such that eigenvalues λnk∈B,λnk<1
2(r+R)and
λn1≥λn2≥λn3≥. . . , lim
k→∞ λnk=r.
196 I. KARPENKO, O. ZAVARZINA
Analogously, the property (3) gives us the existence of distinct nk∈N,k= 0,−1,−2, . . .
such that λnk∈Band
λn1< λn0≤λn−1≤λn−2≤. . . , lim
k→−∞ λnk=R.
Take in Hthe orthonormalized eigenvectors enkcorresponding to λnk(in case of infinite
multiplicity of the minimum we are choosing them to be ONB of Ker(A−λmin), analogously
for the maximum) and extend to an orthonormal basis enin H. Define the linear operator
Tas follows: T en=enfor n∈N\ {nk}k∈Z, and T enk=qλnk
λnk−1
enk−1, for k∈Z. Using the
fact that for a self-adjoint operator Aand x=Pxkenk(here enkare eigenvectors of A):
x+x⊥, A(x+x⊥)=⟨x, Ax⟩+x⊥, Ax⊥, we get that the linear non-expansive operator
Tmaps Eonto itself bijectively but not isometrically.
Now, let us consider the case when the operator has purely continuous spectrum.
Proposition 2. Let µ(t)be an (finite) atomless Borel measure with supp (µ)⊂(α, β),
α > 0. Let A:L2(R, dµ(t)) →L2(R, dµ(t)) be an operator acting by the rule
Af(t) = tf (t).
Consider an ellipsoid E∈L2(R, dµ(t)) generated by this operator A, i.e.
E={f∈L2(R, µ(t)dt): ⟨Af, f ⟩=Zβ
α
t|f(t)|2dµ(t)≤1}.
Eis not LEC-plastic.
Proof. Note that under our assumptions L2(R, dµ(t)) can be identified with L2((α, β), dµ(t)).
Consider some infinite partition of the segment [α, β) = F∞
k=−∞ ∆k,where∆k= [ak, ak+1 ),
a−∞ =α,a∞=β, and such that µ(∆k)>0for all k(we can do this e.g. using any
convergent series). Note that L2((α, β), dµ(t)) = ⊕2L2(∆k, dµk(t)) with µk:= µ|∆k. We will
use the following notation Mµk:= µk(∆k) = µ(∆k). By Theorem 2, the spaces (∆k, µk(t))
are mutually almost isomorphic and we will denote the corresponding isomorphism by Gk:=
Gµk,µk+1 : supp (µk+1)−→ supp (µk).
Introduce the associated operators Hk:L2(∆k, dµk)→L2(∆k+1 , dµk+1)acting as
Hkfk=fk◦Gk·sMµk+1
Mµk
.
Then using Theorem 3 in the first step and Corollary 1 on the second step, we get
Z∆k+1
|fk◦Gk|2Mµk+1
Mµk
dµk+1 =Z∆k
|fk|2Mµk+1
Mµk
d(Gk∗µk+1) = Z∆k
|fk|2dµk.
For the second step note that Gk∗µk+1 =µk+1 ◦Gk=Mµk
Mµk+1
µk. This means that
∥Hkfk∥L2(∆k+1,dµk+1 )=∥fk∥L2(∆k,dµk).
Now, consider an operator T:L2((α, β), dµ(t)) →L2((α, β), dµ(t)) acting as
T fk=gkHkfk,
LINEAR PLASTICITY OF ELLIPSOIDS 197
where gkwill be defined later.
Then we have
⟨f, Af ⟩=X
k∈ZZ∆k
t|f(t)|2dµk(t),
and using again Theorem 3 and Corollary 1, we get
⟨T f, AT f⟩=X
k∈ZZ∆k+1
t|gk(t)|2|(fk◦Gk)(t)|2Mµk+1
Mµk
dµk+1(t) =
=X
k∈ZZ∆k
G−1
k(s)(gk◦G−1
k)(s)
2|fk(s)|2dµ(s).
Let us denote ˆgk(s) := gkG−1
k(s).
We will choose gk(ˆgk(s)) in such way that ⟨f, Af ⟩=⟨T f, AT f ⟩(then T:E−→ Eis
bijective), i.e.
ˆgk(s) := rs
G−1
k(s), s ∈∆k.
It remains to show that Tis non-expansive but not an isometry. Notice that ˆgk(s)2≤1
implies that Tis non-expansive. Then to show that Tis not an isometry it suffices to show
that there are k∈Nand t∈∆ksuch that ˆgk(s)2<1in some neighbourhood of this t.
Indeed, for s∈∆kby construction G−1
k(s)∈∆k+1, which means ˆgk(s)2<1for s∈∆k.
Thus, Tis non-expansive but not an isometry.
Corollary 2. Let Abe a bounded self-adjoint operator. Let σcont ={∅}. Then the ellipsoid
generated by this operator is not LEC-plastic.
Proof. Indeed, due to the Theorem 6 we can split an operator into two pieces: the first piece
corresponds to the continuous measure, the second piece corresponds to the rest. Then we
can apply Theorem 5 to the first piece and get that there is a spectral representation of
A|σcont . One may chose one of those parts where Aacts as multiplication on the independent
variable and the construction described in the previous proposition allows us to obtain an
operator Twhich is a non-expansive bijection, but not an isometry.
Three following lemmas are, in fact, building blocks for the proof of Theorem 9.
Lemma 1. Let T:H→Hbe a linear operator which maps Ebijectively onto itself. Then
Tmaps the whole Hbijectively onto itself and T(S) = S. If, moreover, Tis non-expansive
on E, then ∥T∥ ≤ 1.
Proof. Eis absorbing since it contains a ball (the spectrum is bounded from below by positive
constant). The rest of the proof is as in the article [14]. For convenience of the reader we
give it here.
Eis an absorbing set, so H=∪t>0tE. By linearity Tis injective on every set tE,
consequently it is injective on the whole H. Also, T(H) = ∪t>0T(tE) = ∪t>0tE =Hwhich
gives the surjectivity on H. Finally, S=E\ ∪t∈(0,1)tE, so T(S) = T(E)\ ∪t∈(0,1)T(tE) =
E\ ∪t∈(0,1)tE =S. If, moreover, Tis non-expansive on E, then for every x∈Hthere exists
t > 0such that tx ∈Eand we have ∥T(tx)∥=∥(T(0) −T(tx))∥≤∥tx −0∥=∥tx∥. It
remains to divide by tto obtain that ∥T x∥≤∥x∥for all x∈H.
198 I. KARPENKO, O. ZAVARZINA
Before moving to the next result, let us introduce the following notation:
Ht=Ker(A−t).
Lemma 2. Let σ(A) = σpp(A)⊂(0,+∞)and let the set of eigenvalues of an operator A
contain the minimal element rand let rhave finite multiplicity. Let T:H→Hbe a linear
operator which maps Ebijectively onto itself and whose restriction on Eis non-expansive.
Then T(Hr) = Hr,T(Hr∩E) = Hr∩Eand the restriction of Tonto Hris a bijective
isometry.
Proof. Theorem 7 implies that r= inf||x||=1 ⟨x, Ax⟩= infx∈H, x=0 ⟨x,Ax⟩
||x||2.
Recall that for x∈S, we have ⟨x, Ax⟩= 1. Hence, r= infx∈H, x=0 ⟨x,Ax⟩
||x||2= infx∈S⟨x,Ax⟩
||x||2=
infx∈S1
||x||2(second equality: renormalization using ⟨x, Ax⟩instead of norm). That is, for
x∈Swe have ||x|| ≤ q1
r.
Let x∈Swith ||x|| =q1
r(i.e. by Theorem 8 it is an eigenvector corresponding to r.
Moreover, all eigenvectors can be written in this form after a suitable renormalisation). Since
Tis non-expansive on E, we have ||T−1(x)|| ≥ ||x|| =q1
rand using T−1(S) = S, we get
||T−1(x)|| =q1
r. Hence, by Theorem 8, T−1(x)is an eigenvector corresponding to r. Hence,
T−1(Hr)⊂Hr.
Note that Hris finite dimensional, hence T−1|Hr:Hr7→ Hris surjective iff T−1:Hr7→ Hr
is injective. Injectivity of T−1|Hrfollows from Lemma 1 (i.e. injectivity of T−1). Hence,
T−1|Hr:Hr7→ Hris bijection and T−1(Hr) = Hr(i.e. T(Hr) = Hr). Hence, also T(Hr∩E) =
Hr∩E(combining T(Hr) = Hrwith T(E) = E).
We obtain the last claim using linearity and the fact that for all eigenvectors xcorrespon-
ding to rwith norm q1
r, we have that ||T−1(x)|| =q1
r. (We can also observe that S∩Hr
is a sphere in Hrand map Tfrom S∩Hronto S∩Hris bijective.)
Lemma 3. Let σ(A) = σpp(A)⊂(0,+∞)and let the set of eigenvalues of an operator A
contain the maximal element Rand let Rhave finite multiplicity. Let T:H→Hbe a linear
operator which maps Ebijectively onto itself and whose restriction on Eis non-expansive.
Then T(HR) = HR,T(HR∩E) = HR∩Eand the restriction of Tonto HRis a bijective
isometry.
Proof. The proof is similar to the previous one (alternatively consider T−1instead of Tand
apply the previous result).
The proof of the next theorem partially repeats the proof of Theorem 1 in [14]. To make
this work self-contained we provide all details.
Theorem 9. Let Abe a bounded self-adjoint operator. Then an ellipsoid Egenerated by
Ais LEC-plastic if and only if the following two conditions hold:
1. σcont =∅;
2. every subset of σp(A)that consists of more than one element either has a maximum of
finite multiplicity or has a minimum of finite multiplicity.
LINEAR PLASTICITY OF ELLIPSOIDS 199
Proof. We only need to prove the "if"part of the statement. Note that under our assumptions
the spectrum σ(A) = σp(A) = σpp(A)(i.e. consists of eigenvalues and their limiting points).
Moreover Acannot contain more than one element of infinite multiplicity.
Note that in this case there exists a basis of eigenvectors of A(see Problem 3.26 in [11]).
Claim 1. There is a τ > 0such that A+=σp(A)∩(τ, +∞)is well-ordered with respect
to the ordering ≥(that is every non empty subset of A+has a maximal element), A−=
σp(A)∩(0, τ )is well-ordered with respect to the ordering ≤(that is every non empty subset of
A−has a minimal element), and neither A+nor A−contain elements of infinite multiplicity.
Indeed, if there is an element a∞∈σp(A)of infinite multiplicity, let us take τ=a∞. Let
us demonstrate that (A+,≥)is well-ordered. If A+=∅the statement is clear. In the other
case for every non empty subset Dof A+consider B={τ} ∪ D. Then the minimal element
of Bis τ, which has infinite multiplicity so Bmust have a maximum of finite multiplicity.
This maximum will be also the maximal element of D. The demonstration of well ordering
for (A−,≤)works in the same way.
Now, consider the remaining case of σp(A)consisting only of finite multiplicity elements.
Consider the set Uof all those t∈(0,+∞)that σp(A)∩(t, +∞)is not empty and well-
ordered with respect to the ordering ≥. If Uis not empty, take τ= inf U, if U=∅,
take τ= sup σp(A). Let us demonstrate that this τis what we need. In the first case
A+=σp(A)∩(τ, +∞)and for every t > τ we have σp(A)∩(t, +∞)is not empty and
well-ordered with respect to the ordering ≥. This implies that (A+,≥)is well-ordered. In
the second case A+=∅, which is also well- ordered. So, it remains to demonstrate that
A−=σp(A)∩(0, τ )is well-ordered with respect to the ordering ≤. Assume this is not true.
Then, there is a non empty subset B⊂A−with no minimal element. According to the
conditions of our theorem Bhas a maximal element b. Since b<τ and by definition of τ
the set σp(A)∩(b, +∞)is not well-ordered with respect to the ordering ≥. Consequently,
there is a non empty D⊂σp(A)∩(b, +∞)with no maximal element. Then, B∪Dsatisfies
neither condition (1) nor condition (2) of our theorem. This contradiction completes the
demonstration of Claim 1.
We introduce the following three subspaces:
•H−is the closed linear span of the set of all those eigenvectors, for which the correspon-
ding eigenvalue lies in A−;
•Hτis the closed linear span of the set of all those eigenvectors, for which the correspon-
ding eigenvalue is τif τis eigenvalue or empty otherwise;
•H+is the closed linear span of the set of all those eigenvectors, for which the correspon-
ding eigenvalue lies in A+.
Since eigenvectors of a self-adjoint operator corresponding to different eigenvalues are
orthogonal, using continuity of scalar product, we get that these subspaces are mutually
orthogonal and H=H−⊕Hτ⊕H+(we have equality here since the set of all eigenvectors
of Aspans H). Let T:H7→ Hbe a linear operator which maps Ebijectively onto itself and
whose restriction on E is non-expansive.
Claim 2.T(H−) = H−,T(H+) = H+and the restrictions of Tonto H−and H+are
bijective isometries.
We will demonstrate the part of our claim that speaks about H+: the reasoning about
H−will differ only in the usage of Lemma 2 instead of Lemma 3.
Let us define a subspace H(t)as the closed linear span of the set of all those eigenvectors,
for which the corresponding eigenvalue lies in A+∩[t, +∞).
200 I. KARPENKO, O. ZAVARZINA
If A+=∅there is nothing to do. In the case of A+=∅we are going to demonstrate
by transfinite induction in t∈(A+,≥)the validity for all t∈A+of the following statement
U(t): the subspace H(t)is T-invariant and Tmaps H(t)onto H(t)isometrically. Since the
collection of subspaces H(t),t∈A+is a chain whose union is dense in H+, the continuity
of Twill imply the desired Claim 2.
The base of induction is the statement U(t)for t= max A. This is just the statement of
Lemma 3. We assume now as inductive hypothesis the validity of U(t)for all t>t0∈A+,
and our goal is to prove the statement U(t0). For every x, y in Hlet us introduce a modified
scalar product ⟨⟨x, y⟩⟩ as follows:
⟨⟨x, y⟩⟩ =⟨x, Ay⟩.
Then the norm on Hinduced by this modified scalar product is
|||x||| =p⟨⟨x, x⟩⟩ =p⟨x, Ax⟩.
The ellipsoid Eis the unit ball in this new norm and since Tis linear and maps Eonto E
bijectively, Tis a bijective isometry of (H, ||| · |||)onto itself. Due to [5, Theorem 2, p. 353]
Tis a unitary operator in the modified scalar product and thus Tpreserves the modified
scalar product. In particular, it preserves the orthogonality in the modified scalar product.
Denote
X=[
t>t0
H(t).
In other words, Xis the linear span of the set of all those eigenvectors, for which the
corresponding eigenvalue lies in A+∩(t0,+∞).
The orthogonal complement to Xin the modified scalar product X⊥is the closed linear
span of the set of all those eigenvectors, for which the corresponding eigenvalue lies in
A+∩(0, t0].
Occasionally the orthogonal complement to Xin the original scalar product is the same.
Our inductive hypothesis implies that T(X) = X, consequently T(X⊥) = X⊥and T(X⊥∩
E) = X⊥∩E.
X⊥equipped with the original scalar product is a Hilbert space, X⊥∩Eis an ellipsoid
in X⊥,t0is the maximal eigenvalue of Aand the multiplicity of t0is finite because t0∈A+.
The application of Lemma 3 gives us that T(Ht0) = Ht0and the restriction of Tonto Ht0
is a bijective isometry in the original norm. Now, Tmaps Xonto Xisometrically, maps
Ht0onto Ht0isometrically and H(t0)is the orthogonal direct sum of subspaces Ht0and the
closure of X. This implies that Tmaps H(t0)onto H(t0)isometrically, and the inductive
step is done. This completes the demonstration of Claim 2.
From Claim 2 and mutual orthogonality of H−and H+we deduce that T(H−⊕H+) =
H−⊕H+and Tis an isometry on H−⊕H+. Recalling again that Tpreserves the modified
scalar product and the orthogonal complement to Xin the modified scalar product is Hτ
we obtain that T(Hτ) = Hτand consequently T(Hτ∩E) = Hτ∩E. But Hτ∩Eis equal
to the closed ball (in the original norm) of radius 1
√τcentered at 0, so the equality T(Hτ∩
E) = Hτ∩Eand linearity of Timplies that Tis an isometry on Hτ. Finally, as we know,
H=H−⊕Hτ⊕H+, so Tis an isometry on the whole H.
Acknowledgements. The authors are grateful to Vladimir Kadets and Gerald Teschl for
constant support and useful advices, and to the anonymous referee for the valuable remarks.
The second author was supported by the National Research Foundation of Ukraine funded
by Ukrainian State budget in frames of the project 2020.02/0096 “Operators in infinite-
dimensional spaces: the interplay between geometry, algebra and topology”.
LINEAR PLASTICITY OF ELLIPSOIDS 201
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B. Verkin Institute for Low Temperature Physics and Engineering
Kharkiv, Ukraine
Universitat Wien, Oskar-Morgenstern-Platz 1 Wien, Austria
iryna.karpenko@univie.ac.at.
Department of Mathematics and Informatics
V. N. Karazin Kharkiv National University
Kharkiv, Ukraine
olesia.zavarzina@yahoo.com.
Received 25.02.2022
Revised 11.06.2022
