Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 3, Article 85, 2006
ON SOME APPROXIMATE FUNCTIONAL RELATIONS STEMMING FROM
ORTHOGONALITY PRESERVING PROPERTY
JACEK CHMIELIŃSKI
I NSTYTUT M ATEMATYKI , A KADEMIA P EDAGOGICZNA W K RAKOWIE
P ODCHOR A̧ ŻYCH 2,
30-084 K RAKÓW, P OLAND
jacek@ap.krakow.pl
Received 17 January, 2006; accepted 24 February, 2006
Communicated by K. Nikodem
A BSTRACT. Referring to previous papers on orthogonality preserving mappings we deal with
some relations, connected with orthogonality, which are preserved exactly or approximately. In
particular, we investigate the class of mappings approximately preserving the right-angle. We
show some properties similar to those characterizing mappings which exactly preserve the rightangle. Besides, some kind of stability of the considered property is established. We study also
the property that a particular value c of the inner product is preserved. We compare the case
c 6= 0 with c = 0, i.e., with orthogonality preserving property. Also here some stability results
are given.
Key words and phrases: Orthogonality preserving mappings, Right-angle preserving mappings, Preservation of the inner
product, Stability of functional equations.
2000 Mathematics Subject Classification. 39B52, 39B82, 47H14.
1. P REREQUISITES
Let X and Y be real inner product spaces with the standard orthogonality relation ⊥. For a
mapping f : X → Y it is natural to consider the orthogonality preserving property:
(OP)
∀x, y ∈ X : x⊥y ⇒ f (x)⊥f (y).
The class of solutions of (OP) contains also very irregular mappings (cf. [1, Examples 1 and
2]). On the other hand, a linear solution f of (OP) has to be a linear similarity, i.e., it satisfies
(cf. [1, Theorem 1])
(1.1)
kf (x)k = γkxk,
x∈X
or, equivalently,
(1.2)
hf (x)|f (y)i = γ 2 hx|yi ,
x, y ∈ X
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
The paper was written during the author’s stay at the Silesian University in Katowice.
015-06
2
JACEK C HMIELI ŃSKI
with some γ ≥ 0 (γ > 0 for f 6= 0). (More generally, a linear mapping between real normed
spaces which preserves the Birkhoff-James orthogonality has to satisfy (1.1) – see [5].) Therefore, linear orthogonality preserving mappings are not far from inner product preserving mappings (linear isometries), i.e., solutions of the functional equation:
(1.3)
∀x, y ∈ X : hf (x)|f (y)i = hx|yi .
A property similar to (OP) was introduced by Kestelman and Tissier (see [6]). One says that
f has the right-angle preserving property iff:
(RAP)
∀x, y, z ∈ X : x − z⊥y − z ⇒ f (x) − f (z)⊥f (y) − f (z).
For the solutions of (RAP) it is known (see [6]) that they must be affine, continuous similarities
(with respect to some point).
It is easily seen that if f satisfies (RAP) then, for an arbitrary y0 ∈ Y , the mapping f + y0
satisfies (RAP) as well. In particular, f0 := f − f (0) satisfies (RAP) and f0 (0) = 0.
Summing up we have:
Theorem 1.1. The following conditions are equivalent:
(i) f satisfies (RAP) and f (0) = 0;
(ii) f satisfies (OP) and f is continuous and linear;
(iii) f satisfies (OP) and f is linear;
(iv) f is linear and satisfies (1.1) for some constant γ ≥ 0;
(v) f satisfies (1.2) for some constant γ ≥ 0;
(vi) f satisfies (OP) and f is additive.
Proof. (i)⇒(ii) follows from [6] (see above); (ii)⇒(iii) is trivial; (iii)⇒(iv) follows from [1]
(see above); (iv)⇒(v) by use of the polarization formula and (v)⇒(vi)⇒(i) is trivial.
In particular, one can consider a real vector space X with two inner products h·|·i1 and h·|·i2
and f = id|X a linear and continuous mapping between (X, h·|·i1 ) and (X, h·|·i2 ). Then we
obtain from Theorem 1.1:
Corollary 1.2. Let X be a real vector space equipped with two inner products h·|·i1 and h·|·i2
generating the norms k · k1 , k · k2 and orthogonality relations ⊥1 , ⊥2 , respectively. Then the
following conditions are equivalent:
(i) ∀x, y, z ∈ X : x − z⊥1 y − z ⇒ x − z⊥2 y − z;
(ii) ∀x, y ∈ X : x⊥1 y ⇒ x⊥2 y;
(iii) kxk2 = γkxk1 for x ∈ X with some constant γ > 0;
(iv) hx|yi2 = γ 2 hx|yi1 for x, y ∈ X with some constant γ > 0;
(v) ∀x, y, z ∈ X : x − z⊥1 y − z ⇔ x − z⊥2 y − z;
(vi) ∀x, y ∈ X : x⊥1 y ⇔ x⊥2 y.
For ε ∈ [0, 1) we define an ε-orthogonality by
u⊥ε v :⇔ | hu|vi | ≤ εkukkvk.
(Some remarks on how to extend this definition to normed or semi-inner product spaces can be
found in [2].)
Then, it is natural to consider an approximate orthogonality preserving (a.o.p.) property:
(ε-OP)
∀x, y ∈ X : x⊥y ⇒ f (x)⊥ε f (y)
and the approximate right-angle preserving (a.r.a.p.) property:
(ε-RAP)
∀x, y, z ∈ X : x − z⊥y − z ⇒ f (x) − f (z)⊥ε f (y) − f (z).
J. Inequal. Pure and Appl. Math., 7(3) Art. 85, 2006
http://jipam.vu.edu.au/
A PPROXIMATE F UNCTIONAL R ELATIONS S TEMMING FROM O RTHOGONALITY
3
The class of linear mappings satisfying (ε-OP) has been considered by the author (cf. [1,
3]). In the present paper we are going to deal with mappings satisfying (ε-RAP) and, in the
last section, with mappings which preserve (exactly or approximately) a given value of the
inner product. We will deal also with some stability problems. (For basic facts concerning the
background and main results in the theory of stability of functional equations we refer to [4].)
The following result establishing the stability of equation (1.3) has been proved in [3] and will
be used later on.
Theorem 1.3 ([3], Theorem 2). Let X and Y be inner product spaces and let X be finitedimensional. Then, there exists a continuous mapping δ : R+ → R+ such that limε→0+ δ(ε) = 0
which satisfies the following property: For each mapping f : X → Y (not necessarily linear)
satisfying
| hf (x)|f (y)i − hx|yi | ≤ εkxkkyk,
(1.4)
x, y ∈ X
there exists a linear isometry I : X → Y such that
kf (x) − I(x)k ≤ δ(ε)kxk,
x ∈ X.
2. A DDITIVITY OF A PPROXIMATELY R IGHT- ANGLE P RESERVING M APPINGS
Tissier [6] showed that a mapping f satisfying the (RAP) property has to be additive up to
a constant f (0). Following his idea we will show that a.r.a.p. mappings are, in some sense,
quasi-additive. We start with the following lemma.
Lemma 2.1. Let X be a real inner product space. Let a set of points a, b, c, d, e ∈ X satisfies
the following relations, with ε ∈ [0, 81 ),
(2.1)
a − b⊥ε c − b,
b − c⊥ε d − c,
c − d⊥ε a − d,
d − a⊥ε b − a;
(2.2)
a − e⊥ε b − e,
b − e⊥ε c − e,
c − e⊥ε d − e,
d − e⊥ε a − e.
Then,
a
+
c
e −
≤ δka − ck
2 with δ :=
q
3ε
.
1−4ε
Proof. We have
kc − ek2 = ke − a + a − ck2 = ke − ak2 + ka − ck2 + 2 he − a|a − ci
whence
he − a|a − ci =
kc − ek2 − ka − ek2 − ka − ck2
.
2
Thus
2 2
a
+
c
a
−
c
e −
= e − a +
2 2 1
= ke − ak2 + ka − ck2 + he − a|a − ci
4
1
1
1
1
= ke − ak2 + ka − ck2 + kc − ek2 − ka − ek2 − ka − ck2
4
2
2
2
1
1
1
2
2
2
= ka − ek + kc − ek − ka − ck .
2
2
4
J. Inequal. Pure and Appl. Math., 7(3) Art. 85, 2006
http://jipam.vu.edu.au/
4
JACEK C HMIELI ŃSKI
Finally,
2
e − a + c = 1 2ka − ek2 + 2kc − ek2 − ka − ck2
2
4
(2.3)
and, analogously,
(2.4)
2
e − b + d = 1 2kb − ek2 + 2kd − ek2 − kb − dk2 .
2 4
Adding the equalities:
ka − bk2 = ka − e + e − bk2 = ka − ek2 + ke − bk2 + 2 ha − e|e − bi ,
kb − ck2 = kb − e + e − ck2 = kb − ek2 + ke − ck2 + 2 hb − e|e − ci ,
kc − dk2 = kc − e + e − dk2 = kc − ek2 + ke − dk2 + 2 hc − e|e − di ,
kd − ak2 = kd − e + e − ak2 = kd − ek2 + ke − ak2 + 2 hd − e|e − ai
one gets
(2.5) ka − bk2 + kb − ck2 + kc − dk2 + kd − ak2
= 2ka − ek2 + 2kb − ek2 + 2kc − ek2 + 2kd − ek2
+ 2 ha − e|e − bi + 2 hb − e|e − ci
+ 2 hc − e|e − di + 2 hd − e|e − ai .
Similarly, adding
ka − ck2 = ka − b + b − ck2 = ka − bk2 + kb − ck2 + 2 ha − b|b − ci ,
ka − ck2 = ka − d + d − ck2 = ka − dk2 + kd − ck2 + 2 ha − d|d − ci ,
kb − dk2 = kb − a + a − dk2 = kb − ak2 + ka − dk2 + 2 hb − a|a − di ,
kb − dk2 = kb − c + c − dk2 = kb − ck2 + kc − dk2 + 2 hb − c|c − di
one gets
(2.6) ka − ck2 + kb − dk2 = ka − bk2 + ka − dk2 + kc − bk2 + kc − dk2
+ ha − b|b − ci + ha − d|d − ci
+ hb − a|a − di + hb − c|c − di .
Using (2.3) – (2.6) we derive
2 2
e − a + c + e − b + d 2
2 2
2
2
2
2
2
(2.3),(2.4) 2ka − ek + 2kc − ek + 2kb − ek + 2kd − ek − ka − ck − kb − dk
=
4
(2.5) 1
2
2
2
=
kb − ck + kc − dk + kd − ak + ka − bk2
4
− 2 hb − e|e − ci − 2 hc − e|e − di − 2 hd − e|e − ai
2
2
− 2 ha − e|e − bi − ka − ck − kb − dk
J. Inequal. Pure and Appl. Math., 7(3) Art. 85, 2006
http://jipam.vu.edu.au/
A PPROXIMATE F UNCTIONAL R ELATIONS S TEMMING FROM O RTHOGONALITY
5
1
= − ha − b|b − ci + ha − d|d − ci + hb − a|a − di + hb − c|c − di
4
(2.6)
+ 2 hb − e|e − ci + 2 hc − e|e − di + 2 hd − e|e − ai + 2 ha − e|e − bi .
Thus
2 2
a
+
c
b
+
d
e −
+ e −
2 2 1
≤
| ha − b|b − ci | + | ha − d|d − ci | + | hb − a|a − di |
4
+ | hb − c|c − di | + 2| hb − e|e − ci | + 2| hc − e|e − di |
+ 2| hd − e|e − ai | + 2| ha − e|e − bi | .
Using the assumptions (2.1) and (2.2), we obtain
2 2
a + c
b + d
(2.7)
e − 2 + e − 2 1 ≤ ε ka − bkkb − ck + ka − dkkd − ck + kb − akka − dk
4
+ kb − ckkc − dk + 2kb − ekke − ck + 2kc − ekke − dk
+ 2kd − ekke − ak + 2ka − ekke − bk
1 = ε (kb − ck + ka − dk)(ka − bk + kc − dk)
4
+ 2(kb − ek + kd − ek)(ka − ek + kc − ek) .
Notice, that for ε = 0, (2.7) yields e =
Let
a+c
2
=
b+d
.
2
% := max{ka − bk, kb − ck, kc − dk, kd − ak, ka − ek, kb − ek, kc − ek, kd − ek}.
It follows from (2.7) that
2 2
a
+
c
b
+
d
e −
+ e −
≤ 1 ε(2% · 2% + 2 · 2% · 2%) = 3ε%2 .
2 2 4
Then, in particular
2
a
+
c
e −
≤ 3ε%2 .
2 (2.8)
Since we do not know for which distance the value % is attained, we are going to consider a
few cases.
(1) % ∈ {ka − bk, kb − ck, kc − dk, kd − ak}.
Suppose that % = ka − bk (other possibilities in this case are similar). Then
ka − ck2 = ka − b + b − ck2
= ka − bk2 + kb − ck2 + 2 ha − b|b − ci
≥ %2 + 0 − 2εka − bkkb − ck
≥ %2 − 2ε%2 = (1 − 2ε)%2 .
J. Inequal. Pure and Appl. Math., 7(3) Art. 85, 2006
http://jipam.vu.edu.au/
6
JACEK C HMIELI ŃSKI
Assuming ε < 12 ,
%2 ≤
1
ka − ck2 .
1 − 2ε
Using (2.8) we have
2
e − a + c ≤ 3ε ka − ck2
2 1 − 2ε
whence
r
a
+
c
3ε
e −
≤
ka − ck.
1 − 2ε
2
(2.9)
(2) % ∈ {ka − ek, kc − ek}.
Suppose that % = ka − ek (the other possibility is similar). Then, from (2.3), we have
2
1
a
+
c
2
= 1 ka − ek2 + 1 kc − ek2 ≥ 1 %2 ,
ka − ck + e
−
4
2 2
2
2
whence
2
1
a
+
c
2
2
.
% ≤ ka − ck + 2 e−
2
2 From it and (2.8) we get
2
2
3ε
a
+
c
a
+
c
2
2
≤ 3ε% ≤ ka − ck + 6ε e −
e −
2
2
2 whence (assuming ε < 16 )
s
a
+
c
3ε
e −
≤
(2.10)
ka − ck.
2
2(1 − 6ε)
(3) % ∈ {kb − ek, kd − ek}.
Suppose that % = kb − ek (the other possibility is similar). We have then
kb − ak2 = kb − e + e − ak2
= kb − ek2 + ke − ak2 + 2 hb − e|e − ai
≥ %2 + 0 − 2εkb − ekke − ak
≥ %2 − 2ε%2 = (1 − 2ε)%2 ,
whence
kb − ak2 ≥ (1 − 2ε)%2 .
Using this estimation we have
ka − ck2 = ka − b + b − ck2
= ka − bk2 + kb − ck2 + 2 ha − b|b − ci
≥ (1 − 2ε)%2 + 0 − 2εka − bkkb − ck
≥ (1 − 2ε)%2 − 2ε%2
= (1 − 4ε)%2 ,
whence (for ε < 14 )
%2 ≤
J. Inequal. Pure and Appl. Math., 7(3) Art. 85, 2006
1
ka − ck2 .
1 − 4ε
http://jipam.vu.edu.au/
A PPROXIMATE F UNCTIONAL R ELATIONS S TEMMING FROM O RTHOGONALITY
7
Using (2.8) we get
2
e − a + c ≤ 3ε%2 ≤ 3ε ka − ck2
2 1 − 4ε
and
(2.11)
r
3ε
a
+
c
≤
e −
ka − ck.
1 − 4ε
2
Finally, assuming ε < 81 , we have
s
) r
(r
r
3ε
3ε
3ε
3ε
max
,
=
,
2(1 − 6ε)
1 − 4ε
1 − 4ε
1 − 2ε
and it follows from (2.9) – (2.11)
r
a
+
c
3ε
e −
≤
ka − ck,
2 1 − 4ε
which completes the proof.
Theorem 2.2. Let X and Y be real inner product spaces and let f : X → Y satisfy (ε-RAP)
with ε < 18 . Then f satisfies
x
+
y
f
(x)
+
f
(y)
f
≤ δkf (x) − f (y)k for x, y ∈ X
(2.12)
−
2
2
q
3ε
with δ = 1−4ε
.
Moreover, if additionally f (0) = 0, then f satisfies (ε-OP) and
(2.13)
kf (x + y) − f (x) − f (y)k ≤ 2δ(kf (x + y)k + kf (x) − f (y)k),
for x, y ∈ X.
Proof. Fix arbitrarily x, y ∈ X. The case x = y is obvious. Assume x 6= y. Choose u, v ∈ X
such that x, u, y, v are consecutive vertices of a square with the center at x+y
. Denote
2 x+y
a := f (x),
b := f (u),
c := f (y),
d := f (v),
e := f
.
2
Since x−u⊥y−u, u−y⊥v−y, y−v⊥x−v, v−x⊥u−x and x− x+y
⊥u− x+y
, u− x+y
⊥y− x+y
,
2
2
2
2
x+y
x+y
x+y
x+y
y − 2 ⊥v − 2 , v − 2 ⊥x − 2 , it follows from (ε-RAP) that the conditions (2.1) and (2.2)
are satisfied. The assertion of Lemma 2.1 yields (2.12).
For the second assertion, it is obvious that f satisfies (ε-OP). Inequality (2.13) follows from
(2.12). Indeed, putting y = 0 we get
x
f (x) f
x ∈ X.
2 − 2 ≤ δkf (x)k,
Now, for x, y ∈ X
kf (x + y)−f (x) − f (y)k
f (x + y)
x
+
y
x
+
y
f
(x)
+
f
(y)
= 2
−
f
+
f
−
2
2
2
2
f (x + y)
x
+
y
x
+
y
f
(x)
+
f
(y)
+ 2 f
≤ 2
−
f
−
2
2
2
2
≤ 2δkf (x + y)k + 2δkf (x) − f (y)k.
J. Inequal. Pure and Appl. Math., 7(3) Art. 85, 2006
http://jipam.vu.edu.au/
8
JACEK C HMIELI ŃSKI
For ε = 0 we obtain that if f satisfies (RAP) and f (0) = 0, then f is additive.
The following, reverse in a sense, statement is easily seen.
Lemma 2.3. If f satisfies (ε-OP) and f is additive, then f satisfies (ε-RAP) and f (0) = 0.
Example 2 in [1] shows that it is not possible to omit completely the additivity assumption
in the above lemma. However, the problem arises if additivity can be replaced by a weaker
condition (e.g. by (2.13)). This problem remains open.
3. A PPROXIMATE R IGHT- ANGLE P RESERVING M APPINGS ARE A PPROXIMATE
S IMILARITIES
As we know from [6], a right-angle preserving mappings are similarities. Our aim is to show
that a.r.a.p. mappings behave similarly. We start with a technical lemma.
Lemma 3.1. Let a, x ∈ X and ε ∈ [0, 1). Then
(a − x)⊥ε (−a − x)
(3.1)
if and only if
q
2ε
2
2
kak2 kxk2 − ha|xi2 .
(3.2)
kxk − kak ≤ √
2
1−ε
Moreover, it follows from (3.1) that
r
r
1−ε
1+ε
(3.3)
kak ≤ kxk ≤
kak.
1+ε
1−ε
Proof. The condition (3.1) is equivalent to:
| ha + x|a − xi | ≤ εka + xkka − xk,
p
p
kak2 − kxk2 ≤ ε kak2 + 2 ha|xi + kxk2 kak2 − 2 ha|xi + kxk2 ,
kak2 − kxk2
2
≤ ε2 (kak2 + kxk2 + 2 ha|xi)(kak2 + kxk2 − 2 ha|xi)
= ε2 (kak2 + kxk2 )2 − 4 ha|xi2
2
2
2
2 2
2
2
= ε (kak − kxk ) + 4kak kxk − 4 ha|xi ,
and finally
(1 − ε2 )(kak2 − kxk2 )2 ≤ 4ε2 kak2 kxk2 − ha|xi2
which is equivalent to (3.2).
Inequality (3.2) implies
kxk2 − kak2 ≤ √ 2ε kakkxk,
1 − ε2
which yields
kxk kak 2ε
(3.4)
kak − kxk ≤ √1 − ε2
(we assume x 6= 0 and a 6= 0, otherwise the assertion of the lemma is trivial). Denoting
2ε
t := kxk
> 0 and α := √1−ε
2 , the inequality (3.4) can be written in the form
kak
|t − t−1 | ≤ α
J. Inequal. Pure and Appl. Math., 7(3) Art. 85, 2006
http://jipam.vu.edu.au/
A PPROXIMATE F UNCTIONAL R ELATIONS S TEMMING FROM O RTHOGONALITY
with a solution
−α +
√
α2 + 4
≤t≤
α+
9
√
α2 + 4
.
2
2
r
r
1−ε
kxk
1+ε
≤
≤
1+ε
kak
1−ε
Therefore,
whence (3.3) is satisfied.
Theorem 3.2. If f : X → Y is homogeneous and satisfies (ε-RAP), then with some k ≥ 0:
3
3
1+ε 2
1−ε 2
kxk ≤ kf (x)k ≤ k ·
kxk,
x ∈ X.
(3.5)
k·
1−ε
1+ε
Proof. 1. For arbitrary x, y ∈ X we have
kxk = kyk ⇔ x − y⊥ − x − y.
It follows from (ε-RAP) and the oddness of f that
kxk = kyk ⇒ f (x) − f (y)⊥ε − f (x) − f (y).
Lemma 3.1 yields
r
r
1−ε
1+ε
(3.6)
kxk = kyk ⇒
kf (x)k ≤ kf (y)k ≤
kf (x)k.
1+ε
1−ε
2. Fix arbitrarily x0 6= 0 and define for r ≥ 0, ϕ(r) := f kxr0 k x0 . Using (3.6) we have
r
r
1−ε
1+ε
kxk = r ⇒
ϕ(r) ≤ kf (x)k ≤
ϕ(r),
1+ε
1−ε
whence
r
r
1−ε
1+ε
ϕ(kxk) ≤ kf (x)k ≤
ϕ(kxk),
x ∈ X.
(3.7)
1+ε
1−ε
3. For t ≥ 0 and kxk = r we have
"r
#
r
1−ε
1+ε
ϕ(tr),
kf (tx)k ∈
ϕ(tr)
1+ε
1−ε
and
"r
tkf (x)k ∈
1−ε
tϕ(r),
1+ε
r
#
1+ε
tϕ(r) .
1−ε
Since kf (tx)k = tkf (x)k (homogeneity of f ),
# "r
#
"r
r
r
1−ε
1+ε
1−ε
1+ε
ϕ(tr),
ϕ(tr) ∩
tϕ(r),
tϕ(r) 6= ∅.
1+ε
1−ε
1+ε
1−ε
hq
q i
1−ε
Thus there exist λ, µ ∈
, 1+ε
such that λϕ(tr) = µtϕ(r), whence
1+ε
1−ε
1−ε
1+ε
tϕ(r) ≤ ϕ(tr) ≤
tϕ(r).
1+ε
1−ε
In particular, for r = 1 and k := ϕ(1) we get
1−ε
1+ε
(3.8)
kt ≤ ϕ(t) ≤
kt,
t ≥ 0.
1+ε
1−ε
J. Inequal. Pure and Appl. Math., 7(3) Art. 85, 2006
http://jipam.vu.edu.au/
10
JACEK C HMIELI ŃSKI
4. Using (3.7) and (3.8), we get
r
kf (x)k ≤
and
r
kf (x)k ≥
1+ε
ϕ(kxk) ≤
1−ε
r
1+ε 1+ε
·
kkxk
1−ε 1−ε
1−ε
ϕ(kxk) ≥
1+ε
r
1−ε 1−ε
·
kkxk
1+ε 1+ε
whence (3.5) is proved.
For ε = 0 we get kf (x)k = kkxk for x ∈ X and, from (2.13), f is additive hence linear.
Thus f is a similarity.
4. S OME S TABILITY P ROBLEMS
The stability of the orthogonality preserving property has been studied in [3]. We present the
main result from that paper which will be used in this section.
Theorem 4.1 ([3, Theorem 4]). Let X, Y be inner product spaces and let X be finite-dimensional.
Then, there exists a continuous function δ : [0, 1) → [0, +∞) with the property limε→0+ δ(ε) =
0 such that for each linear mapping f : X → Y satisfying (ε-OP) one finds a linear, orthogonality preserving one T : X → Y such that
kf − T k ≤ δ(ε) min{kf k, kT k}.
The mapping δ depends, actually, only on the dimension of X. Immediately, we have from
the above theorem:
Corollary 4.2. Let X, Y be inner product spaces and let X be finite-dimensional. Then, for
each δ > 0 there exists ε > 0 such that for each linear mapping f : X → Y satisfying ( ε-OP)
one finds a linear, orthogonality preserving one T : X → Y such that
kf − T k ≤ δ min{kf k, kT k}.
We start our considerations with the following observation.
Proposition 4.3. Let f : X → Y satisfy (RAP) and f (0) = 0. Suppose that g : X → Y
satisfies
kf (x) − g(x)k ≤ M kf (x)k,
x∈X
(M +2)
with some constant M < 41 . Then g satisfies (ε-OP) with ε := M(1−M
and
)2
√
(4.1)
kg(x + y) − g(x) − g(y)k ≤ 2 ε(kg(x)k + kg(y)k),
x, y ∈ X;
(4.2)
√
kg(λx) − λg(x)k ≤ 2 ε|λ|kg(x)k,
x ∈ X, λ ∈ R.
Proof. It follows from Theorem 1.1 that for some γ ≥ 0, f satisfies (1.2) and (1.1). Thus we
have for arbitrary x, y ∈ X:
| hg(x)|g(y)i − γ 2 hx|yi |
= | hg(x) − f (x)|g(y) − f (y)i + hg(x) − f (x)|f (y)i
+ hf (x)|g(y) − f (y)i |
≤ kg(x) − f (x)kkg(y) − f (y)k + kg(x) − f (x)kkf (y)k
+ kf (x)kkg(y) − f (y)k
2
≤ M kf (x)kkf (y)k + M kf (x)kkf (y)k + M kf (x)kkf (y)k
= M (M + 2)γ 2 kxkkyk.
J. Inequal. Pure and Appl. Math., 7(3) Art. 85, 2006
http://jipam.vu.edu.au/
A PPROXIMATE F UNCTIONAL R ELATIONS S TEMMING
FROM
O RTHOGONALITY
11
Using [1, Lemma 2], we get, for arbitrary x, y ∈ X, λ ∈ R:
p
kg(x + y) − g(x) − g(y)k ≤ 2 M (M + 2)γ(kxk + kyk)
p
= 2 M (M + 2)(kf (x)k + kf (y)k);
p
kg(λx) − λg(x)k ≤ 2 M (M + 2)γ|λ|kxk
p
= 2 M (M + 2)|λ|kf (x)k.
Since kf (x)k − kg(x)k ≤ M kf (x)k,
kf (x)k ≤
kg(x)k
.
1−M
Therefore
| hg(x)|g(y)i − γ 2 hx|yi | ≤
Putting ε :=
M (M +2)
(1−M )2
M (M + 2)
kg(x)kkg(y)k.
(1 − M )2
we have x⊥y ⇒ g(x)⊥ε g(y) and
√
kg(x + y) − g(x) − g(y)k ≤ 2 ε(kg(x)k + kg(y)k)
√
kg(λx) − λg(x)k ≤ 2 ε|λ|kg(x)k.
Corollary 4.4. If f satisfies (RAP), f (0) = 0 (whence f is linear) and g : X → Y is a linear
mapping satisfying, with M < 14 ,
kf − gk ≤ M kf k,
(4.3)
then g is (ε-OP) (and linear) whence (ε-RAP).
The above result yields a natural question if the reverse statement is true. Namely, we may
ask if for a linear mapping g : X → Y satisfying (ε-RAP) (with some ε > 0) there exists a
(linear) mapping f satisfying (RAP) such that an estimation of the (4.3) type holds.
A particular solution to this problem follows easily from Theorem 4.1.
Theorem 4.5. Let X be a finite-dimensional inner product space and Y an arbitrary one. There
exists a mapping δ : [0, 1) → R+ satisfying limε→0+ δ(ε) = 0 and such that for each linear
mapping satisfying (ε-RAP) g : X → Y there exists f : X → Y satisfying (RAP) and such that
(4.4)
kf − gk ≤ δ(ε) min{kf k, kgk}.
Proof. If g is linear and satisfies (ε-RAP), then g is linear and satisfies (ε-OP). It follows then
from Theorem 4.1 that there exists f linear and satisfying (OP), whence (RAP), such that (4.4)
holds.
Corollary 4.6. Let X be a finite-dimensional inner product space and Y an arbitrary one.
Then, for each δ > 0 there exists ε > 0 such that for each linear and satisfying (ε-RAP)
mapping g : X → Y there exists f : X → Y satisfying (RAP) and such that
kf − gk ≤ δ min{kf k, kgk}.
It is an open problem to verify if the above result remains true in the infinite dimensional case
or without the linearity assumption.
J. Inequal. Pure and Appl. Math., 7(3) Art. 85, 2006
http://jipam.vu.edu.au/
12
JACEK C HMIELI ŃSKI
5. O N M APPINGS P RESERVING
A
PARTICULAR VALUE
OF THE I NNER
P RODUCT
The following considerations have been inspired by a question of L. Reich during the 43rd
ISFE. In this section X and Y are inner product spaces over the field K of real or complex
numbers. Let f : X → Y and suppose that, for a fixed number c ∈ K, f preserves this
particular value of the inner product, i.e.,
(5.1)
∀x, y ∈ X : hx|yi = c ⇒ hf (x)|f (y)i = c.
If c = 0, the condition (5.1) simply means that f preserves orthogonality, i.e., that f satisfies
(OP).
We will show that the solutions of (5.1) behave differently for c = 0 and for c 6= 0.
Obviously, if f satisfies (1.3) then f also satisfies (5.1), with an arbitrary c. The converse
is not true, neither with c = 0 (cf. examples mentioned in the Introduction) nor with c 6= 0.
Indeed, if c > 0, then fixing x0 ∈ X such that kx0 k2 = c, a constant mapping f (x) = x0 ,
x ∈ X satisfies (5.1) but not (1.3). Another example: let X = Y = C and let 0 6= c ∈ C.
Define
c
f (z) := , z ∈ C \ {0}; f (0) := 0.
z
Then, for z, w ∈ C \ {0}
|c|2
hf (z)|f (w)i =
hz|wi
and, in particular, if hz|wi = c, then hf (z)|f (w)i = c. Thus f satisfies (5.1) but not (1.3).
Let us restrict our investigations to the class of linear mappings. As we will see below (Corollary 5.2), a linear solution of (5.1), with c 6= 0, satisfies (1.3).
Let us discuss a stability problem. For fixed 0 6= c ∈ K and ε ≥ 0 we consider the condition
(5.2)
∀x, y ∈ X : hx|yi = c ⇒ | hf (x)|f (y)i − c| ≤ ε.
Theorem 5.1. For a finite-dimensional inner product space X and an arbitrary inner product
space Y there exists a continuous mapping δ : R+ → R+ satisfying limε→0+ δ(ε) = 0 and
such that for each linear mapping f : X → Y satisfying (5.2) there exists a linear isometry
I : X → Y such that
kf − Ik ≤ δ(ε).
Proof. Let 0 6= d ∈ K. If hx|yi = d, then dc x|y = c and hence, using (5.2) and homogeneity
|c|
of f , |d|
| hf (x)|f (y)i − d| ≤ ε. Therefore we have (for d 6= 0)
(5.3)
hx|yi = d ⇒ | hf (x)|f (y)i − d| ≤
|d|
ε.
|c|
Now, D
let d = 0. Let
E 0 6= dn ∈ K and limn→∞ dn = 0. Suppose that hx|yi = d = 0 and y 6= 0.
dn y
Then x + kyk2 |y = dn and thus, from (5.3) and linearity of f ,
f (x) + dn f (y)|f (y) − dn ≤ |dn | ε.
2
kyk
|c|
Letting n → ∞ we obtain hf (x)|f (y)i = 0. For y = 0 the latter equality is obvious.
Summing up, we obtain that
| hf (x)|f (y)i − hx|yi | ≤
J. Inequal. Pure and Appl. Math., 7(3) Art. 85, 2006
| hx|yi |
ε
|c|
http://jipam.vu.edu.au/
A PPROXIMATE F UNCTIONAL R ELATIONS S TEMMING
FROM
O RTHOGONALITY
13
whence also
| hf (x)|f (y)i − hx|yi | ≤
(5.4)
ε
kxkkyk.
|c|
Let δ 0 : R+ → R+ be a mapping from the assertion of Theorem 1.3. Define δ(ε) := δ 0 (ε/|c|)
and notice that limε→0+ δ(ε) = 0. Then it follows from (5.4) that there exists a linear isometry
I : X → Y such that
ε
0
kf − Ik ≤ δ
= δ(ε).
|c|
For ε = 0 we obtain from the above result (we can omit the assumption concerning the
dimension of X in this case, considering a subspace spanned on given vectors x, y ∈ X):
Corollary 5.2. Let f : X → Y be linear and satisfy (5.1), with some 0 6= c ∈ K. Then f
satisfies (1.3).
Notice that for c = 0, the condition (5.2) has the form
hx|yi = 0 ⇒ | hf (x)|f (y)i | ≤ ε.
If hx|yi = 0, then also hnx|yi = 0 for all n ∈ N. Thus | hf (nx)|f (y)i | ≤ ε, whence
| hf (x)|f (y)i | ≤ nε for n ∈ N. Letting n → ∞ one gets
∀x, y ∈ X : hx|yi = 0 ⇒ hf (x)|f (y)i = 0,
i.e, f is a linear, orthogonality preserving mapping.
Now, let us replace the condition (5.2) by
(5.5)
∀x, y ∈ X : hx|yi = c ⇒ | hf (x)|f (y)i − c| ≤ εkf (x)kkf (y)k.
For c = 0, (5.5) states that f satisfies (ε-OP).
Let us consider the class of linear mappings satisfying (5.5) with c 6= 0.
c We proceed similarly as in the proof of Theorem 5.1. Let 0 6= d ∈ K. If hx|yi = d, then
x|y = c and hence, using (5.5) and homogeneity of f , we obtain
d
|c|
|c|
| hf (x)|f (y)i − d| ≤ ε kf (x)kkf (y)k.
|d|
|d|
Therefore we have (for d 6= 0)
(5.6)
hx|yi = d ⇒ | hf (x)|f (y)i − d| ≤ εkf (x)kkf (y)k.
D
Now, suppose that hx|yi = d = 0. Let 0 6= dn ∈ K and limn→∞ dn = 0. Then x +
dn and thus, from (5.6) and linearity of f ,
d
d
n
n
f (x) +
≤ ε f (x) +
kf (y)k.
f
(y)|f
(y)
−
d
f
(y)
n
kyk2
kyk2
dn y
|y
kyk2
E
=
Letting n → ∞ we obtain | hf (x)|f (y)i | ≤ εkf (x)kkf (y)k. So we have
hx|yi = 0 ⇒ | hf (x)|f (y)i | ≤ εkf (x)kkf (y)k.
Summing up, we obtain that
(5.7)
| hf (x)|f (y)i − hx|yi | ≤ εkf (x)kkf (y)k,
x, y ∈ X.
for x ∈ X, which gives
Putting in the above inequality x = y we get kf (x)k ≤ √kxk
1−ε
ε
| hf (x)|f (y)i − hx|yi | ≤
kxkkyk,
x, y ∈ X,
1−ε
J. Inequal. Pure and Appl. Math., 7(3) Art. 85, 2006
http://jipam.vu.edu.au/
14
JACEK C HMIELI ŃSKI
ε
i.e., f satisfies (1.4) with the constant 1−ε
.
Therefore, applying Theorem 1.3, we get
Theorem 5.3. If X is a finite-dimensional inner product space and Y an arbitrary inner product
space, then there exists a continuous mapping δ : R+ → R+ with limε→0+ δ(ε) = 0 such that
for a linear mapping f : X → Y satisfying (5.5), with c 6= 0, there exists a linear isometry
I : X → Y such that
kf − Ik ≤ δ(ε).
A converse theorem is also true, even with no restrictions concerning the dimension of X.
Let I : X → Y be a linear isometry and f : X → Y a mapping, not necessarily linear, such
that
kf (x) − I(x)k ≤ δkxk,
x∈X
q
− 1 (for a given ε ≥ 0). Reasoning similarly as in the proof of Proposition 4.3
with δ = 1+2ε
1+ε
one can show that
| hf (x)|f (y)i − hx|yi | ≤ δ(δ + 2)kxkkyk,
which implies
1
kxk ≤ p
kf (x)k
1 − δ(δ + 2)
and finally
δ(δ + 2)
|hf (x)|f (y)i − hx|yi| ≤
kf (x)kkf (y)k
1 − δ(δ + 2)
= εkf (x)kkf (y)k.
Thus f satisfies (5.5) with an arbitrary c.
Remark 5.4. From Theorems 5.1 and 5.3 one can derive immediately the stability results formulated as in Corollaries 4.2 and 4.6.
R EFERENCES
[1] J. CHMIELIŃSKI, Linear mappings approximately preserving orthogonality, J. Math. Anal. Appl.,
304 (2005), 158–169.
[2] J. CHMIELIŃSKI, On an ε-Birkhoff orthogonality, J. Inequal. Pure Appl. Math., 6(3) (2005), Art.
79, 7pp. [ONLINE: http://jipam.vu.edu.au/article.php?sid=552].
[3] J. CHMIELIŃSKI, Stability of the orthogonality preserving property in finite-dimensional inner
product spaces, J. Math. Anal. Appl., 318 (2006), 433–443.
[4] D.H. HYERS, G. ISAC AND Th.M. RASSIAS, Stability of Functional Equations in Several Variables, Progress in Nonlinear Differential Equations and Their Applications vol. 34, Birkhäuser,
Boston-Basel-Berlin, 1998.
[5] A. KOLDOBSKY, Operators preserving orthogonality are isometries, Proc. Roy. Soc. Edinburgh
Sect. A, 123 (1993), 835–837.
[6] A. TISSIER, A right-angle preserving mapping (a solution of a problem proposed in 1983 by H.
Kestelman), Advanced Problem 6436, Amer. Math. Monthly, 92 (1985), 291–292.
J. Inequal. Pure and Appl. Math., 7(3) Art. 85, 2006
http://jipam.vu.edu.au/