The next theorem is the main result of this paper.
Proof.
Let
𝒰
⊂
l
(
X
)
be nonempty and let (16) be valid. Write
X
0
:
=
X
∖
{
0
}
. Note that (15), with
x
replaced by
u
′
x
and
y
=
u
x
, gives
(19)
d
(
f
(
x
)
,
f
(
u
′
x
)
+
f
(
u
x
)
)
≤
H
(
u
′
x
,
u
x
)
x
∈
X
0
,
u
∈
𝒰
.
Given
u
∈
𝒰
, we define operators
𝒯
u
:
E
X
0
→
E
X
0
and
Λ
u
:
ℝ
+
X
0
→
ℝ
+
X
0
by
(20)
𝒯
u
ξ
(
x
)
:
=
ξ
(
u
′
x
)
+
ξ
(
u
x
)
x
∈
X
0
,
ξ
∈
E
X
0
,
u
∈
𝒰
,
(21)
Λ
u
δ
(
x
)
:
=
δ
(
u
′
x
)
+
δ
(
u
x
)
x
∈
X
0
,
δ
∈
ℝ
+
X
0
,
u
∈
𝒰
.
It is easily seen that, for each
u
∈
𝒰
,
Λ
:
=
Λ
u
has form (9) with
Z
:
=
X
0
,
f
1
(
x
)
=
u
′
x
, and
f
2
(
x
)
=
u
x
. Moreover, (19) can be written in the following way
(22)
d
(
f
(
x
)
,
𝒯
u
f
(
x
)
)
≤
H
(
u
′
x
,
u
x
)
=
:
ε
u
(
x
)
x
∈
X
0
,
u
∈
𝒰
.
(Here and in the sequel, the restriction of
f
to the set
X
0
is also denoted by
f
; we believe that this will not cause any confusion.) And
(23)
d
(
𝒯
u
ξ
(
x
)
,
𝒯
u
μ
(
x
)
)
=
d
(
ξ
(
u
′
x
)
+
ξ
(
u
x
)
,
μ
(
u
′
x
)
+
μ
(
u
x
)
)
≤
d
(
ξ
(
u
′
x
)
,
μ
(
u
′
x
)
)
+
d
(
ξ
(
u
x
)
,
μ
(
u
x
)
)
for every
ξ
,
μ
∈
E
X
0
,
x
∈
X
0
, and
u
∈
𝒰
. Consequently, for each
u
∈
𝒰
, also (8) is valid with
Z
:
=
X
0
,
Y
:
=
E
, and
𝒯
:
=
𝒯
u
.
Note that, in view of the definition of
λ
(
u
)
,
(24)
H
(
u
x
,
u
y
)
≤
λ
(
u
)
H
(
x
,
y
)
u
∈
𝒰
,
x
,
y
∈
X
0
.
So, it is easy to show by induction on
n
that
(25)
Λ
u
n
ε
u
(
x
)
≤
(
λ
(
u
′
)
+
λ
(
u
)
)
n
H
(
u
′
x
,
u
x
)
,
for
x
∈
X
0
,
n
∈
ℕ
0
(nonnegative integers), and
u
∈
𝒰
. Hence,
(26)
ε
u
*
(
x
)
:
=
∑
n
=
0
∞
(
Λ
u
n
ε
u
)
(
x
)
≤
H
(
u
′
x
,
u
x
)
∑
n
=
0
∞
(
λ
(
u
′
)
+
λ
(
u
)
)
n
=
H
(
u
′
x
,
u
x
)
1

λ
(
u
)

λ
(
u
′
)
x
∈
X
0
,
u
∈
𝒰
.
Now, we can use Theorem 4 with
Z
=
X
0
,
Y
=
E
,
ε
:
=
ε
u
, and
φ
=
f
. According to it, the limit
(27)
T
u
′
(
x
)
:
=
lim
n
→
∞
(
𝒯
u
n
f
)
(
x
)
exists for each
x
∈
X
0
and
u
∈
𝒰
,
(28)
d
(
f
(
x
)
,
T
u
(
x
)
)
≤
H
(
u
′
x
,
u
x
)
1

λ
(
u
)

λ
(
u
′
)
x
∈
X
0
,
u
∈
𝒰
,
and the function
T
u
:
X
→
E
defined by
(29)
T
u
(
0
)
=
0
,
T
u
(
x
)
:
=
T
u
′
(
x
)
x
∈
X
0
is a solution of the equation
(30)
T
(
x
)
=
T
(
u
′
x
)
+
T
(
u
x
)
,
because
T
u
′
is a fixed point of
𝒯
u
.
Now we show that
(31)
d
(
𝒯
u
n
f
(
x
+
y
)
,
𝒯
u
n
f
(
x
)
+
𝒯
u
n
f
(
y
)
)
≤
(
λ
(
u
′
)
+
λ
(
u
)
)
n
H
(
x
,
y
)
for every
x
,
y
∈
X
0
,
x
+
y
≠
0
,
n
∈
ℕ
0
, and
u
∈
𝒰
.
Since the case
n
=
0
is just (15), take
k
∈
ℕ
0
and assume that (31) holds for
n
=
k
and every
x
,
y
∈
X
0
,
x
+
y
≠
0
, and
u
∈
𝒰
. Then, by (24),
(32)
d
(
𝒯
u
k
+
1
f
(
x
+
y
)
,
𝒯
u
k
+
1
f
(
x
)
+
𝒯
u
k
+
1
f
(
y
)
)
=
d
(
+
𝒯
u
k
f
(
u
y
)
+
𝒯
u
k
f
(
u
y
)
)
𝒯
u
k
f
(
u
′
(
x
+
y
)
)
+
𝒯
u
k
f
(
u
(
x
+
y
)
)
,
𝒯
u
k
f
(
u
′
x
)
+
𝒯
u
k
f
(
u
x
)
+
𝒯
u
k
f
(
u
′
y
)
+
𝒯
u
k
f
(
u
y
)
(
+
𝒯
u
k
f
(
u
y
)
+
𝒯
u
k
f
(
u
y
)
)
𝒯
u
k
f
(
u
′
(
x
+
y
)
)
+
𝒯
u
k
f
(
u
(
x
+
y
)
)
,
)
≤
d
(
𝒯
u
k
f
(
u
′
x
+
u
′
y
)
,
𝒯
u
k
f
(
u
′
x
)
+
𝒯
u
k
f
(
u
′
y
)
)
+
d
(
𝒯
u
k
f
(
u
x
+
u
y
)
,
𝒯
u
k
f
(
u
x
)
+
𝒯
u
k
f
(
u
y
)
)
≤
(
λ
(
u
′
)
+
λ
(
u
)
)
k
H
(
u
′
x
,
u
′
y
)
+
(
λ
(
u
′
)
+
λ
(
u
)
)
k
H
(
u
x
,
u
y
)
≤
(
λ
(
u
′
)
+
λ
(
u
)
)
k
+
1
H
(
x
,
y
)
w
w
w
w
w
w
w
w
w
w
w
w
w
w
x
,
y
∈
X
0
,
x
+
y
≠
0
,
u
∈
𝒰
.
Thus, by induction, we have shown that (31) holds for every
x
,
y
∈
X
0
,
x
+
y
≠
0
,
n
∈
ℕ
0
, and
u
∈
𝒰
. Letting
n
→
∞
in (31), we obtain the equality
(33)
T
u
(
x
+
y
)
=
T
u
(
x
)
+
T
u
(
y
)
i
w
x
,
y
∈
X
0
,
x
+
y
≠
0
,
u
∈
𝒰
.
From this we can deduce that
T
u
is additive for each
u
∈
𝒰
. The reasoning is very simple, but for the convenience of readers we present it here.
In view of (33), it is only enough to consider the situation
y
=

x
. So take
u
∈
𝒰
and
x
∈
X
0
(the case
x
=
0
is trivial). Then, by (33),
(34)
T
u
(
x
)
=
T
u
(
x
+
x

x
)
=
T
u
(
2
x
)
+
T
u
(

x
)
=
2
T
u
(
x
)
+
T
u
(

x
)
,
which yields
T
u
(
x
)
+
T
u
(

x
)
=
0
and consequently
T
u
(
x

x
)
=
T
u
(
0
)
=
0
=
T
u
(
x
)
+
T
u
(

x
)
.
Next, we prove that each additive
T
:
X
→
Y
satisfying the inequality
(35)
d
(
f
(
x
)
,
T
(
x
)
)
≤
L
H
(
v
x
,
v
′
x
)
x
∈
X
0
,
with some
L
>
0
and
v
∈
𝒰
, is equal to
T
w
for each
w
∈
𝒰
. To this end fix
v
,
w
∈
𝒰
,
L
>
0
, and an additive
T
:
X
→
Y
satisfying (35). Note that, by (28) and (35), there is
L
0
>
0
such that
(36)
d
(
T
(
x
)
,
T
w
(
x
)
)
≤
d
(
T
(
x
)
,
f
(
x
)
)
+
d
(
f
(
x
)
,
T
w
(
x
)
)
≤
L
0
(
H
(
v
′
x
,
v
x
)
+
H
(
w
′
x
,
w
x
)
)
×
∑
n
=
0
∞
(
λ
(
w
′
)
+
λ
(
w
)
)
n
for
x
∈
X
0
. Observe yet that
T
and
T
w
are solutions to (30) for all
u
∈
𝒰
, because they are additive.
We show that, for each
j
∈
ℕ
0
,
(37)
d
(
T
(
x
)
,
T
w
(
x
)
)
≤
L
0
(
H
(
v
′
x
,
v
x
)
+
H
(
w
′
x
,
w
x
)
)
×
∑
n
=
j
∞
(
λ
(
w
′
)
+
λ
(
w
)
)
n
x
∈
X
0
.
The case
j
=
0
is exactly (36). So fix
l
∈
ℕ
0
and assume that (37) holds for
j
=
l
. Then, in view of (24),
(38)
d
(
T
(
x
)
,
T
w
(
x
)
)
=
d
(
T
(
w
x
)
+
T
(
w
′
x
)
,
T
w
(
w
x
)
+
T
w
(
w
′
x
)
)
≤
d
(
T
(
w
x
)
,
T
w
(
w
x
)
)
+
d
(
T
(
w
′
x
)
,
T
w
(
w
′
x
)
)
≤
L
0
(
H
(
v
′
w
x
,
v
w
x
)
+
H
(
w
′
w
x
,
w
w
x
)
)
×
∑
n
=
j
∞
(
λ
(
w
′
)
+
λ
(
w
)
)
n
+
L
0
(
H
(
v
′
w
′
x
,
v
w
′
x
)
+
H
(
w
′
w
′
x
,
w
w
′
x
)
)
×
∑
n
=
j
∞
(
λ
(
w
′
)
+
λ
(
w
)
)
n
≤
L
0
(
H
(
v
′
x
,
v
x
)
+
H
(
w
′
x
,
w
x
)
)
(
λ
(
w
)
+
λ
(
w
′
)
)
×
∑
n
=
j
∞
(
λ
(
w
′
)
+
λ
(
w
)
)
n
=
L
0
(
H
(
v
′
x
,
v
x
)
+
H
(
w
′
x
,
w
x
)
)
×
∑
n
=
l
+
1
∞
(
λ
(
w
′
)
+
λ
(
w
)
)
n
x
∈
X
0
.
Thus we have shown (37). Now, letting
j
→
∞
in (37), we get
(39)
T
(
x
)
=
T
w
(
x
)
x
∈
X
0
.
Since
T
and
T
w
are additive, we have
T
=
T
w
.
In this way, we also have proved that
T
u
=
T
w
for each
u
∈
𝒰
(on account of (28)), which yields
(40)
d
(
f
(
x
)
,
T
w
(
x
)
)
≤
H
(
u
′
x
,
u
x
)
1

λ
(
u
′
)

λ
(
u
)
x
∈
X
0
,
u
∈
𝒰
.
This implies (17) with
T
:
=
T
w
; clearly, equality (39) means the uniqueness of
T
, as well.
Thus we have completed the proof of Theorem 5.