Luận án Tính chính quy mêtric phi tuyến của ánh xạ đa trị trên một tập hợp và ứng dụng

lar
on W with constant κ. In addition, suppose that Y is a Fre´chet smooth Banach
space, the conclusion is true if in (2.21), ∆(x, y) is replaced by τ(x, y) defined as in
Lemma 2.3.1.
Remark 2.3.1. (i) The sufficient conditions for the local metric regularity
involving some coderivatives were well established in the literature (see, e.g.,
[11, 39, 48, 56, 63, 66]). The preceding theorem gives a non-local version of
sufficient conditions based on coderivatives.
58
(ii) In the case of X, Y being finite dimensional spaces, obviously the quantity
τ(x, y) is determined more simply as follows.
τ(x, y) = inf
{
µ′(‖y − v‖)‖x∗‖ : x∗ ∈ D∗F (x, v)
(
v − y
‖v − y‖
)
, v ∈ ProjF (x)y
}
,
where ProjF (x)y is the projection of y on F (x) defined by
{z ∈ F (x) : ‖y − z‖ = d(y, F (x))} ,
and D∗F (x, v)(y∗) denotes the limiting coderivative of F at (x, v) ∈ GraphF .
Below we give some illustrated examples.
Example 2.3.1. Let us consider F : Rn ⇒ R defined by F (x) = f(x) + R+,
where f : Rn → R given by f(x) = ‖x‖2, x ∈ Rn. Using Theorem 2.3.1, we shall
verify that F is (µ(t) = t1/2,+∞)-regular on W = Rn × [0,+∞) with modulus
τ = 1. Firstly, for the gauge γ(x) := ‖x‖, for any (x, y) ∈ W with y /∈ F (x), i.e.,
0 ≤ y < ‖x‖2, ϕy(x)1/2 = (‖x‖2 − y)1/2 ≤ ‖x‖, as ProjF (x)y = f(x), and
D∗F (x, f(x))
(
f(x)− y
|f(x)− y|
)
= D∗F (x, f(x)) (1) = ∇f(x) = 2x,
one has
τ(x, y) =
1
2(‖x‖2 − y)1/2 2‖x‖ ≥ 1,
for all (x, y) ∈ W, with 0 < ϕy(x)1/2 ≤ γ(x) = ‖x‖. Thus according to Theorem 2.3.1,
F is (t1/2, ‖ · ‖)-metrically regular on W with modulus 1, that is,
d(x, F−1(y)) ≤ d(y, F (x)), (2.22)
for all (x, y) ∈ W with 0 < d(y, F (x))1/2 < ‖x‖. On the other hand, for (x, y) ∈ W,
with d(y, F (x))1/2 ≥ ‖x‖, as 0 ∈ F−1(y), one has
d(x, F−1(y)) ≤ ‖x− 0‖ ≤ d(y, F (x))1/2,
that is (2.22) holds for all (x, y) ∈ W, so F is (t1/2,+∞)-metrically regular on W.
59
Example 2.3.2. Consider the function f : Rn → R defined by
f(x) =
 ‖x‖2 sin
1
‖x‖ if x 6= 0,
0 if x = 0.
(2.23)
Clearly this function is locally surjective around (0, 0) ∈ Rn × R. We shall verify
the following two statements.
(i) The function f is not (µ, γ)-metrically regular around (0, 0) for any modulus
function µ : R+ → R+, and any gauge function γ : Rn → R+ being positive
on a neighborhood of 0 ∈ Rn.
Indeed, for any ε > 0, pick an integer k sufficiently large such that 1/(2kpi) < ε.
Consider the function α : R→ R, defined by
α(t) =
 t2 sin
1
|t| if t 6= 0,
0 if t = 0.
(2.24)
Then α′(0) = 0, and for t > 0,
α′(t) = 2t sin(1/t)− cos(1/t);
α′′(t) = (2− 1/t2) sin(1/t)− (2/t) cos(1/t).
Since α′(1/(2kpi)) = −1 0, there
is some t¯ ∈ (1/(2kpi + pi/2), 1/2kpi) such that α′(t¯) = 0. Moreover, since
α′′(t) < 0 for all [1/(2kpi+ pi/2), 1/(2kpi)], the point t¯ is a strict maximizer of
the function α(·) on [1/(2kpi+pi/2), 1/(2kpi)]. Hence for x¯ ∈ Rn with ‖x¯‖ = t¯,
and any y > f(x¯) = α(t¯), one has
d(x¯, f−1(y)) ≥ min{(2kpi)−1 − t¯, t¯− (2kpi + pi/2)−1}.
Thus for any sequence (yj)→ f(x¯) with yj > f(x¯) for all j, for any modulus
function µ : R+ → R+, one has
lim
j→∞
µ(|yj − f(x¯)|)
d(x¯, f−1(yj))
= 0,
showing that f is not (µ, γ)-metrically regular on εBRn×[−ε, ε], for any ε > 0,
any modulus function µ, any gauge function γ being positive on εBRn.
60
(ii) f is (t1/2,+∞)-metrically regular on
W = {(x, y) ∈ Rn × R : ‖x‖ ≤ 1, y ≥ y0, |y| ≤ ‖x‖2/
√
2},
where y0 = min{f(x) : ‖x‖ ≤ 1}.
Firstly, invoking Theorem 2.3.1 and Remark 2.3.1, we show that f is
(t1/2, ‖ · ‖)-metrically regular on W. Indeed, for x 6= 0, y∗ ∈ R,
D∗f(x)(y∗) = y∗x
(
2 sin
1
‖x‖ −
1
‖x‖ cos
1
‖x‖
)
.
Thus for y∗ either 1 or −1,
‖D∗f(x)(y∗)‖ = ‖x‖
∣∣∣∣2 sin 1‖x‖ − 1‖x‖ cos 1‖x‖
∣∣∣∣ .
Observe that for any δ > 0 sufficiently small, for any x ∈ Rn \ {0} with
‖x‖ ≤ 1 + δ, and | sin(1/‖x‖)| < √2/2 + 2δ, there is γ = pi/4 + O(δ), such
that 1/‖x‖ ≥ pi − γ. Therefore,∣∣∣∣2 sin 1‖x‖ − 1‖x‖ cos 1‖x‖
∣∣∣∣ ≥ (3pi/4−O(δ))(1− (√2/2 + 2δ)2)1/2 − 2(√2/2 + 2δ)
≥ 3pi/4− 11/5−O(δ), (2.25)
for all x ∈ Rn with ‖x‖ ≤ 1 + δ, | sin(1/‖x‖)| < √2/2 + 2δ. For given
δ ∈ (0, 1/8) for y ∈ PRW ; d(x,Wy) < δ/2 · ‖x‖ with 0 < |y − f(x)|1/2 <
δ/2 · ‖x‖, then obviously 0 < ‖x‖ < 1 + δ, and |y| ≤ (1 + δ/2)2‖x‖2/√2, so
|f(x)| ≤ |y|+ δ2/4 · ‖x‖2 ≤ (
√
2/2 + 2δ)‖x‖2,
which implies sin(1/‖x‖) ≤ √2/2 + 2δ. Hence by (2.25),
1
2|y − f(x)|1/2‖D
∗f(x)(y∗)‖ ≥
∣∣∣∣2 sin 1‖x‖ − 1‖x‖ cos 1‖x‖
∣∣∣∣ δ−1
≥ (3pi/4− 11/5−O(δ))δ−1
tending to +∞, as δ → 0+. This implies that f is (t1/2, ‖·‖)-metrically regular
on W, in view of Theorem 2.3.1. That is, there are κ > 0, r > 0 such that
d(x, f−1(y)) ≤ κ|y − f(x)|1/2, (2.26)
61
for all (x, y) ∈ W, with |y − f(x)|1/2 < r−1‖x‖. Let now (x, y) ∈ W with
|y − f(x)|1/2 ≥ r−1‖x‖. Set y1 = max{f(x) : ‖x‖ ≤ 1}. We prove the
following claim.
Claim. There is τ > 0 such that for all y ∈ [y0, y1], one has
d(0, f−1(y)) ≤ τ |y|1/2.
For y ∈ [y0, y1], then f−1(y)∩BRn 6= ∅, due to the continuity of the function
f. So for |y| ≥ 1/(2pi + pi/2)2, one has d(0, f−1(y)) ≤ 1 ≤ 12pi+pi/2 |y|1/2. For
0 < |y| < 1/(2pi + pi/2)2, let xy ∈ f−1(y) ∩ BRn such that
‖xy‖ = min{‖x‖ : x ∈ f−1(y) ∩ BRn}.
Consider the following two cases:
Case 1. 0 < y < 1(2pi+pi/2)2 . As f(0) = 0 < y <
1
(2pi+pi/2)2 = f
(
1
2pi+pi/2
)
, there
exists x ∈ Rn with ‖x‖ < 12pi+pi/2 such that x ∈ f−1(y). Then 0 < ‖xy‖ < 12pi+pi/2
and sin 1‖xy‖ > 0, so there is k ∈ N∗, the set of positive integers, such that
‖xy‖ ∈
(
1
2kpi+pi ,
1
2kpi
)
. As f
(
1
2kpi+2pi
)
= 0; f
(
1
2kpi+5pi/2
)
= 1(2kpi+5pi/2)2 , by the
definition of xy, it implies y >
1
(2kpi+5pi/2)2 , since if it would be 0 < y <
1
(2kpi+5pi/2)2 ,
by the continuity of f, we can find x ∈ Rn, with ‖x‖ ∈] 12kpi+5pi/2 , 12kpi+2pi [ such
that f(x) = y, which contradicts the definition of xy. Therefore,
sin(1/‖xy‖) > 1‖xy‖2(2kpi + 5pi/2)2 >
(2k)2
(2k + 5/2)2
≥ 16
81
,
so
|y|1/2 = ‖xy‖(sin(1/‖xy‖))1/2 ≥ 4
9
‖xy‖ ≥ 4
9
d(0, f−1(y)),
or d(0, f−1(y)) ≤ 94 |y|1/2.
Case 2. − 1(2pi+pi/2)2 < y < 0. Similarly, as f
(
2
3pi
)
= − 1(3pi)2 < y < f(0),
‖xy‖ < 23pi , and sin 1‖xy‖ < 0, therefore there is k ∈ N∗ such that
‖xy‖ ∈
(
1
2kpi ,
1
2kpi−pi
)
. Since f
(
1
2kpi+pi
)
= 0; f
(
1
2kpi+3pi/2
)
= − 1(2kpi+3pi/2)2 , one
has y < − 1(2kpi+3pi/2)2 . Thus
| sin(1/‖xy‖)| > 1‖xy‖2(2kpi + 3pi/2)2 >
(2k − 1)2
(2k + 3/2)2
≥ 4
49
,
62
and one obtains d(0, f−1(y)) ≤ 72 |y|1/2. The claim is proved.
For (x, y) ∈ W with |y − f(x)|1/2 ≥ r−1‖x‖, since PRW ⊆ [y0, y1], and
|y| ≤ ‖x‖2/√2, one has
d(x, f−1(y)) ≤ ‖x‖+ d(0, f−1(y)) ≤ ‖x‖+ τ |y|1/2
≤ (1 + 2−1/4τ)‖x‖
≤ r(1 + 2−1/4τ)|y − f(x)|1/2.
This together with relation (2.26) show that f is (t1/2,+∞)-metrically regular
on W.
63
Chapter 3
Perturbation stability of
Milyutin-type regularity and
applications
The main results of this chapter concentrate on two following issues. The
first one is the theorems giving the estimation of the metric regularity modulus
of perturbed mapping under composite perturbation of mappings between metric
spaces which contain as their particular case about additive perturbation of a
set-valued mapping between Banach spaces by a Lipschitz mapping. Secondly,
a fixed point theorem for a composition of a couple of mappings: one from X to
Y and the other from Y to X is obtained by the use of the above perturbation
theorems. The results mentioned here were appeared in the papers [69,79,81].
Let X, Y be metric spaces and W be a nonempty subset of X × Y . In this
chapter, we consider the definitions of γ-metrically regular and (γ, κ)-metrically
regular in which the regularity horizon function γ : X → R+ satisfies the following
two properties:
1. γ(·) is Lipschitz with constant 1; (3.1)
2. γ(x) ≤ d(x,X\PXW ) for all x ∈ X. (3.2)
64
It is easy to see that the regularity horizon function associated to the Milyutin
regularity m(x) := d(x,X\PXW ) satisfies (3.1) and (3.2).
3.1 Perturbation stability of Milyutin-type
regularity
Firstly, we recall the definitions of γ-metrically regular and (γ, κ)-metrically
regular as in Definitions 2.1.3, 2.1.4 for the function γ mentioned above.
Definition 3.1.1. A set-valued mapping F : X ⇒ Y is said to be γ-Milyutin
regular on W with constant κ if there is a real number r > 0 such that
d(x, F−1(y)) ≤ κd(y, F (x)), (3.3)
for all (x, y) ∈ W with 0 < rd(y, F (x)) < γ(x). The lower bound regγ F (W ) of κ
in (3.3) is the modulus of γ-Milyutin regularity of F on W . If no such κ exists, set
regγ F (W ) =∞.
Definition 3.1.2. A set-valued mapping F : X ⇒ Y is said to be (γ, κ)-Milyutin
regular on W with constant κ > 0 if
d(x, F−1(y)) ≤ κd(y, F (x)),
for all (x, y) ∈ W with 0 < κd(y, F (x)) < γ(x). The infimum of all above constant
κ denoted by reg(γ,κ) F (W ) is the modulus of (γ, κ)-Milyutin regularity of F on W .
If no such κ exists, we set reg(γ,κ) F (W ) =∞.
The difference between these two concepts is that in Definition 3.1.1, the
constant κ in (3.3) and the constant r in the gauge condition 0 < rd(y, F (x)) < γ(x)
could be different; meanwhile, in Definition 3.1.2, they are equal.
Before stating main results, it is essential to suggest the definition of closure
of a mapping as follows.
65
Definition 3.1.3. Let X, Y be metric spaces and F : X ⇒ Y . Denote by F the
closure of the mapping F which has GraphF = GraphF , i.e., F : X ⇒ Y defined
by
F (x) = {y ∈ Y : ∃{xn} → x, yn ∈ F (xn) such that yn → y}.
Then, the lower semicontinuous envelop function of the distance function
associated to F defined by
ϕFy (x) := lim inf
u→x d(y, F (u)),
is agreed with the one associated to F as in the following lemma.
Lemma 3.1.1. Let X, Y be metric spaces and F : X ⇒ Y . Then, for each y ∈ Y ,
ϕFy (x) = ϕ
F
y (x), for all x ∈ X.
Proof. Take (x, y) ∈ X × Y . Then, ϕFy (x) ≤ ϕFy (x) due to F (x) ⊂ F (x). If
ϕFy (x) = +∞ then ϕFy (x) ≤ ϕFy (x) is obvious. In the case of ϕFy (x) < +∞, there
exists {un} → x, vn ∈ F (un) such that d(y, vn) → ϕFy (x). From the definition of
F , for each n ∈ N∗, there exist k(n) ≥ n, zk(n) ∈ X and wk(n) ∈ F (zk(n)) such that
d(zk(n), un) <
1
n and d(wk(n), vn) <
1
n . It follows that
d(zk(n), x) ≤ d(zk(n), un) + d(un, x) ≤ 1
n
+ d(un, x).
Letting n→∞, one obtains zk(n) → x. Furthermore, one has
d(y, F (zk(n))) ≤ d(y, wk(n)) ≤ d(y, vn) + d(vn, wk(n)) ≤ d(y, vn) + 1
n
.
By taking lim inf on both sides of the inequality in above when n tends to infinity,
we deduce that ϕFy (x) ≤ ϕFy (x).
Under the assumptions of the function γ, all results in the preceding section
can be applied to the γ-Milyutin regularity and (γ, κ)-Milyutin regularity. The
results of this section are given in the following theorems.
66
Now, given a function G : X × Y → Z, U ⊂ X, V ⊂ Y , and ε > 0. We set
W ε := {(x, z) ∈ X × Z : x ∈ U,B(G−1x (z), εγ(x)) ⊂ V },
here,
G−1x (z) := {y ∈ Y : z ∈ G(x, y)}, and
B(G−1x (z), εγ(x)) := {y ∈ Y : d(y,G−1x (z)) < εγ(x)}.
Theorem 3.1.1. Let X be a complete metric space and Y, Z be metric spaces. Let
U and V be open subsets of X and Y , respectively. Consider a set-valued mapping
F : X ⇒ Y with closed graph and a single-valued mapping G : X × Y → Z. Let
γ : X → R+ satisfy assumptions (3.1) and (3.2) for W = U × V . We assume that
the following statements hold true:
(a) F is γ-Milyutin regular on U×V with constant τ , i.e., there exists r > 0 such
that for all (x, y) ∈ U × V with 0 < rd(y, F (x)) < γ(x),
d(x, F−1(y)) ≤ τd(y, F (x));
(b) G(x, ·) is metrically regular on Y ×Z with modulus λ > 0 uniformly in x ∈ U ,
i.e.,
d(y,G−1x (z)) ≤ λd(z,G(x, y)), ∀x ∈ U,∀(y, z) ∈ Y × Z;
(c) G(·, y) satisfies the Lipschitz condition with constant ` > 0 such that `λτ < 1
on U for all y ∈ Y , i.e.,
d(G(x, y), G(u, y)) ≤ `d(x, u), ∀y ∈ Y, ∀x, u ∈ U.
Set Φ(x) = G(x, F (x)). Then, for every ε > 0, Φ is γ-Milyutin regular on W λ`ε
with regγ Φ(W
λ`ε) ≤ ((λτ)−1 − `)−1.
Proof. Firstly, we shall prove that
lim
δ↓0
inf
{
|ΓϕΦz |(x) :
d(x,W λ`εz )
γ(x)
< δ, z ∈ PZW λ`ε, 0 < ϕ
Φ
z (x)
γ(x)
< δ
}
≥ (λτ)−1 − `.
67
Indeed, choose δ such that
δ
1− δ < min{1, ε} and 0 < rλδ < 1. Take (x, z) ∈ X×Z
such that d(x,W λ`εz ) < δγ(x), z ∈ PZW λ`ε, and 0 < ϕΦz (x) < δγ(x). Then there
exists u ∈ W λ`εz such that d(x, u) < δγ(x). So, u ∈ U , B(G−1u (z), λ`εγ(u)) ⊂ V ,
and since γ is Lipschitz with constant 1, it follows that d(x, u) < δγ(u) + δd(x, u).
By the choice of δ, one has
d(x, u) <
δ
1− δγ(u) < γ(u), (3.4)
which gives x ∈ U .
Let now {un} ⊂ X be such that un → x and d(z,Φ(un))→ ϕΦz (x) as n→∞.
Thus, there exists n0 ∈ N such that for all n ≥ n0, 0 < d(z,Φ(un)) < δγ(un), and
as un → x ∈ U , we have un ∈ U due to the openness of U .
For each n ∈ N∗, for any yn ∈ G−1un (z), i.e., z = G(un, yn), from the continuity
of distance function and by (3.4), it follows that d(un, u) <
δ
1− δγ(u) for all n ≥ n0.
Then, from (b), (c), un, u ∈ U , and by the choice of δ, when n is sufficiently large,
we find that
d(yn, G
−1
u (z)) ≤ λd(z,G(u, yn)) = λd(G(un, yn), G(u, yn))
≤ λ`d(un, u) < λ` δ
1− δγ(u) < λ`εγ(u),
Hence, one has yn ∈ B(G−1u (z), λ`εγ(u)) ⊂ V , for any yn ∈ G−1un (z). In addition,
by taking zn ∈ F (un) such that d(z,G(un, zn)) ≤ (1 + 1
n
)d(z,G(un, F (un))) and by
(b), one sees that
d(zn, G
−1
un (z)) ≤ λd(z,G(un, zn)) ≤ λ(1 +
1
n
)d(z,G(un, F (un))),
and by taking yn ∈ G−1un (z) such that d(yn, zn) ≤ (1 +
1
n
)d(zn, G
−1
un (z)), we get
d(yn, F (un)) ≤ d(yn, zn) ≤ (1 + 1
n
)d(zn, G
−1
un (z))
≤ λ(1 + 1
n
)(1 +
1
n
)d(z,G(un, F (un))) (3.5)
< λ(1 +
1
n
)(1 +
1
n
)δγ(un) < r
−1γ(un),
68
where the last inequality is followed from the choice of δ when n is sufficiently
large. Then from the fact that F is γ-Milyutin regular on U×V with the regularity
modulus not greater than τ , we obtain
d(un, F
−1(yn)) ≤ τd(yn, F (un)), for all n ≥ n0.
We now choose some vn ∈ F−1(yn) (i.e., yn ∈ F (vn)) such that
d(un, vn) ≤ (τ + 1
n
)d(yn, F (un)). (3.6)
Then by the choice of δ, for all n ≥ n0, one has
d(un, vn) < (τ +
1
n
)λ(1 +
1
n
)δγ(un) < γ(un).
This yields vn ∈ U , and thus from (c), we have
d(G(un, yn), G(vn, yn)) ≤ `d(un, vn). (3.7)
Since ϕΦz (x) = ϕ
Φ
z (x) > 0, the closedness of Φ, and lim
n→∞un = x then
lim inf
n→∞ d(un, vn) > 0. Note that d(z,G(un, yn)) = 0 since z = G(un, yn), and by
(3.5), (3.6), (3.7) one concludes that
|ΓϕΦz |(x) ≥ lim sup
n→∞
ϕΦz (x)− ϕΦz (vn)
d(x, vn)
≥ lim sup
n→∞
d(z,Φ(un))− d(z,Φ(vn))
d(un, vn)
= lim sup
n→∞
d(z,G(un, F (un)))− d(z,G(vn, F (vn)))
d(un, vn)
≥ lim sup
n→∞
d(z,G(un, F (un)))− d(z,G(vn, yn))
d(un, vn)
= lim sup
n→∞
d(z,G(un, F (un)))− d(G(un, yn), G(vn, yn))
d(un, vn)
≥ lim sup
n→∞
d(z,G(un, F (un)))
d(un, vn)
− `
≥ lim sup
n→∞
1
λ
(
1 + 1n
)2 (
τ + 1n
) − ` = (λτ)−1 − `.
Hence,
lim
δ↓0
inf
{
|ΓϕΦz |(x) :
d(x,W λ`εz )
γ(x)
< δ, z ∈ PZW λ`ε, 0 < ϕ
Φ
z (x)
γ(x)
< δ
}
≥ (λτ)−1 − `,
69
which follows from Lemma 3.1.1 that
lim
δ↓0
inf
{
|ΓϕΦz |(x) :
d(x,W λ`εz )
γ(x)
< δ, z ∈ PZW λ`ε, 0 < ϕ
Φ
z (x)
γ(x)
< δ
}
≥ (λτ)−1 − `.
Applying Theorem 2.2.1, the conclusion of the theorem is satisfied.
In the case of the function G in Theorem 3.1.1 has the special form
G(x, F (x)) = g(x) + F (x), here g is a Lipschitz function, the result of stability
under sum perturbation in [79] is considered as a consequence of the above theorem,
and then the linear structure of range space Y is required as a normed space.
Given a function g : X → Y , U ⊂ X and ε > 0, we set
Ωε := {(x, y) ∈ X × Y : x ∈ U, B(y − g(x), εγ(x)) ⊂ V } .
Theorem 3.1.2. Let X be a complete metric space and Y be a normed space. Let
U and V be open subsets of X and Y , respectively. Let F : X ⇒ Y be a closed
set-valued mapping. Suppose that γ : X → R+ satisfies assumptions (3.1) and (3.2)
for W = U × V . If F is γ-Milyutin regular on U × V with constant τ , i.e.,
d(x, F−1(y)) ≤ τd(y, F (x))
for all (x, y) ∈ U × V, 0 < rd(y, F (x)) < γ(x), and g : X → Y is Lipschitz on U
with constant λ > 0 such that τλ 0, one has
regγ(F + g)(Ω
λε) ≤ (τ−1 − λ)−1.
Proof. Based on Theorem 2.2.1, we only need to prove that
lim
δ↓0
inf
{
|ΓϕF+gy |(x) :
d(x,Ωλεy )
γ(x)
< δ, y ∈ PY Ωλε, 0 <
ϕF+gy (x)
γ(x)
< δ
}
≥ τ−1 − λ.
Indeed, choose 0 < δ < min{ 11+λ , τ−1}, δ(1+δ)1−δ < ε, δ1−δ < 1, and take
(x, y) ∈ X × Y such that d(x,Ωλεy ) < δγ(x), y ∈ PY Ωλε and 0 < ϕF+gy (x) < δγ(x).
Then there exists u ∈ Ωλεy such that
d(x, u) < δγ(x). (3.8)
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It follows that d(x, u) < δγ(u) + δd(x, u), which implies that
d(x, u) <
δ
1− δγ(u) < γ(u).
So, u ∈ U , B(y − g(u), λεγ(u)) ⊂ V and x ∈ U .
Let now {un} ⊂ X such that un → x and
d(y, (F + g)(un))→ ϕF+gy (x) as n→∞.
Thus, there exists n0 ∈ N such that for all n ≥ n0,
0 < d(y − g(un), F (un)) < δγ(un) and un ∈ U( since un → x ∈ U - open).
Using (3.8), the Lipschitz property of g and γ, we get
‖g(un)− g(u)‖ ≤ λd(un, u)
< λδγ(un)
≤ λδ(1 + δ)γ(x)( since γ is continuous)
<
λδ(1 + δ)
1− δ γ(u)
< λεγ(u).
Consequently, y − g(un) ∈ B(y − g(u), λεγ(u)) ⊂ V . Furthermore, since
0 < d(y − g(un), F (un)) < δγ(un) < τ−1γ(un)
then from the Milyutin regularity of F on U × V with modulus τ , one obtains
d(un, F
−1(y − g(un))) ≤ τd(y − g(un), F (un)), for all n ≥ n0.
We now choose zn ∈ F−1(y − g(un)), i.e., y − g(un) ∈ F (zn) such that
d(un, zn) ≤
(
τ +
1
n
)
d(y − g(un), F (un)). (3.9)
It follows that for all n ≥ n0
d(un, zn) ≤
(
τ +
1
n
)
τ−1γ(un) =
(
1 +
τ−1
n
)
γ(un).
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From here, one has zn ∈ U and thus,
‖g(zn)− g(un)‖ ≤ λd(un, zn). (3.10)
Since ϕF+gy (x) > 0 and lim
n→∞un = x then lim infn→∞ d(un, zn) > 0. Combining (3.9),
and (3.10), one gets
|ΓϕF+gy |(x) ≥ lim sup
n→∞
ϕF+gy (x)− ϕF+gy (zn)
d(x, zn)
≥ lim sup
n→∞
d(y, (F + g)(un))− d(y, (F + g)(zn))
d(un, zn)
≥ lim sup
n→∞
d(y − g(un), F (un))− d(y − g(un), F (zn))− ‖g(zn)− g(un)‖
d(un, zn)
≥ lim sup
n→∞
d(y − g(un), F (un))− d(y − g(un), F (zn))
d(un, zn)
− λ
≥ lim sup
n→∞
d(y − g(un), F (un))
d(un, zn)
− λ
≥ lim sup
n→∞
1
τ + 1n
− λ = τ−1 − λ > 0.
So,
lim
δ↓0
inf
{
|ΓϕF+gy |(x) :
d(x,Ωλεy )
γ(x)
< δ, y ∈ PY Ωλε, 0 <
ϕF+gy (x)
γ(x)
< δ
}
≥ τ−1 − λ.
The proof is completed.
Analysis similar to that in the proof of Theorem 3.1.1 actually shows that the
version of the perturbation theorem for (γ, κ)-Milyutin regularity is also achieved.
Theorem 3.1.3. Let X be a complete metric space, Y, Z be metric spaces, and U, V
be open subsets of X, Y , respectively. Let a closed set-valued mapping F : X ⇒ Y
and a single-valued mapping G : X × Y → Z be given. Let γ : X → R+ satisfy
assumptions (3.1) and (3.2) for W = U × V . We assume that the following
statements are fulfilled:
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(a) F is (γ, τ)-Milyutin regular on U × V with constant τ , i.e., for all
(x, y) ∈ U × V with 0 < τd(y, F (x)) < γ(x),
d(x, F−1(y)) ≤ τd(y, F (x));
(b) G(x, ·) is metrically regular on Y ×Z with modulus λ > 0 uniformly in x ∈ U ,
i.e.,
d(y,G−1x (z)) ≤ λd(z,G(x, y)), ∀x ∈ U,∀(y, z) ∈ Y × Z;
(c) G(·, y) satisfies the Lipschitz condition with constant ` > 0 such that `λτ < 1
on U for all y ∈ Y , i.e.,
d(G(x, y), G(u, y)) ≤ `d(x, u), ∀y ∈ Y, ∀x, u ∈ U.
Set Φ(x) = G(x, F (x)). Then, Φ is (γ, κ)-Milyutin regular on W λ` with
reg(γ,κ) Φ(W
λ`) ≤ κ := ((λτ)−1 − `)−1, where
W λ` := {(x, z) ∈ X × Z : x ∈ U,B(G−1x (z), λ`γ(x)) ⊂ V }.
Proof. Set (W λ`z )γ := ∪u∈Wλ`z B(u, γ(u)). Now, we shall prove that for any
x ∈ (W λ`z )γ, z ∈ PZW λ` with 0 < ((λτ)−1 − `)−1ϕΦz (x) < γ(x),
|ΓϕΦz |(x) ≥ (λτ)−1 − `.
Indeed, ta