A NOTE ON GLOBAL IMPLICIT FUNCTION THEOREM

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Global Implicit Function Theorem Mihai Cristea

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A NOTE ON GLOBAL IMPLICIT FUNCTION THEOREM

MIHAI CRISTEA

University of Bucharest Faculty of Mathematics Str. Academiei 14, R-010014, Bucharest, Romania

EMail:mcristea@fmi.unibuc.ro

Received: 21 April, 2006

Accepted: 01 May, 2006

Communicated by: G. Kohr 2000 AMS Sub. Class.: 26B10, 46C05.

Key words: Global implicit function, Boundary behaviour of a maximal implicit function.

Abstract: We study the boundary behaviour of some certain maximal implicit function. We give estimates of the maximal balls on which some implicit functions are defined and we consider some cases when the implicit function is globally defined. We extend in this way an earlier result from [3] concerning an inequality satisfied by the partial derivatives ^∂h_∂xand ^∂h_∂y of the maphwhich verifies the global implicit function problem

h(t, x) =h(a, b), x(a) =b.

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The implicit function theorem is a classical result in mathematical analysis. Local versions can be found in [1],[8], [10], [13], [15], [17], [18] and some papers deal with some global versions (see [2], [3], [9], [16]).

We first give some local versions of the implicit function theorem, using our local homeomorphism theorem from [4].

Theorem 1. Let E, F be Banach spaces, dimF < ∞, U ⊂ E open, V ⊂ F open, h : U ×V → F continuous such that there exists K ⊂ U ×V countable such thathis differentiable on(U ×V) \K and ^∂h_∂y(x, y)∈Isom (F, F)for every (x, y) ∈ (U ×V) \K and let A = Pr₁K ⊂ U. Then, for every (a, b) ∈ U ×V there existsr, δ > 0and a unique continuous mapϕ:B(a, r)→B(b, δ)such that

ϕ(a) =bandh(x, ϕ(x)) =h(a, b) for everyx∈B(a, r) andϕis differentiable onB(a, r)\A.

Proof. Let(a, b)∈U ×V be fixed andf :U×V →E×F be defined by f(x, y) = (x, h(x, y) +b−h(a, b)) for (x, y)∈U ×V.

Also, letT :U×V →E×F,

T (x, y) = (0, y−h(x, y) +h(a, b)−b) for (x, y)∈U ×V.

Then ImT ⊂ F, hence T is compact and we see that f = I −T, f is differentiable on (U ×V) \ K and f⁰(x, y) ∈ Isom (E×F, E×F) for every (x, y) ∈ (U×V) \K. Using the local inversion theorem from [4], we see thatf is a local homeomorphism onU ×V.LetW ∈ V((a, b))andδ >0be such that

f|B(a, δ)×B(b, δ) :B(a, δ)×B(b, δ)→W

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is a homeomorphism and let

g = (g₁, g₂) :W →B(a, δ)×B(b, δ)

be its inverse. We take ` > 0such that Q = B(a, b)×(b, `) ⊂ W and let r = min{`, δ}. We have

(x, z) = f(g(x, z))

=f(g₁(x, z), g₂(x, z))

= (g₁(x, z), h(g₁(x, z), g₂(x, z)) +b−h(a, b)) for every(x, z)∈Q, hence

x=g₁(x, z), h(x, g₂(x, z)) =z+h(a, b)−b forx∈B(a, r), z ∈B(b, `).

We define now ϕ : B(a, r) → B(b, δ) by ϕ(x) = g₂(x, b) for every x ∈ B(a, r) and we see that h(x, ϕ(x)) = h(a, b) for every x ∈ B(a, r). We have f(a, b) = (a, b) =f(a, ϕ(a))and using the injectivity off onB(a, δ)×B(b, δ), we see thatϕ(a) = b. Also, ifψ : B(a, r) → B(b, δ) is continuous andψ(a) = b, h(x, ψ(x)) = h(a, b) for every x ∈ B(a, r), then f(x, ϕ(x)) = (x, b) = f(x,Ψ (x)) for every x ∈ B(a, r) and using again the injectivity of the map f onB(a, δ)×B(b, δ), we find thatϕ =ψ onB(a, r).

Let nowx₀ ∈ B(a, r)\A. Then (x₀, b) = f(x₀, β), with(x₀, β) ∈ (B(a, r)× B(b, δ)\K),hencef is differentiable in(x0, β), f⁰(x0, β)∈ Isom (E×F, E×F) and sincef is a homeomorphism onB(a, r)×B(b, δ), it results thatg is also differentiable in(x₀, b) =f(x₀, β)andg⁰(x₀, b) = [f⁰(x₀, β)⁻¹], and we see thatϕis differentiable inx₀.

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Theorem 2. LetE be an infinite dimensional Banach space,dimF < ∞, U ⊂ E, V ⊂F be open sets,h:U×V →F be continuous such that there existsK ⊂U×V,

K =

∞

[

p=1

K_p

withK_p compact sets forp ∈ Nsuch thathis differentiable on(U ×V)\K, there exists ^∂h_∂y onU ×V and ^∂h_∂y(x, y) ∈ Isom (F, F)for every(x, y) ∈ U ×V and let A = P_r₁K. Then, for every (a, b) ∈ U × V there exists r, δ > 0 and a unique continuous implicit function ϕ : B(a, r) → B(b, δ)differentiable on B(a, r)\A such thatϕ(a) =bandh(x, ϕ(x)) =h(a, b)for everyx∈B(a, r).

Proof. We apply Theorem 11 of [8]. We see that in an infinite dimensional Banach space E, a set K which is a countable union of compact sets is a "thin" set, i.e., intK = φ andB\K is connected and simply connected for every ball B fromE.

Also, sinceAis a countable union of compact sets, we see thatintA=φ.

IfE, F are Banach spaces andA∈ L(E, F), we let kAk= sup

kxk=1

kA(x)k and

`(A) = inf

kxk=1kA(x)k and ifD⊂E, λ >0, we let

λD={x∈E|there existsy ∈Dsuch thatx=λy}.

IfX, Y are Banach spaces,D⊂X is open,x∈Dandf :D→Y is a map, we let D⁺f(x) = lim sup

y→x

kf(y)−f(x)k ky−xk

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and we say thatf is a light map if for everyx∈Dand everyU ∈ V(x), there exists Q∈ V(x)such thatQ⊂U andf(x)∈/ f(∂Q).

Remark 1. We can replace in Theorem 1 and Theorem 2the condition "dimF <

∞" by "There exists ^∂h_∂y onU ×V and it is continuous onU ×V and ^∂h_∂y(x, y) ∈ Isom (F, F)for every(x, y)∈U×V" to obtain the same conclusion, and this is the classical implicit function theorem. Also, keeping the notations from Theorem1and Theorem2, we see that if(α, β)∈B(a, r)×B(b, δ)is such thath(α, β) = h(a, b), thenβ =ϕ(α).

We shall use the following lemma from [7].

Lemma 3. Leta >0, f : [0, a]→[0,∞)be continuous and letω: [0,∞)→[0,∞) be continuous such thatω >0on(0,∞)and

|f(b)−f(c)| ≤ Z c

b

ω(f(t))dtfor every0< b < c ≤a.

Then, if

m= inf

t∈[0,a]f(t), M = sup

t∈[0,a]

f(t), it results that

Z M

m

ds ω(s) ≤a.

We obtain now the following characterization of the boundary behaviour of the solutions of some differential inequalities.

Theorem 4. LetE, F be Banach spaces,U ⊂ E a domain,K ⊂ U at most count- able,ϕ :U →F continuous onU and differentiable onU\Ksuch that there exists

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ω : [0,∞) → [0,∞)continuous withkϕ⁰(x)k ≤ ω(kϕ(x)k)for everyx ∈ U\K.

Then, ifα∈∂U andC ⊂U is convex such thatα ∈C,either there exists limx→α

x∈C

ϕ(x) = `∈F or lim

x→α x∈C

kϕ(x)k=∞ or, ifω >0on(1,∞)and

Z ∞

1

ds

ω(s) =∞, there exists

limx→α x∈C

ϕ(x) = `∈F.

Ifω >0on(0,1)and

Z 1

0

ds

ω(s) =∞

and there exists α ∈ U such that ϕ(α) = 0, it results that ϕ(x) = 0, for every x∈U.

Proof. Replacing, if necessary, ω by ω +λ for someλ > 0, we can suppose that ω >0on[0,∞).Letα∈∂U, C ⊂Uconvex such thatα∈Cand letq: [0,1)→C be a path such that

limt→1q(t) = α

and there existsL >0such thatD⁺_q (t)≤Lfor everyt∈[0,1). Then kq(s)−q(t)k ≤L·(s−t)

for every0≤t < s <1and let0≤c < d <1be fixed, A= co (q([c, d]))

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andε >0. Letg :A→Rbe defined by

g(z) =ω(kϕ(z)k) for everyz ∈A.

ThenA is compact and convex andg is uniformly continuous onA, hence we can findδ_ε⁰ >0such that

|g(z₁)−g(z₂)| ≤εforz₁, z₂ ∈A

with kz₁−z₂k ≤ δ⁰_ε. Since q : [c, d] → C is uniformly continuous, we can find δ_ε > 0such thatkq(t)−q(s)k ≤ δ_ε⁰ ifs, t ∈ [c, d]are such that |s−t| ≤ δ_ε. Let now

∆ = (c=t0 < t1 <· · ·< tm =d)∈ D([c, d]) be such thatk∆k ≤δ_ε. Using Denjoi-Bourbaki’s theorem we have

|kϕ(q(d))k − kϕ(q(c))k| ≤ kϕ(q(d))−ϕ(q(c))k

≤

m−1

X

k=0

ϕ(q(t_k+1))−ϕ(q(t_k))

≤

m−1

X

k=0

k(q(t_k+1)−q(t_k))k · sup

z∈[q(t_k),q(tk+1)]\K

kϕ⁰(z)k

≤L·

m−1

X

k=0

(t_l+1−t_k)· sup

z∈[q(t_k),q(tk+1)]

ω(kϕ(z)k)

≤L·

m−1

X

k=0

(t_l+1−t_k)·(ω(kϕ(q(t_k)k+ε).

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Lettingk∆k →0and thenε→0, we obtain

| kϕ(q(d))k − kϕ(q(c))k| ≤ kϕ(q(d))−ϕ(q(c))k

≤L· Z d

c

ω(kϕ(q(t))k)dtfor0≤c < d <1.

(1) If

m= inf

t∈[0,1)kϕ(q(t))k, M = sup

t∈[0,1)

kϕ(q(t))k, we obtain from Lemma3and (1)

(2)

Z M

m

ds

ω(s) ≤L.

Let nowz_p →αbe such that

kz_p−αk ≤ 1

2^P, z_p ∈Cforp∈N

and suppose that there existsρ >0such thatkϕ(z_p)k ≤ρfor everyp∈N. We take 0 = t₀ < t₁ <· · ·< t_k< t_k+1 <· · ·<1

such thatt_k %1and we defineq: [ 0,1)→Cby q(t) = z_k(t_k+1−t) +z_k+1(t−t_k)

t_k+1−t_k fort∈[t_k, t_k+1], k ∈N. Then

D⁺q(t) =ck = kz_k+1−z_kk

t_k+1−t_k fort∈[tk, tk+1]

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and takingt_k= _k+1^k fork∈N, we see thatc_k →0.Then αp = sup

t∈[t_p,1)

D⁺q(t) = sup

k≥p

ck→0 and let

a_p = inf

t∈[t_p,1)kϕ(q(t))k, b_p = sup

t∈[tp,1)

kϕ(q(t))k forp∈N. Using (2) we obtain that

Z bp

ap

ds

ω(s) ≤α_p forp∈N and letp₀ ∈Nbe such that

α_p <

Z ∞

ρ

ds

ω(s) forp≥p₀.

Suppose that there existsp≥ p₀ such thatb_p =∞. Then, sinceq(t_k) = z_k, we see that

a_k≤ kϕ(q(t_k))k=kϕ(z_k)k ≤ρfork∈N, hence

0<

Z ∞

ρ

ds ω(s) ≤

Z bp

ap

ds

ω(s) ≤α_p <

Z ∞

ρ

ds ω(s) and we have reached a contradiction.

It results thatbp <∞forp≥p0and let Kp = sup

t∈[0,bp]

ω(t)

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forp≥p₀. ThenK_p <∞and we see from (1) that

kϕ(q(d))−ϕ(q(c))k ≤α_p ·K_p · |d−c|

fort_p ≤c < d <1andp≥p₀, and this implies that limt→1ϕ(q(t)) = `∈F.

It results that

p→∞limϕ(z_p) = lim

p→∞ϕ(q(t_p)) = `∈F.

Now, if the case

limx→α x∈C

kϕ(x)k=∞

does not hold, there existsρ >0andx_p →α, x_p ∈Cwith

||ϕ(x_p)|| ≤ρand||x_p−α|| ≤ 1 2^p

for everyp∈N, and from what we have proved before, it results that

p→∞limϕ(x_p) = `∈F.

Ifa_p ∈C, ka_p−αk ≤ ₂¹p for everyp∈N, then

p→∞limϕ(a_p) = `₁ ∈F.

Letz_2p =x_p, z_2p+1 =a_pforp∈N. We see that

p→∞limϕ(z_p) =`₂ ∈F,

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hence

`= lim

p→∞ϕ(x_p) = lim

p→∞ϕ(z_2p) =`₂ and

`1 = lim

p→∞ϕ(ap) = lim

p→∞ϕ(z2p+1) =`2,

hence` =`1 =`2. We have proved that ifap ∈C, kap−αk ≤ ₂¹p for everyp∈N, then

p→∞limϕ(a_p) =`.

We show now that ifa_p ∈C, a_p →α, then

p→∞limϕ(a_p) =`.

Indeed, if this is false, there existsε >0and(a_p_k)_k∈

Nsuch that kϕ(a_p_k)−`k> ε

for everyk ∈N. Let a_p_kq

q∈N

be a subsequence such that

a_p_kq −α < 1

2^q

for everyq∈N. From what we have proved before it results that

q→∞limϕ a_p_kq

=` and we have reached a contradiction.

We have therefore proved that either limx→α x∈C

ϕ(x) = `∈F,

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or

limx→α x∈C

kϕ(x)k=∞.

Suppose now

Z ∞

1

ds

ω(s) =∞

and letα∈∂U. We takex∈Cand letq: [0,1)→C be defined by q(t) = (1−t)x+tα

fort ∈[0,1)and

m= inf

t∈[0,1)kϕ(q(t))k, M = sup

t∈[0,1)

kϕ(q(t))k. SinceD⁺q(t) = kx−αkfor everyt ∈(0,1), we see from (2)

Z M

kxk

ds ω(s) ≤

Z M

m

ds

ω(s) ≤ kx−αk and this implies thatM <∞. Let

b= sup

t∈[0,M]

ω(t).

Thenb <∞and using (1), we see that

kϕ(q(d))−ϕ(q(c))k ≤b· kx−αk · |d−c|

for0≤c < d <1and this implies that limx→α

x∈C

ϕ(x) = `∈F.

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It results that the case

limx→α x∈C

kϕ(x)k=∞ cannot hold, hence

x→αlimϕ(x) = `∈F.

Suppose now that

Z 1

0

ds

ω(s) =∞

and that there existsα ∈ U such thatϕ(α) = 0. Letr = d(α, ∂U), y ∈ B(α, r) and q : [0,1] → B(α, r), q(t) = (1−t)α+ty for t ∈ [0,1]. ThenD⁺q(t) = ky−αkfort∈[0,1]and let

m = inf

t∈[0,1]kϕ(q(t))k and

M = sup

t∈[0,1]

kϕ(q(t))k. Thenm= 0and we see from (2) that

Z M

0

ds

ω(s) ≤ ky−αk.

This implies thatM = 0and henceϕ(y) = 0. We proved thatϕ ≡ 0onB(α, r) and sinceU is a domain, we see thatϕ≡0onU.

Remark 2. We proved that ifϕis as in Theorem2and Z ∞

1

ds

ω(s) =∞, thenϕhas angular limits in every pointα∈∂U.

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We now obtain the following characterization of the boundary behaviour of some implicit function.

Theorem 5. LetE, F be Banach spaces,U ⊂ E a domain,K ⊂ U ×F such that A = Pr₁K is at most countable and leth :U ×F → F be continuous onU ×F, differentiable on(U ×F)\K such that

` ∂h

∂y (x, y)

>0on (U ×F)\K and there existsω: [0,∞)→[0,∞)continuous such that

∂h

∂x (x, y)

` ∂h

∂y(x, y)

≤ω(kyk)

for every (x, y) ∈ (U ×F)\K. Suppose that ϕ : U → F is continuous on U, differentiable onU\A, ϕ(a) =band

h(x, ϕ(x)) = h(a, b) for everyx∈U.

Then, ifα∈∂U andC⊂U is convex such thatα∈C, either limx→α

x∈C

kϕ(x)k=`∈F, or

limx→α x∈C

kϕ(x)k=∞.

Also, ifω >0on(1,∞)and

Z ∞

1

ds

ω(s) =∞,

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then

limx→α x∈C

ϕ(x) = `∈F.

Proof. We see that if x ∈ U \ A, then (x, ϕ(x)) ∈ (U ×F) \ K, hence h is differentiable in(x, ϕ(x))and we have

∂h

∂x(x, ϕ(x)) + ∂h

∂y (x, ϕ(x))·ϕ⁰(x) = 0 and we see that

kϕ⁰(x)k ·` ∂h

∂y (x, ϕ(x))

≤

∂h

∂y (x, ϕ(x)) (ϕ⁰(x))

=

∂h

∂x(x, ϕ(x)) . It results that

kϕ⁰(x)k ≤

∂h

∂x(x, ϕ(x))

` ∂h

∂y (x, ϕ(x))

≤ω(kϕ(x)k) for everyx∈U \Aand we now apply Theorem4.

Remark 3. IfEis an infinite dimensional Banach space and K =

∞

[

p=1

Kp withKp ⊂E are compact sets for everyp∈Nandy ∈E, then the set

M(K, y) ={w∈E|there existst >0andx∈K such thatw=tx}

is also a countable union of compact sets and hence a "thin" set. Keeping the notations from Theorem4, we see that the basic inequality

(3) kϕ(z₁)−ϕ(z₂)k ≤ sup

z∈[z1,z2]

ω(kϕ(z)k) if [z₁, z₂]⊂U

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is also valid forK a countable union of compact sets andϕas in Theorem4.

If dimE = n and K ⊂ E has a σ-finite (n−1)-dimensional measure (i.e.

K =S∞

p=1K_p,withmn−1(K_p)<∞for everyp∈N, wherem_qis theq-Hausdorff measure fromRⁿ), a theorem of Gross shows that if H ⊂ E is a hyperplane and P :E →His the projection onH, thenP⁻¹(z)∩K is at most countable with the possible exception of a setB ⊂ H,with mn−1(B) = 0. Applying as in Theorem 4the theorem of Denjoi and Bourbaki on each interval fromP⁻¹(z)∩K for every z ∈H \B and using a natural limiting process, we see that ifdimE =nandK ⊂ E has aσ-finite(n−1)-dimensional measure, then the inequality (3) also holds. It is easy see now that Theorem4and Theorem5hold if the setK, respectively the set A = Pr1K are chosen to be a countable union of compact sets ifdimE = ∞and havingσ-finite(n−1)-dimensional measure ifdimE =n.

The following theorem is the main theorem of the paper and it gives some cases when the implicit function is globally defined or some estimates of the maximal balls on which some implicit function is defined.

We say that a domainDfrom a Banach space is starlike with respect to the point a ∈ Dif[a, x] ⊂ Dfor everyx ∈ D, and ifDis a domain in the Banach space E anda∈D. We set

D_a={x∈D|[a, x]⊂D}.

Theorem 6. Let E, F be Banach spaces, dimF < ∞, D ⊂ E a domain, K ⊂ D×F at most countable, andA= Pr₁K.Also, leth : D×F → F be continuous onD×F and differentiable on(D×F)\K such that

` ∂h

∂y (x, y)

>0on (D×F)\K.

In addition, there existsω : [0,∞) →[0,∞)continuous such thatω >0on(0,∞)

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and

∂h

∂x (x, y)

` ∂h

∂y(x, y)

≤ω(kyk) for every(x, y)∈(D×F)\K.Then, if(a, b)∈D×F and

Q_a,b=D_a∩B

a, Z ∞

kbk

ds ω(s)

,

there exists a unique continuous mapϕ : Q_a,b → F, differentiable onQ_a,b\Asuch thath(x, ϕ(x)) = h(a, b)for everyx∈Q_a,b.IfDis starlike with respect toaand

Z ∞

1

ds

ω(s) =∞, thenQ_a,b =Dandϕ :D→F is globally defined onD.

Proof. Letz ∈ Q_a,b and letB = {x ∈ [a, z]| there exists an open, convex domain D_x ⊂ Q_a,b such that[a, x]⊂ D_x}and a continuous implicit functionϕ_x :D_x →F, differentiable on D_x\A such thatϕ_x(a) = b andh(u, ϕ_x(u)) = h(a, b)for every u ∈ D_x.We see that B is open, and from Theorem1, B 6= ∅. We show thatB is a closed set. Let x_p ∈ B, x_p → x, and we can suppose that x_p ∈ [a, x)for every p∈N,and let

Q=

∞

[

p=1

D_x_p.

We defineΨ :Q→F byΨ (u) =ϕ_x_p(u)foru∈D_x_pandp∈Nand the definition is correct. Indeed, ifp, q ∈N, p6=qletU_p_q =D_x_p ∩D_x_q.ThenU_pq is a nonempty, open and convex set, hence it is a nonempty domain. If

Vpq =

u∈Upq|ϕxq(u) = ϕxp(u) ,

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we see thata ∈V_pq, henceV_pq 6=∅and we see thatV_pqis a closed set inU_pq. Using the property of the local unicity of the implicit function from Theorem1, we obtain that V_pq is also an open set. Since U_pq is a domain, it results that U_pq = V_pq and hence thatΨis correctly defined. We also see immediately thatΨ (a) =b, andΨis continuous onQand differentiable onQ\A.

Letq: [0,1]→Qbe defined by

q(t) = (1−t)a+txfort ∈[0,1]. Then

D⁺q(t) = kx−ak<

Z ∞

||b|

ds ω(s). Let

m= inf

t∈[0,1)kΨ (q(t))k, M = sup

t∈[0,1)

kΨ (q(t))k. As in Theorem5, we see that

kΨ⁰(u)k ≤ω(kΨ (u)k) for everyu∈Q\A, hence

kΨ (z₁)−Ψ (z₂)k ≤ sup

z∈[z1,z2]

ω(kΨ (z)k)

if[z₁, z₂]⊂Q.This implies that relations (1) and (2) from Theorem4also hold and we see that

Z M

m

ds

ω(s) ≤ kx−ak. Then

Z M

kbk

ds ω(s) ≤

Z M

m

ds

ω(s) ≤ kx−ak

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and this implies thatM <∞, hence

`= sup

t∈[0,M]

ω(t)<∞.

Using (1) and Theorem4, we see that

kΨ (q(d))−Ψ (q(c))k ≤`· kx−ak · |d−c| for every0≤c < d <1 and this implies that

limu→x u∈Imq

Ψ (u) = w∈F.

Using Theorem1, we can findr, δ > 0and a unique continuous implicit function Ψ_x :B(x, r)→B(w, δ), differentiable onB(x, r) \Asuch that

Ψ_x(x) =wandh(u,Ψ_x(u)) = h(a, b) for everyu∈B(x, r). Let0< ε < randp_ε∈Nbe such that kx_p−xk< εand kΨ (x_p)−wk< δ forp≥pεand letp≥pεbe fixed. Since

ϕ_x_p(x_p) = Ψ (x_p)∈B(w, δ),

we see from Remark1thatϕ_x_p(x_p) = Ψ_x(x_p)and hence the set U =

u∈D_x_p∩B(x, ε)|ϕ_x_p(u) = Ψ_x(u)

is nonempty. We also see thatD_x_p∩B(x, ε)is an open, nonempty, convex set, hence it is a domain andU is an open, closed and nonempty subset ofD_x_p ∩B(x, ε), and this implies that

U =Dxp∩B(x, ε).

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LetU₀ =D_x_p∪B(x, ε). We can now correctly defineΦ :U₀ →F by Φ (u) = ϕxp(u) ifu∈Dxp

and

Φ (u) = Ψx(u) ifu∈B(x, ε)

and we see that Φ is continuous on U₀, differentiable on U₀ \ A, Φ (a) = b and h(u,Φ (u)) = h(a, b) for everyu ∈ U₀. It results that x ∈ B, hence B is also a closed set and since[a, z]is a connected set, we see thatB = [a, z].

We have therefore proved that for everyz ∈Qa,bthere exists a convex domainDz

such that [a, z] ⊂ Dz and a unique continuous implicit function ϕz : Dz → F, differentiable onD_z \Asuch that

ϕz(a) =b, h(u, ϕz(u)) =h(a, b) for everyu∈Dz.

We now defineϕ :Q_a,b → F byϕ(x) = ϕ_z(x)forx ∈ D_z and we see, as before, that the definition is correct, thatϕ(a) = b, ϕis continuous on Q_a,b, differentiable onQa,b\Aandh(x, ϕ(x)) =h(a, b)for everyx∈Qa,b.

Remark 4. The result from Theorem 6extends a global implicit function theorem from [3]. The result from [3] also involves an inequality containing

∂h

∂x(x, y)

and ` ∂h

∂y (x, y)

and it says that ifE, F are Banach spaces,h : E ×F → F is aC¹ map such that

∂h

∂y(x, y) ∈ Isom (F, F)for every (x, y) ∈ E ×F and there exists ω : [0,∞) → (0,∞)continuous such that

1 +

∂h

∂x(x, y)

` ∂h

∂y (x, y)

≤ω(max (kxk,kyk))

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for every(x, y)∈E×F, then, for(x_0,y₀)∈E×F, z₀ =h(x₀, y₀)and r=

Z ∞

max(kx₀k,ky₀k)

ds 1 +ω(s),

there exists a C¹ mapϕ : B(x₀, r)×B(z₀, r) → F such that h(x, ϕ(x, z)) = z for every (x, z) ∈ B(x₀, r)× B(z₀, r). The main advantage of our new global implicit function theorems is that these theorems hold even if the maph is defined on a proper subset ofE×F,namely, on a setD×F ⊂E×F, whereD⊂Eis an open starlike domain.

Example 1. A known global inversion theorem of Hadamard, Lévy and John says that if E, F are Banach spaces, f : E → F is a C¹ map such that f⁰(x) ∈ Isom (E, F) for every x ∈ E and there exists ω : [0,∞) → (0,∞) continuous such that

Z ∞

1

ds

ω(s) =∞ and

f⁰(x)⁻¹

≤ω(kxk)

for everyx ∈ E, then it results thatf : E → F is aC¹ diffeomorphism (see [11], [14], [12], [3], [7]). If E = F = Rⁿ or if dimE = ∞ and f = I − T with T compact, we can drop the continuity of the derivative onEand we can impose the essential condition "f⁰(x)∈Isom (F, F)" with the possible exception of a "thin" set (see [4], [5],[6]) and we will still obtain thatf :E →F is a homeomorphism.

Now, letE, F be Banach spaces,D ⊂ E a domain,a ∈ D, b ∈F, g :D → F be differentiable on D, f : F → F be differentiable on F such that f⁰(y) ∈ Isom (F, F) for every y ∈ F and there exists ω : [0,∞) → (0,∞) continuous such that

f⁰(y)⁻¹

≤ω(kyk) for everyy∈F.

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Let h : D×F → F be defined byh(x, y) = f(y)−g(x) for x ∈ D, y ∈ F, r₀ =d(a, ∂D)and suppose that

M_r = sup

x∈B(a,r)

kg⁰(x)k<∞for every0< r≤r₀. Then

∂h

∂x(x, y)

` ∂h

∂y (x, y)

≤ kg⁰(x)k ·

f⁰(y)⁻¹

≤Mr·ω(kyk) if(x, y)∈B(a, r)×F and0< r≤r₀ and let

δ_r= min

r, 1 M_r ·

Z ∞

kbk

ds ω(s)

for0< r≤r₀.

Using Theorem6, we see that there exists a unique differentiable mapϕ:B(a, δr)→ F such that ϕ(a) = b and h(x, ϕ(x)) = h(a, b) for every x ∈ B(a, δ_r), i.e.

f(ϕ(x)) = g(x) +h(a, b)for everyx∈B(a, r)and every0< r≤r₀. If

r0·Mr0 <

Z ∞

kbk

ds ω(s),

then ϕ is defined on B(a, r₀). Additionally, if D = B(a, r₀), then ϕ is globally defined onDandf ◦ϕ =g +h(a, b)onD. In the special caseD =E, g(x) = x for everyx∈E,

Z ∞

1

ds

ω(s) =∞ and b =f(a),

thenf(ϕ(x)) =xfor everyx∈E, andϕis defined onEand is the inverse off and it results thatf :F →F is a homeomorphism. In this way we obtain an alternative proof of the Hadamard-Lévy-John theorem.

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Remark 5. The global implicit function problem

h(t, x) =h(a, b), x(a) = b considered before has two basic properties:

1. It satisfies the differential inequalitykϕ⁰(x)k ≤ω(kϕ(x)k).

2. It has the property of the local existence and local unicity of the solutions around each point(t0, x0).

This shows that by considering some other conditions of local existence and local unicity of the implicit function instead of the conditions from Theorem 1, we can produce corresponding global implicit function results. Using the conditions of local existence and local unicity from Theorem 11 of [8], we obtain the following corresponding version of Theorem6.

Theorem 7. Let E, F be Banach spaces, dimE = ∞, dimF < ∞, D ⊂ E a domain,K ⊂D×F,

K =

∞

[

p=1

K_p

withK_p compact sets for everyp ∈N,A = Pr₁K, h: D×F → F continuous on D×F and differentiable on(D×F)\K such that

` ∂h

∂y (x, y)

>0on (D×F)\K,

and there existsω: [0,∞)→[0,∞)continuous such thatω >0on(0,∞)and

∂h

∂x(x, y)

` ∂h

∂y (x, y)

≤ω(kyk)

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for every(x, y) ∈ (D×F)\K. Suppose that the mapy → h(x, y)is a light map onF for everyx∈D. Then, ifa, b∈D×F and

Q_a,b=D_a∩B

a, Z ∞

kbk

ds ω(s)

,

there exists a unique continuous implicit function ϕ : Q_a,b → F, differentiable on Q_a,b\Asuch thatϕ(a) = bandh(x, ϕ(x)) = h(a, b)for everyx ∈ Q_a,b and ifD is starlike with respect toaand

Z ∞

1

ds

Remark 6. The condition "the mapy→h(x, y)is a light map onF for everyx∈D"

is satisfied if ^∂h_∂y exists onD×F and

` ∂h

∂y (x, y)

>0for every (x, y)∈D×F.

Using the conditions of local existence and local unicity from Theorem 7 of [8], we obtain the following global implicit function theorem.

Theorem 8. Letn≥2, D⊂Rⁿbe a domain,h:D×R^m →R^m be differentiable and letK ⊂D×R^m,

K =

∞

[

p=1

Kp

with K_p closed sets such that mn−2(Pr₁K_p) = 0 for every p ∈ N, A = Pr₁K, such that ^∂h_∂y(x, y)∈ Isom (R^m,R^m)for every(x, y) ∈(D×R^m)\K and the map

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y → h(x, y) is a light map on R^m for every x ∈ D. Suppose that there exists ω: [0,∞)→[0,∞)continuous such thatω >0on(0,∞)and

∂h

∂x(x, y)

` ∂h

∂y (x, y)

≤ω(kyk) for every (x, y)∈(D×F)\K.

Then, if(a, b)∈D×F and

Qa,b=Da∩B

a, Z ∞

kbk

ds ω(s)

,

there exists a unique continuous implicit function ϕ : Qa,b → F, differentiable on Q_a,b\Asuch thatϕ(a) =b andh(x, ϕ(x)) = h(a, b)for everyx ∈Q_a,b,and ifD is starlike with respect toaand

Z ∞

1

ds

Proof. We see thatm_m+n−2(K_p) = 0 for everyp ∈ N andAhasσ-finite(n−1)- dimensional measure. We now apply the local implicit function theorem from The- orem 7 of [8], Remark3and the preceding arguments.

Using the classical implicit function theorem, we obtain the following global implicit function theorem

Theorem 9. Let E, F be Banach spaces,D ⊂ E a domain, h : D×F → F be continuous such that ^∂h_∂y exists onD×F, it is continuous onD×F and

` ∂h

∂y(x, y)

>0for every (x, y)∈D×F.

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Also, let K ⊂ D× F be such thatA = Pr₁K is a countable union of compact sets if dimE = ∞ and has σ-finite (n−1)-dimensional measure if dimE = n.

Additionally, suppose that h is differentiable on (D×F)\K and there exists ω : [0,∞)→[0,∞)continuous such thatω > 0on(0,∞)and

∂h

∂x(x, y)

` ∂h

∂y (x, y)

≤ω(kyk) for every(x, y)∈(D×F)\K. Then, if(a, b)∈D×F and

Q_a,b=D_a∩B

a, Z ∞

kbk

ds ω(s)

,

there exists a unique continuous implicit functionϕ : Q_a.b → F,differentiable on Q_a,b\A such thatϕ(a) = b andh(x, ϕ(x)) = h(a, b)for everyx ∈ Q_a,b. IfD is starlike with respect toaand

Z ∞

1

ds

ω(s) =∞, thenQ_a,b =Dandϕ :D→F is globally defined.

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