Volume 8 (2007), Issue 3, Article 100, 15 pp.
A NOTE ON GLOBAL IMPLICIT FUNCTION THEOREM
MIHAI CRISTEA
U NIVERSITY OF B UCHAREST
FACULTY OF M ATHEMATICS
S TR . ACADEMIEI 14, R-010014,
B UCHAREST, ROMANIA
mcristea@fmi.unibuc.ro
Received 21 April, 2006; accepted 01 May, 2006
Communicated by G. Kohr
A BSTRACT. 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
∂h
from [3] concerning an inequality satisfied by the partial derivatives ∂h
∂x and ∂y of the map h
which verifies the global implicit function problem
h (t, x) = h (a, b) ,
x (a) = b.
Key words and phrases: Global implicit function, Boundary behaviour of a maximal implicit function.
2000 Mathematics Subject Classification. 26B10, 46C05.
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, dim F < ∞, U ⊂ E open, V ⊂ F open, h :
U ×V → F continuous such that there exists K ⊂ U ×V countable such that h is differentiable
(x, y) ∈ Isom (F, F ) for every (x, y) ∈ (U × V ) \ K and let A =
on (U × V ) \ K and ∂h
∂y
Pr1 K ⊂ U . Then, for every (a, b) ∈ U × V there exists r, δ > 0 and a unique continuous map
ϕ : B (a, r) → B (b, δ) such that
ϕ (a) = b and h (x, ϕ (x)) = h (a, b) for every x ∈ B (a, r)
and ϕ is differentiable on B (a, r) \A.
Proof. Let (a, b) ∈ U × V be fixed and f : U × V → E × F be defined by
f (x, y) = (x, h (x, y) + b − h (a, b)) for (x, y) ∈ U × V.
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M IHAI C RISTEA
Also, let T : 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 f 0 (x, y) ∈ Isom (E × F, E × F ) for every (x, y) ∈ (U × V ) \ K. Using
the local inversion theorem from [4], we see that f is a local homeomorphism on U × V. Let
W ∈ V ((a, b)) and δ > 0 be such that
f |B (a, δ) × B (b, δ) : B (a, δ) × B (b, δ) → W
is a homeomorphism and let
g = (g1 , g2 ) : W → B (a, δ) × B (b, δ)
be its inverse. We take ` > 0 such 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
for x ∈ 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 of f on B (a, δ) × B (b, δ) , we see that ϕ (a) = b. Also, if ψ :
B (a, r) → B (b, δ) is continuous and ψ (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 on B (a, δ) × B (b, δ), we find that ϕ = ψ on B (a, r).
Let now x0 ∈ B (a, r) \A. Then (x0 , b) = f (x0 , β), with (x0 , β) ∈ (B(a, r) × B(b, δ)\K),
hence f is differentiable in (x0 , β) , f 0 (x0 , β) ∈ Isom (E × F, E × F ) and since f is a homeomorphism on B(a, r) × B(b, δ), it results that g is also differentiable in (x0 , b) = f (x0 , β) and
g 0 (x0 , b) = [f 0 (x0 , β)−1 ], and we see that ϕ is differentiable in x0 .
Theorem 2. Let E be an infinite dimensional Banach space, dim F < ∞, U ⊂ E, V ⊂ F be
open sets, h : U × V → F be continuous such that there exists K ⊂ U × V ,
K=
∞
[
Kp
p=1
with Kp compact sets for p ∈ N such that h is differentiable on (U × V ) \K, there exists ∂h
∂y
(x,
y)
∈
Isom
(F,
F
)
for
every
(x,
y)
∈
U
×
V
and
let
A
=
P
K.
Then,
on U × V and ∂h
r
1
∂y
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 that ϕ (a) = b and h (x, ϕ (x)) = h (a, b)
for every x ∈ 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., int K = φ and B\K is
connected and simply connected for every ball B from E. Also, since A is a countable union
of compact sets, we see that int A = φ.
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 100, 15 pp.
http://jipam.vu.edu.au/
A N OTE ON G LOBAL I MPLICIT F UNCTION T HEOREM
3
If E, F are Banach spaces and A ∈ L (E, F ), we let
kAk = sup kA (x)k
kxk=1
and
` (A) = inf kA (x)k
kxk=1
and if D ⊂ E, λ > 0, we let
λD = {x ∈ E| there exists y ∈ D such that x = λy}.
If X, Y are Banach spaces, D ⊂ X is open, x ∈ D and f : D → Y is a map, we let
D+ f (x) = lim sup
y→x
kf (y) − f (x)k
ky − xk
and we say that f is a light map if for every x ∈ D and every U ∈ V (x), there exists Q ∈ V (x)
such that Q ⊂ U and f (x) ∈
/ f (∂Q) .
Remark 3. We can replace in Theorem 1 and Theorem 2 the condition "dim F < ∞ " by
"There exists ∂h
on U × V and it is continuous on U × V and ∂h
(x, y) ∈ Isom (F, F ) for
∂y
∂y
every (x, y) ∈ U × V " to obtain the same conclusion, and this is the classical implicit function
theorem. Also, keeping the notations from Theorem 1 and Theorem 2, we see that if (α, β) ∈
B (a, r) × B (b, δ) is such that h (α, β) = h (a, b), then β = ϕ (α) .
We shall use the following lemma from [7].
Lemma 4. Let a > 0, f : [0, a] → [0, ∞) be continuous and let ω : [0, ∞) → [0, ∞) be
continuous such that ω > 0 on (0, ∞) and
Z c
|f (b) − f (c)| ≤
ω (f (t)) dt for every 0 < b < c ≤ a.
b
Then, if
m = inf f (t) ,
M = sup f (t) ,
t∈[0,a]
t∈[0,a]
it results that
Z
M
m
ds
≤ a.
ω (s)
We obtain now the following characterization of the boundary behaviour of the solutions of
some differential inequalities.
Theorem 5. Let E, F be Banach spaces, U ⊂ E a domain, K ⊂ U at most countable, ϕ :
U → F continuous on U and differentiable on U \K such that there exists ω : [0, ∞) → [0, ∞)
continuous with kϕ0 (x)k ≤ ω (kϕ (x)k) for every x ∈ U \K. Then, if α ∈ ∂U and C ⊂ U is
convex such that α ∈ C, either there exists
lim ϕ (x) = ` ∈ F or x→α
lim kϕ (x)k = ∞
x→α
x∈C
x∈C
or, if ω > 0 on (1, ∞) and
Z
∞
1
ds
= ∞,
ω (s)
there exists
lim ϕ (x) = ` ∈ F.
x→α
x∈C
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 100, 15 pp.
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4
M IHAI C RISTEA
If ω > 0 on (0, 1) and
Z
1
ds
=∞
0 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 some λ > 0, we can suppose that ω > 0 on
[0, ∞). Let α ∈ ∂U, C ⊂ U convex such that α ∈ C and let q : [0, 1) → C be a path such that
lim q (t) = α
t→1
and there exists L > 0 such that
Dq+
(t) ≤ L for every t ∈ [0, 1). Then
kq (s) − q (t)k ≤ L · (s − t)
for every 0 ≤ t < s < 1 and let 0 ≤ c < d < 1 be fixed,
A = co (q ([c, d]))
and ε > 0. Let g : A → R be defined by
g (z) = ω (kϕ (z)k) for every z ∈ A.
Then A is compact and convex and g is uniformly continuous on A, hence we can find δε0 > 0
such that
|g (z1 ) − g (z2 )| ≤ ε for z1 , z2 ∈ A
with kz1 − z2 k ≤ δε0 . Since q : [c, d] → C is uniformly continuous, we can find δε > 0 such
that kq (t) − q (s)k ≤ δε0 if s, t ∈ [c, d] are such that |s − t| ≤ δε . Let now
∆ = (c = t0 < t1 < · · · < tm = d) ∈ D ([c, d])
be such that k∆k ≤ δε . 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
≤
ϕ (q (tk+1 )) − ϕ (q (tk ))
k=0
≤
m−1
X
k(q (tk+1 ) − q (tk ))k ·
k=0
≤L·
m−1
X
(tl+1 − tk ) ·
kϕ0 (z)k
m−1
X
sup
ω (kϕ (z)k)
z∈[q(tk ),q(tk+1 )]
k=0
≤L·
sup
z∈[q(tk ),q(tk+1 )] \ K
(tl+1 − tk ) · (ω (kϕ(q (tk )k + ε) .
k=0
Letting k∆k → 0 and then ε → 0, we obtain
(1)
| kϕ (q (d))k − kϕ (q (c))k| ≤ kϕ (q (d)) − ϕ (q (c))k
Z d
≤L·
ω (kϕ (q (t))k) dt for 0 ≤ c < d < 1.
c
If
m = inf kϕ (q (t))k ,
t∈[0,1)
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 100, 15 pp.
M = sup kϕ (q (t))k ,
t∈[0,1)
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A N OTE ON G LOBAL I MPLICIT F UNCTION T HEOREM
5
we obtain from Lemma 4 and (1)
Z
M
(2)
m
ds
≤ L.
ω (s)
Let now zp → α be such that
1
, zp ∈ C for p ∈ N
2P
and suppose that there exists ρ > 0 such that kϕ (zp )k ≤ ρ for every p ∈ N. We take
kzp − αk ≤
0 = t0 < t1 < · · · < tk < tk+1 < · · · < 1
such that tk % 1 and we define q : [ 0, 1) → C by
q (t) =
Then
zk (tk+1 − t) + zk+1 (t − tk )
for t ∈ [tk , tk+1 ] ,
tk+1 − tk
k ∈ N.
kzk+1 − zk k
for t ∈ [tk , tk+1 ]
tk+1 − tk
for k ∈ N, we see that ck → 0. Then
D+ q (t) = ck =
and taking tk =
k
k+1
αp = sup D+ q (t) = sup ck → 0
k≥p
t∈[tp ,1)
and let
ap = inf kϕ (q (t))k ,
bp = sup kϕ (q (t))k for p ∈ N.
t∈[tp ,1)
t∈[tp ,1)
Using (2) we obtain that
Z
bp
ap
and let p0 ∈ N be such that
ds
≤ αp for p ∈ N
ω (s)
Z
∞
ds
for p ≥ p0 .
ω (s)
ρ
Suppose that there exists p ≥ p0 such that bp = ∞. Then, since q (tk ) = zk , we see that
αp <
ak ≤ kϕ (q (tk ))k = kϕ (zk )k ≤ ρ for k ∈ N,
hence
Z bp
Z ∞
ds
ds
ds
0<
≤
≤ αp <
ω (s)
ω (s)
ρ
ap ω (s)
ρ
and we have reached a contradiction.
It results that bp < ∞ for p ≥ p0 and let
Z
∞
Kp = sup ω (t)
t∈[0,bp ]
for p ≥ p0 . Then Kp < ∞ and we see from (1) that
kϕ (q (d)) − ϕ (q (c))k ≤ αp · Kp · |d − c|
for tp ≤ c < d < 1 and p ≥ p0 , and this implies that
lim ϕ (q (t)) = ` ∈ F.
t→1
It results that
lim ϕ (zp ) = lim ϕ (q (tp )) = ` ∈ F.
p→∞
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 100, 15 pp.
p→∞
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6
M IHAI C RISTEA
Now, if the case
lim kϕ (x)k = ∞
x→α
x∈C
does not hold, there exists ρ > 0 and xp → α, xp ∈ C with
1
2p
for every p ∈ N, and from what we have proved before, it results that
||ϕ(xp )|| ≤ ρ and ||xp − α|| ≤
lim ϕ (xp ) = ` ∈ F.
p→∞
If ap ∈ C, kap − αk ≤
1
2p
for every p ∈ N, then
lim ϕ (ap ) = `1 ∈ F.
p→∞
Let z2p = xp , z2p+1 = ap for p ∈ N. We see that
lim ϕ (zp ) = `2 ∈ F,
p→∞
hence
` = lim ϕ (xp ) = lim ϕ (z2p ) = `2
p→∞
p→∞
and
`1 = lim ϕ (ap ) = lim ϕ (z2p+1 ) = `2 ,
p→∞
p→∞
hence ` = `1 = `2 . We have proved that if ap ∈ C, kap − αk ≤
1
2p
for every p ∈ N, then
lim ϕ (ap ) = `.
p→∞
We show now that if ap ∈ C, ap → α, then
lim ϕ (ap ) = `.
p→∞
Indeed, if this is false, there exists ε > 0 and (apk )k∈N such that
kϕ (apk ) − `k > ε
for every k ∈ N. Let apkq
be a subsequence such that
q∈N
1
apkq − α < q
2
for every q ∈ N. From what we have proved before it results that
lim ϕ apkq = `
q→∞
and we have reached a contradiction.
We have therefore proved that either
lim ϕ (x) = ` ∈ F,
x→α
x∈C
or
lim kϕ(x)k = ∞.
x→α
x∈C
Suppose now
Z
∞
1
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 100, 15 pp.
ds
=∞
ω (s)
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A N OTE ON G LOBAL I MPLICIT F UNCTION T HEOREM
7
and let α ∈ ∂U . We take x ∈ C and let q : [0, 1) → C be defined by
q (t) = (1 − t) x + tα
for t ∈ [0, 1) and
m = inf kϕ (q (t))k ,
M = sup kϕ (q (t))k .
t∈[0,1)
t∈[0,1)
+
Since D q (t) = kx − αk for every t ∈ (0, 1), we see from (2)
Z M
Z M
ds
ds
≤
≤ kx − αk
m ω (s)
kxk ω (s)
and this implies that M < ∞. Let
b = sup ω (t) .
t∈[0,M ]
Then b < ∞ and using (1), we see that
kϕ (q (d)) − ϕ (q (c))k ≤ b · kx − αk · |d − c|
for 0 ≤ c < d < 1 and this implies that
lim ϕ (x) = ` ∈ F.
x→α
x∈C
It results that the case
lim kϕ (x)k = ∞
x→α
x∈C
cannot hold, hence
lim ϕ (x) = ` ∈ F.
x→α
Suppose now that
1
Z
ds
=∞
0 ω (s)
and that there exists α ∈ U such that ϕ (α) = 0. Let r = d (α, ∂U ) , y ∈ B (α, r) and
q : [0, 1] → B (α, r) , q (t) = (1 − t) α + ty for t ∈ [0, 1]. Then D+ q (t) = ky − αk for
t ∈ [0, 1] and let
m = inf kϕ (q (t))k
t∈[0,1]
and
M = sup kϕ (q (t))k .
t∈[0,1]
Then m = 0 and we see from (2) that
Z
0
M
ds
≤ ky − αk .
ω (s)
This implies that M = 0 and hence ϕ (y) = 0. We proved that ϕ ≡ 0 on B (α, r) and since U
is a domain, we see that ϕ ≡ 0 on U.
Remark 6. We proved that if ϕ is as in Theorem 2 and
Z ∞
ds
= ∞,
ω (s)
1
then ϕ has angular limits in every point α ∈ ∂U.
We now obtain the following characterization of the boundary behaviour of some implicit
function.
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 100, 15 pp.
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8
M IHAI C RISTEA
Theorem 7. Let E, F be Banach spaces, U ⊂ E a domain, K ⊂ U × F such that A = Pr1 K
is at most countable and let h : U × F → F be continuous on U × F , differentiable on
(U × F ) \K such that
∂h
`
(x, y) > 0 on (U × F ) \K
∂y
and there exists ω : [0, ∞) → [0, ∞) continuous such that
∂h
∂h
∂x (x, y) ` ∂y (x, y) ≤ ω (kyk)
for every (x, y) ∈ (U × F ) \K. Suppose that ϕ : U → F is continuous on U , differentiable on
U \A, ϕ (a) = b and
h (x, ϕ (x)) = h (a, b)
for every x ∈ U.
Then, if α ∈ ∂U and C ⊂ U is convex such that α ∈ C, either
lim kϕ (x)k = ` ∈ F,
x→α
x∈C
or
lim kϕ (x)k = ∞.
x→α
x∈C
Also, if ω > 0 on (1, ∞) and
Z
∞
1
ds
= ∞,
ω (s)
then
lim ϕ (x) = ` ∈ F.
x→α
x∈C
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
∂h
(x, ϕ (x)) +
(x, ϕ (x)) · ϕ0 (x) = 0
∂x
∂y
and we see that
0
kϕ (x)k · `
∂h
∂h
∂h
0
(x, ϕ (x)) ≤ (x, ϕ (x)) (ϕ (x))
=
(x, ϕ (x))
.
∂y
∂y
∂x
It results that
∂h
∂h
0
kϕ (x)k ≤ (x, ϕ (x)) `
(x, ϕ (x)) ≤ ω (kϕ (x)k)
∂x
∂y
for every x ∈ U \ A and we now apply Theorem 5.
Remark 8. If E is an infinite dimensional Banach space and
K=
∞
[
Kp with Kp ⊂ E
p=1
are compact sets for every p ∈ N and y ∈ E, then the set
M (K, y) = {w ∈ E| there exists t > 0 and x ∈ K such that w = tx}
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 100, 15 pp.
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A N OTE ON G LOBAL I MPLICIT F UNCTION T HEOREM
9
is also a countable union of compact sets and hence a "thin" set. Keeping the notations from
Theorem 5, we see that the basic inequality
(3)
kϕ (z1 ) − ϕ (z2 )k ≤ sup ω (kϕ (z)k)
if [z1 , z2 ] ⊂ U
z∈[z1 ,z2 ]
is also valid for K a countable union of compact sets and ϕ as in Theorem 5.
S
If dim E = n and K ⊂ E has a σ-finite (n − 1)-dimensional measure (i.e. K = ∞
p=1 Kp ,
with mn−1 (Kp ) < ∞ for every p ∈ N, where mq is the q-Hausdorff measure from Rn ), a
theorem of Gross shows that if H ⊂ E is a hyperplane and P : E → H is the projection on
H, then P −1 (z) ∩ K is at most countable with the possible exception of a set B ⊂ H, with
mn−1 (B) = 0. Applying as in Theorem 5 the theorem of Denjoi and Bourbaki on each interval
from P −1 (z) ∩ K for every z ∈ H \ B and using a natural limiting process, we see that if
dim E = n and K ⊂ E has a σ-finite (n − 1)-dimensional measure, then the inequality (3)
also holds. It is easy see now that Theorem 5 and Theorem 7 hold if the set K, respectively
the set A = Pr1 K are chosen to be a countable union of compact sets if dim E = ∞ and
having σ-finite (n − 1)-dimensional measure if dim E = 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 domain D from a Banach space is starlike with respect to the point a ∈ D if
[a, x] ⊂ D for every x ∈ D, and if D is a domain in the Banach space E and a ∈ D. We set
Da = {x ∈ D| [a, x] ⊂ D} .
Theorem 9. Let E, F be Banach spaces, dim F < ∞, D ⊂ E a domain, K ⊂ D × F at most
countable, and A = Pr1 K.Also, let h : D × F → F be continuous on D × F and differentiable
on (D × F ) \K such that
∂h
`
(x, y) > 0 on (D × F ) \K.
∂y
In addition, there exists ω : [0, ∞) → [0, ∞) continuous such that ω > 0 on (0, ∞) and
∂h
∂h
∂x (x, y) ` ∂y (x, y) ≤ ω (kyk)
for every (x, y) ∈ (D × F ) \K. Then, if (a, b) ∈ D × F and
Z ∞
ds
Qa,b = Da ∩ B a,
,
kbk ω (s)
there exists a unique continuous map ϕ : Qa,b → F , differentiable on Qa,b \A such that
h (x, ϕ (x)) = h (a, b) for every x ∈ Qa,b . If D is starlike with respect to a and
Z ∞
ds
= ∞,
ω (s)
1
then Qa,b = D and ϕ : D → F is globally defined on D.
Proof. Let z ∈ Qa,b and let B = {x ∈ [a, z]| there exists an open, convex domain Dx ⊂ Qa,b
such that [a, x] ⊂ Dx } and a continuous implicit function ϕx : Dx → F, differentiable on
Dx \A such that ϕx (a) = b and h (u, ϕx (u)) = h (a, b) for every u ∈ Dx . We see that B is
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 100, 15 pp.
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10
M IHAI C RISTEA
open, and from Theorem 1, B 6= ∅. We show that B is a closed set. Let xp ∈ B, xp → x, and
we can suppose that xp ∈ [a, x) for every p ∈ N, and let
∞
[
Q=
Dxp .
p=1
We define Ψ : Q → F by Ψ (u) = ϕxp (u) for u ∈ Dxp and p ∈ N and the definition is correct.
Indeed, if p, q ∈ N, p 6= q let Upq = Dxp ∩ Dxq . Then Upq is a nonempty, open and convex set,
hence it is a nonempty domain. If
Vpq = u ∈ Upq |ϕxq (u) = ϕxp (u) ,
we see that a ∈ Vpq , hence Vpq 6= ∅ and we see that Vpq is a closed set in Upq . Using the property
of the local unicity of the implicit function from Theorem 1, we obtain that Vpq is also an open
set. Since Upq is a domain, it results that Upq = Vpq and hence that Ψ is correctly defined. We
also see immediately that Ψ (a) = b, and Ψ is continuous on Q and differentiable on Q \ A.
Let q : [0, 1] → Q be defined by
q (t) = (1 − t) a + tx for t ∈ [0, 1] .
Then
Z
+
∞
D q (t) = kx − ak <
||b|
ds
.
ω (s)
Let
m = inf kΨ (q (t))k ,
M = sup kΨ (q (t))k .
t∈[0,1)
t∈[0,1)
As in Theorem 7, we see that
kΨ0 (u)k ≤ ω (kΨ (u)k) for every u ∈ Q \A,
hence
kΨ (z1 ) − Ψ (z2 )k ≤ sup ω (kΨ (z)k)
z∈[z1 ,z2 ]
if [z1 , z2 ] ⊂ Q. This implies that relations (1) and (2) from Theorem 5 also hold and we see that
Z M
ds
≤ kx − ak .
m ω (s)
Then
Z
M
ds
≤
kbk ω (s)
and this implies that M < ∞, hence
Z
M
m
ds
≤ kx − ak
ω (s)
` = sup ω (t) < ∞.
t∈[0,M ]
Using (1) and Theorem 5, we see that
kΨ (q (d)) − Ψ (q (c))k ≤ ` · kx − ak · |d − c| for every 0 ≤ c < d < 1
and this implies that
lim Ψ (u) = w ∈ F.
u→x
u∈Im q
Using Theorem 1, we can find r, δ > 0 and a unique continuous implicit function Ψx :
B (x, r) → B (w, δ), differentiable on B (x, r) \ A such that
Ψx (x) = w and h (u, Ψx (u)) = h (a, b)
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 100, 15 pp.
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A N OTE ON G LOBAL I MPLICIT F UNCTION T HEOREM
11
for every u ∈ B (x, r). Let 0 < ε < r and pε ∈ N be such that
kxp − xk < ε and kΨ (xp ) − wk < δ
for p ≥ pε and let p ≥ pε be fixed. Since
ϕxp (xp ) = Ψ (xp ) ∈ B (w, δ) ,
we see from Remark 3 that ϕxp (xp ) = Ψx (xp ) and hence the set
U = u ∈ Dxp ∩ B (x, ε) |ϕxp (u) = Ψx (u)
is nonempty. We also see that Dxp ∩ B (x, ε) is an open, nonempty, convex set, hence it is a
domain and U is an open, closed and nonempty subset of Dxp ∩ B (x, ε), and this implies that
U = Dxp ∩ B (x, ε) .
Let U0 = Dxp ∪ B (x, ε). We can now correctly define Φ : U0 → F by
Φ (u) = ϕxp (u) if u ∈ Dxp
and
Φ (u) = Ψx (u) if u ∈ 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 every u ∈ U0 . It results that x ∈ B, hence B is also a closed set and since [a, z] is a
connected set, we see that B = [a, z] .
We have therefore proved that for every z ∈ Qa,b there exists a convex domain Dz such that
[a, z] ⊂ Dz and a unique continuous implicit function ϕz : Dz → F , differentiable on Dz \ A
such that
ϕz (a) = b, h (u, ϕz (u)) = h (a, b) for every u ∈ Dz .
We now define ϕ : Qa,b → F by ϕ (x) = ϕz (x) for x ∈ Dz and we see, as before, that the
definition is correct, that ϕ (a) = b, ϕ is continuous on Qa,b , differentiable on Qa,b \ A and
h (x, ϕ (x)) = h (a, b) for every x ∈ Qa,b .
Remark 10. The result from Theorem 9 extends a global implicit function theorem from [3].
The result from [3] also involves an inequality containing
∂h
∂h
∂x (x, y) and ` ∂y (x, y)
(x, y) ∈
and it says that if E, F are Banach spaces, h : E × F → F is a C 1 map such that ∂h
∂y
Isom (F, F ) for every (x, y) ∈ E × F and there exists ω : [0, ∞) → (0, ∞) continuous such
that
∂h
∂h
1+
`
(x, y) ≤ ω (max (kxk , kyk))
∂x (x, y)
∂y
for every (x, y) ∈ E × F , then, for (x0, y0 ) ∈ E × F, z0 = h (x0 , y0 ) and
Z ∞
ds
r=
,
max(kx0 k,ky0 k) 1 + ω (s)
there exists a C 1 map ϕ : 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 map h is defined on a proper subset of E × F, namely,
on a set D × F ⊂ E × F , where D ⊂ E is an open starlike domain.
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 100, 15 pp.
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12
M IHAI C RISTEA
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 C 1 map such that f 0 (x) ∈ Isom (E, F ) for every x ∈ E
and there exists ω : [0, ∞) → (0, ∞) continuous such that
Z ∞
ds
= ∞ and f 0 (x)−1 ≤ ω (kxk)
ω (s)
1
for every x ∈ E, then it results that f : E → F is a C 1 diffeomorphism (see [11], [14],
[12], [3], [7]). If E = F = Rn or if dim E = ∞ and f = I − T with T compact, we can
drop the continuity of the derivative on E and we can impose the essential condition "f 0 (x) ∈
Isom (F, F )" with the possible exception of a "thin" set (see [4], [5],[6]) and we will still obtain
that f : E → F is a homeomorphism.
Now, let E, 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 f 0 (y) ∈ Isom (F, F ) for every
y ∈ F and there exists ω : [0, ∞) → (0, ∞) continuous such that
0 −1 f (y) ≤ ω (kyk) for every y ∈ F.
Let h : D × F → F be defined by h (x, y) = f (y) − g (x) for x ∈ D, y ∈ F , r0 = d (a, ∂D)
and suppose that
Mr = sup kg 0 (x)k < ∞ for every 0 < r ≤ r0 .
x∈B(a,r)
Then
∂h
0 −1 ∂h
0
≤ Mr · ω (kyk)
∂x (x, y) ` ∂y (x, y) ≤ kg (x)k · f (y)
if (x, y) ∈ B (a, r) × F and 0 < r ≤ r0 and let
Z ∞
1
ds
δr = min r,
·
for 0 < r ≤ r0 .
Mr kbk ω (s)
Using Theorem 9, we see that there exists a unique differentiable map ϕ : B (a, δr ) → F such
that ϕ (a) = b and h (x, ϕ (x)) = h (a, b) for every x ∈ B (a, δr ), i.e. f (ϕ (x)) = g (x)+h (a, b)
for every x ∈ B (a, r) and every 0 < r ≤ r0 .
If
Z ∞
ds
r0 · Mr 0 <
,
kbk ω (s)
then ϕ is defined on B (a, r0 ). Additionally, if D = B (a, r0 ), then ϕ is globally defined on D
and f ◦ ϕ = g + h (a, b) on D. In the special case D = E, g (x) = x for every x ∈ E,
Z ∞
ds
= ∞ and b = f (a) ,
ω (s)
1
then f (ϕ (x)) = x for every x ∈ E, and ϕ is defined on E and is the inverse of f and it
results that f : F → F is a homeomorphism. In this way we obtain an alternative proof of the
Hadamard-Lévy-John theorem.
Remark 11. 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 inequality kϕ0 (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 ) .
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 100, 15 pp.
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A N OTE ON G LOBAL I MPLICIT F UNCTION T HEOREM
13
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 Theorem 9.
Theorem 12. Let E, F be Banach spaces, dim E = ∞, dim F < ∞, D ⊂ E a domain,
K ⊂ D × F,
∞
[
K=
Kp
p=1
with Kp compact sets for every p ∈ N, A = Pr1 K, h : D × F → F continuous on D × F and
differentiable on (D × F ) \K such that
∂h
(x, y) > 0 on (D × F ) \K,
`
∂y
and there exists ω : [0, ∞) → [0, ∞) continuous such that ω > 0 on (0, ∞) and
∂h
∂h
∂x (x, y) ` ∂y (x, y) ≤ ω (kyk)
for every (x, y) ∈ (D × F ) \K. Suppose that the map y → h (x, y) is a light map on F for
every x ∈ D. Then, if a, b ∈ D × F and
Z ∞
ds
Qa,b = Da ∩ B a,
,
kbk ω (s)
there exists a unique continuous implicit function ϕ : Qa,b → F , differentiable on Qa,b \A such
that ϕ (a) = b and h (x, ϕ (x)) = h (a, b) for every x ∈ Qa,b and if D is starlike with respect to
a and
Z ∞
ds
= ∞,
ω (s)
1
then Qa,b = D and ϕ : D → F is globally defined on D.
Remark 13. The condition "the map y → h (x, y) is a light map on F for every x ∈ D" is
satisfied if ∂h
exists on D × F and
∂y
∂h
`
(x, y) > 0 for every (x, y) ∈ D × F.
∂y
Using the conditions of local existence and local unicity from Theorem 7 of [8], we obtain
the following global implicit function theorem.
Theorem 14. Let n ≥ 2, D ⊂ Rn be a domain, h : D × Rm → Rm be differentiable and let
K ⊂ D × Rm ,
∞
[
K=
Kp
p=1
with Kp closed sets such that mn−2 (Pr1 Kp ) = 0 for every p ∈ N, A = Pr1 K, such that
∂h
(x, y) ∈ Isom (Rm , Rm ) for every (x, y) ∈ (D × Rm ) \K and the map y → h (x, y) is a
∂y
light map on Rm for every x ∈ D. Suppose that there exists ω : [0, ∞) → [0, ∞) continuous
such that ω > 0 on (0, ∞) and
∂h
∂h
`
(x, y) ≤ ω (kyk) for every (x, y) ∈ (D × F ) \K.
∂x (x, y)
∂y
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 100, 15 pp.
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14
M IHAI C RISTEA
Then, if (a, b) ∈ D × F and
Qa,b
Z
= Da ∩ B a,
∞
kbk
ds
ω (s)
,
there exists a unique continuous implicit function ϕ : Qa,b → F , differentiable on Qa,b \A such
that ϕ (a) = b and h (x, ϕ (x)) = h (a, b) for every x ∈ Qa,b , and if D is starlike with respect to
a and
Z ∞
ds
= ∞,
ω (s)
1
then Qa,b = D and ϕ : D → F is globally defined on D.
Proof. We see that mm+n−2 (Kp ) = 0 for every p ∈ N and A has σ-finite (n − 1)-dimensional
measure. We now apply the local implicit function theorem from Theorem 7 of [8], Remark 8
and the preceding arguments.
Using the classical implicit function theorem, we obtain the following global implicit function theorem
Theorem 15. Let E, F be Banach spaces, D ⊂ E a domain, h : D × F → F be continuous
such that ∂h
exists on D × F , it is continuous on D × F and
∂y
∂h
`
(x, y) > 0 for every (x, y) ∈ D × F.
∂y
Also, let K ⊂ D×F be such that A = Pr1 K is a countable union of compact sets if dim E = ∞
and has σ-finite (n − 1)-dimensional measure if dim E = n. Additionally, suppose that h is
differentiable on (D × F ) \K and there exists ω : [0, ∞) → [0, ∞) continuous such that ω > 0
on (0, ∞) and
∂h
∂h
`
(x, y) ≤ ω (kyk)
∂x (x, y)
∂y
for every (x, y) ∈ (D × F ) \K. Then, if (a, b) ∈ D × F and
Z ∞
ds
Qa,b = Da ∩ B a,
,
kbk ω (s)
there exists a unique continuous implicit function ϕ : Qa.b → F, differentiable on Qa,b \A such
that ϕ (a) = b and h (x, ϕ (x)) = h (a, b) for every x ∈ Qa,b . If D is starlike with respect to a
and
Z ∞
ds
= ∞,
ω (s)
1
then Qa,b = D and ϕ : D → F is globally defined.
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