On a diffeomorphism between two Banach spaces

On a di¤eomorphism between two Banach spaces - a variational approach Marek Galewski LUT, Institute of Mathematics

June 29, 2017

M. G. (LUT, Institute of Mathematics)

Di¤eomorphisms via MPT

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Main aims of the talk

Provide su¢ cient conditions for a C 1 mapping f : X ! Z to be a di¤eomorphism using local di¤eomorphism theorem and mountain pass theorem;



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Provide su¢ cient conditions for a C 1 mapping f : X ! Z to be a di¤eomorphism using local di¤eomorphism theorem and mountain pass theorem; Examine whether the assumption that f is C 1 can be weakened; …nite dimensional case and non-smooth critical point theory methods;



June 29, 2017

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Provide su¢ cient conditions for a C 1 mapping f : X ! Z to be a di¤eomorphism using local di¤eomorphism theorem and mountain pass theorem; Examine whether the assumption that f is C 1 can be weakened; …nite dimensional case and non-smooth critical point theory methods; Examine what happens when f is only locally Lipschitz continuous describe problems which appear here in the context of non-smooth di¤erentiation; …nite dimensional case.



June 29, 2017

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Provide su¢ cient conditions for a C 1 mapping f : X ! Z to be a di¤eomorphism using local di¤eomorphism theorem and mountain pass theorem; Examine whether the assumption that f is C 1 can be weakened; …nite dimensional case and non-smooth critical point theory methods; Examine what happens when f is only locally Lipschitz continuous describe problems which appear here in the context of non-smooth di¤erentiation; …nite dimensional case. Provide outline of the applications and state the mains di¢ culties appearing there



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Existing results 1 Theorem (D.I. A.S. S.W. Adv. Nonlinear Stud. 12, No. 1, 89-100 (2012).)Let X be a real Banach space, H - a real Hilbert space. If f : X ! H is a C 1 -mapping such that (i) for any y 2 H the functional ϕ : X ! R given by the formula ϕ (x ) =

1 kf (x ) 2

y k2 =

1 hf (x ) 2

y , f (x )

yi

satis…es the Palais-Smale condition, 0 (ii) for any x 2 X , f (x )X = H and there exists a constant αx > 0 such that 0 f (x )h αx kh k then f is a di¤eomorphism. M. G. (LUT, Institute of Mathematics)


June 29, 2017

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Existing results 2

Theorem (D.I. Discrete Contin. Dyn. Syst. Ser. B, 19 (8)

(2014)). Let X , Y be a

real Banach space, H - a real Hilbert space. Assume that F : X Y ! H is a C 1 -mapping such that (i) for any y 2 H the functional ϕ : X ! R given by the formula ϕ (x ) =

1 kF (x, y )k2 2

satis…es the P-S condition, 0 (ii) for any (x, y ) 2 X Y the di¤erential Fx (x, y ) : X ! H is a bijection. Then there exists a unique function f : Y ! X such that equations F (x, y ) = 0 and x = f (y ) are equivalent on X Y .



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Main ideas and problems

replace a Hilbert space with a Banach space - to extend applicability



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replace a Hilbert space with a Banach space - to extend applicability use weak version of MPT - in order to relax some proof steps



June 29, 2017

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replace a Hilbert space with a Banach space - to extend applicability use weak version of MPT - in order to relax some proof steps …nd a suitable formula for ϕ - so that the arguments from existing sources work



June 29, 2017

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Preliminary material. P-S condition X (re‡exive) Banach space J : X ! R satis…es the Palais-Smale condition if every sequence fun g such that fJ (un )g is bounded and J 0 (un ) ! 0, has a convergent subsequence;



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Preliminary material. P-S condition X (re‡exive) Banach space J : X ! R satis…es the Palais-Smale condition if every sequence fun g such that fJ (un )g is bounded and J 0 (un ) ! 0, has a convergent subsequence; a C 1 functional bounded from below and satisfying the P-S condition has an argument of a minimum-valid also when derivative replaced with Clarke derivative with genuinely same proof;



June 29, 2017

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Preliminary material. P-S condition X (re‡exive) Banach space J : X ! R satis…es the Palais-Smale condition if every sequence fun g such that fJ (un )g is bounded and J 0 (un ) ! 0, has a convergent subsequence; a C 1 functional bounded from below and satisfying the P-S condition has an argument of a minimum-valid also when derivative replaced with Clarke derivative with genuinely same proof; a C 1 functional bounded from below and satisfying the P-S condition is coercive - converse true only in a …nite dimensional setting



June 29, 2017

6 / 28

Preliminary material. P-S condition X (re‡exive) Banach space J : X ! R satis…es the Palais-Smale condition if every sequence fun g such that fJ (un )g is bounded and J 0 (un ) ! 0, has a convergent subsequence; a C 1 functional bounded from below and satisfying the P-S condition has an argument of a minimum-valid also when derivative replaced with Clarke derivative with genuinely same proof; a C 1 functional bounded from below and satisfying the P-S condition is coercive - converse true only in a …nite dimensional setting a coercive and w.l.s.c. functional has an argument of a minimum;



June 29, 2017

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Preliminary material. P-S condition X (re‡exive) Banach space J : X ! R satis…es the Palais-Smale condition if every sequence fun g such that fJ (un )g is bounded and J 0 (un ) ! 0, has a convergent subsequence; a C 1 functional bounded from below and satisfying the P-S condition has an argument of a minimum-valid also when derivative replaced with Clarke derivative with genuinely same proof; a C 1 functional bounded from below and satisfying the P-S condition is coercive - converse true only in a …nite dimensional setting a coercive and w.l.s.c. functional has an argument of a minimum; examples: f (x ) = x 2 + x 4 Yes and f (x ) = exp f (x ) = exp (x ) No



x2 ,

June 29, 2017

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Preliminary material. P-S condition X (re‡exive) Banach space J : X ! R satis…es the Palais-Smale condition if every sequence fun g such that fJ (un )g is bounded and J 0 (un ) ! 0, has a convergent subsequence; a C 1 functional bounded from below and satisfying the P-S condition has an argument of a minimum-valid also when derivative replaced with Clarke derivative with genuinely same proof; a C 1 functional bounded from below and satisfying the P-S condition is coercive - converse true only in a …nite dimensional setting a coercive and w.l.s.c. functional has an argument of a minimum; examples: f (x ) = x 2 + x 4 Yes and f (x ) = exp x 2 , f (x ) = exp (x ) No R1 R1 H01 (0, 1), J (x ) = 12 0 x˙ 2 (t ) dt 14 0 x 4 (t ) dt satis…es PS and is not coercive M. G. (LUT, Institute of Mathematics)


June 29, 2017

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Preliminary material- MPT

Lemma (Version Solimini- de Figuereido, 1984) Let X be a Banach space and J 2 C 1 (X , R) satisfy the P-S condition. Assume that inf J (x )

kx k=r

maxfJ (0), J (e )g,

where 0 < r < ke k and e 2 X . Then J has a non-zero critical point x0 and J (x0 ) maxfJ (0), J (e )g. (G. Bonanno, 2014). Additionally, there is another critical point (not necessarily non-trivial) which is an argument of a local minimum. (no localization).



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Di¤eomorphism Theorem - …rst (hammer) version Theorem (M.G., E.G., E.S., Electron. J. Di¤. Equ., 99 (2014)) X , B real B-spaces; 1 1 f 2 C (X , B ), η 2 C (B, R+ ), η (0) = 0 () x = 0 and 0 0 η (0) = 0 () x = 0, η : B ! R has a cont. inverse. f is a di¤eomorphism if (i)ϕ : X ! R, ϕ (x ) = η (f (x ) y ) satis…es the P-S cond. for y 2 B 0 (ii) for any x 2 X , f (x )X = B and 0

f (x )h

αx khk for some αx > 0

(iii) there exist α, c, M > 0 such that η (x )


c kx kα for kx k


M

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Di¤eomorphism Theorem - second (more re…ned) version Theorem (M.G. M.K. Stud. Sci. Math. Hung., 52

(2015)) Let X , B be B-spaces.

Let the duality mapping A : B ! B be such that its potential u ! is C 1 . If f : X ! B is a C 1 -mapping such that: 1

ku k p p

for any y 2 B the functional ϕ : X ! R ϕ (x ) =

1 kf (x ) p

y kp

satis…es the P-S condition; 2

for any x 2 X , f 0 (x )X = B, there exists αx > 0 such that f 0 (x )h

αx kh k ,

(1)

then f is a di¤eomorphism. M. G. (LUT, Institute of Mathematics)


June 29, 2017

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On a duality mapping Let p > 1. A duality mapping on X corresponding to ϕ (t ) = t p operator A : X ! 2X such that for all u 2 X , y 2 A (u )

ky k = ku kp

1

1

is an

, hy , u i = ky k ku k .

Lemma X is a re‡exive B-space with a strictly convex dual X . Then A is single u p valued and u ! kp k is Gâteaux di¤erentiable with A being its derivative. u!

ku k p p

=

R ku k 0


ϕ (t ) dt has a "convex subdi¤erential"


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ky k = ku kp

1

1

is an


Lemma X is a re‡exive B-space with a strictly convex dual X . Then A is single u p valued and u ! kp k is Gâteaux di¤erentiable with A being its derivative.

R ku k = 0 ϕ (t ) dt has a "convex subdi¤erential" X strictly convex means A is single valued u!

ku k p p



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ky k = ku kp

1

1

is an



R ku k = 0 ϕ (t ) dt has a "convex subdi¤erential" X strictly convex means A is single valued A : X ! X is demicontinous (norm to weak) u!

ku k p p



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ky k = ku kp

1

1

is an



R ku k = 0 ϕ (t ) dt has a "convex subdi¤erential" X strictly convex means A is single valued A : X ! X is demicontinous (norm to weak) u!

ku k p p

since A demincontinuous, u ! M. G. (LUT, Institute of Mathematics)

ku k p p

is di¤erentiable


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Duality mapping- example ˜ 1,p ([0, 1], E) contains AC fncs x : [0, 1] ! Rn , x (0) = 0 and W 0 x 0 2 Lp ([0, 1] , E);E smooth, uniformly convex, separable



June 29, 2017

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Duality mapping- example ˜ 1,p ([0, 1], E) contains AC fncs x : [0, 1] ! Rn , x (0) = 0 and W 0 x 0 2 Lp ([0, 1] , E);E smooth, uniformly convex, separable ˜ 1,p ([0, 1], E) uniformly convex, with strictly convex dual W 0



June 29, 2017

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Duality mapping- example ˜ 1,p ([0, 1], E) contains AC fncs x : [0, 1] ! Rn , x (0) = 0 and W 0 x 0 2 Lp ([0, 1] , E);E smooth, uniformly convex, separable ˜ 1,p ([0, 1], E) uniformly convex, with strictly convex dual W 0 ˜ W 1,p is considered with a usual norm 0

jjx jjW˜ 1,p = ( 0


Z1 0

x 0 (t )

p

˜ 1,p . dt )1/p , x 2 W 0


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jjx jjW˜ 1,p = ( 0

ϕ:

˜ 1,p W 0

Z1

x 0 (t )

0

p

˜ 1,p . dt )1/p , x 2 W 0

! R is continuously di¤erentiable if ϕ (u ) =


Z 1 0

u 0 (t )


p

dt

(2)

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jjx jjW˜ 1,p = ( 0

ϕ:

˜ 1,p W 0

Z1

x 0 (t )

p

0

˜ 1,p . dt )1/p , x 2 W 0

! R is continuously di¤erentiable if ϕ (u ) =

Z 1 0

u 0 (t )

˜ 1,p ! W ˜ 1,p A duality mapping A : W 0 0

hAu, v i = M. G. (LUT, Institute of Mathematics)

Z 1 0

u 0 (t )

p 2


p

dt

(2)

such that u 0 (t )v 0 (t )dt June 29, 2017

11 / 28

Sketch of the proof

1

Main aim: show that f is onto and one-one - local di¤eomorphism holds since f 0 (x )X = B and there id αx > 0 s.t. f 0 (x )h


αx kh k ,


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Sketch of the proof

1


αx kh k ,

2

ϕ (x ) =

1 kf (x ) p

y kp

is C 1 as a composition of C 1 mappings and has a (global) minimizer since it is bounded below and satis…es the P-S



June 29, 2017

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Sketch of the proof

1


αx kh k ,

2

ϕ (x ) =

1 kf (x ) p

y kp

is C 1 as a composition of C 1 mappings and has a (global) minimizer since it is bounded below and satis…es the P-S 3

if the minimizer is unique, we are done



June 29, 2017

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Sketch of the proof

1


αx kh k ,

2

ϕ (x ) =

1 kf (x ) p

y kp

is C 1 as a composition of C 1 mappings and has a (global) minimizer since it is bounded below and satis…es the P-S 3

if the minimizer is unique, we are done

4

if not, argue by contradiction using MPT



June 29, 2017

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Contradiction the argmin x which satis…es the Fermat’s rule 0

0 = ϕ (x ) = A (f (x )



y ) f 0 (x ).

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y ) f 0 (x ).

0 = ϕ (x ) = A (f (x )

y ) = 0; kA (u )k = ku kp

f 0 (x ) is invertible we see that A (f (x )

kA (f (x )

y )k = kf (x )

f (x )

y = 0.

y kp

1

1

.

Thus f is surjective. e = x1

x2

g (x ) = f (x + x2 ) ψ (x ) =


1 kf (x + x2 ) p


a, a kp

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y ) f 0 (x ).

0 = ϕ (x ) = A (f (x )

y ) = 0; kA (u )k = ku kp

f 0 (x ) is invertible we see that A (f (x )

kA (f (x )

y )k = kf (x )

f (x )

y = 0.

y kp

1

1

.

Thus f is surjective. e = x1

x2

g (x ) = f (x + x2 )

a,

1 kf (x + x2 ) akp p There is a number ρ > 0, ρ < min fke k , M g such that 1 αx kx k kg (x )k for x 2 B (0, ρ). 2 ψ (x ) =



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Mountain geometry. Conclusion ψ satis…es the P-S condition



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Mountain geometry. Conclusion ψ satis…es the P-S condition ψ (e ) = ψ (0) = 0



June 29, 2017

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Mountain geometry. Conclusion ψ satis…es the P-S condition ψ (e ) = ψ (0) = 0 take any 0 < r < ρ and observe inf ψ(x )

kx k=r


1 p

1 αx r 2


p

>0

June 29, 2017

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kx k=r

1 p

1 αx r 2

p

>0

ψ has a critical point 0

v 6= 0 and ψ (v ) = A(f (v + x1 )



0

a) f (v + x1 ) = 0

June 29, 2017

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kx k=r

1 p

1 αx r 2

p

>0


v 6= 0 and ψ (v ) = A(f (v + x1 ) either A(f (v + x1 )


0

a) f (v + x1 ) = 0

0

a) = 0 or f (v + x1 ) = 0 - not possible


June 29, 2017

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kx k=r

1 p

1 αx r 2

p

>0


v 6= 0 and ψ (v ) = A(f (v + x1 ) 0


either A(f (v + x1 ) thus A(f (v + x1 )


0

a) f (v + x1 ) = 0

a) = 0 and so f (v + x1 )


a=0

June 29, 2017

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kx k=r

1 p

1 αx r 2

p

>0


v 6= 0 and ψ (v ) = A(f (v + x1 )

0

a) f (v + x1 ) = 0

0


either A(f (v + x1 ) thus A(f (v + x1 )

a) = 0 and so f (v + x1 )

a=0

This means the equality ψ(v ) = 0 holds which contradicts ψ(v ) > 0. Contradiction with MPT M. G. (LUT, Institute of Mathematics)


June 29, 2017

14 / 28

Application Based on J.D. M.G. M.K. E.S. An application of a di¤eomorphism theorem to Volterra integral operator, accepted to Di¤erential and Integral Equations. Consider

V (x )(t ) = x (t ) +

Zt

v (t, τ, x (τ ))d τ, x (0) = 0,

(3)

0

˜ 1,p ([0, 1], Rn ) space consisting of AC fncs where p > 1; x 2 W 0 x : [0, 1] ! Rn s.t. x (0) = 0 and x 0 2 Lp (0, 1). The associated Volterra integral equation x (t ) +

Zt 0

v (t, τ, x (τ ))d τ = y (t ) , for t 2 [0, 1], x (0) = 0,

(4)

˜ 1,p in case V : W ˜ 1,p ! W ˜ 1,p is a is uniquely solvable for any y 2 W 0 0 0 di¤eomorphism. M. G. (LUT, Institute of Mathematics)


June 29, 2017

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Assumptions P∆ = f(t, τ ) 2 [0, 1]

[0, 1]; τ

t g , v : P∆

Rn ! Rn

c1 v and its partial derivatives satisfy some technical standard assumptions (no real growth) c2

jv (t, τ, x )j

c1 (t, τ )jx j + d1 (t, τ )

for a.e (t, τ ) 2 P∆ , x 2 Rn and 1

2

(p 1 ) p

jjc1 jjLp (P ∆ ,R)

> 0,

(5)

c3 vt (t, , x ) is measurable on [0, 1] for all (t, x ) 2 G and there exist functions c2 , d2 2 Lp (P∆ , R+ ) such that

jvt (t, τ, x )j

c2 (t, τ )jx j + d2 (t, τ )

for a.e. (t, τ ) 2 P∆ , x 2 Rn , M. G. (LUT, Institute of Mathematics)


June 29, 2017

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Applications - main di¢ culties 1

1. Examination of Volterra operator- well posedness, di¤erentiability. 2. De…nition of ϕ lack of Hilbert space structure 1

(p ϕ(x )) p = jjx ( ) jjx jjW˜ 1,p 0

1

2

(p 1 ) p


y ( )+

R 0

jjy jjW˜ 1,p 0

jj

R 0

v ( , τ, x (τ ))d τ jjW˜ 1,p 0

v ( , τ, x (τ ))d τ jjW˜ 1,p

jjc1 jjLp (P ∆ ,Rn ) jjx jjW˜ 1,p ([0,1 ],Rn ) 0


(6)

0

c˜

June 29, 2017

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Applications - main di¢ culties 2 3. PS-condition. Convergence of p ( ϕ0 (xn )

ϕ0 (x0 ))(xn

x0 ) ! 0

should provide that xn ! x0

˜ 1,p in order to have the PS condition satis…ed. Technical! Uses in W 0 coercivity and properties of Lebesgue spaces. 4. Solvability of h (t ) +

Zt 0

vx (t, τ, x (τ ))h(τ )d τ = g (t )fort 2 [0, 1] .

Classical successive approximation scheme but in Lp instead of L2 . M. G. (LUT, Institute of Mathematics)


June 29, 2017

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Example

Proposition ˜ 1,p ! W ˜ 1,p de…ned pointwisely for The integral Volterra operator V : W 0 0 all t 2 [0, 1] by V (x )(t ) = x (t ) +

Zt

21

p

(t

τ )1/2 ln(1 + (t

τ )4 x (τ )2 )d τ, x (0) = 0

0

de…nes a global di¤eomorphism. Thus the assciated Volteraa equation is uniquely solvable.



June 29, 2017

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Main result-without cont. di¤erentiability

Theorem (M.G. Quaestiones Mathematicae, 39 (2016), 683-688) If f : Rn ! Rn is a Fréchet-di¤erentiable mapping such that (b1) for any y 2 Rn the functional ϕ : Rn ! R de…ned by ϕ (x ) =

1 kf (x ) 2

y k2

is coercive; (b2) for any x 2 Rn we have det[f 0 (x )] 6= 0; (b3) f : Rn ! Rn is strictly di¤erentiable, then f is a di¤eomorphism.



June 29, 2017

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Some background

Lemma (R¼adulescu Twins) Let D be an open set of Rn and let f : D ! Rn be a di¤erentiable map and the following condition holds: det[f 0 (x )] 6= 0 for every x 2 D.Then f is a local di¤eomorphism. Problems - ϕ is not C 1 , MPT does not work, no results for di¤erentiable only fncs. f : D ! Rn is strictly di¤erentiable at x0 , if there exists f 0 (x0 ) 2 Rn lim

w !x0 ,t !0 +

f (w + tz ) t

f (w )

= f 0 (x0 ), z

for all z 2 Rn

provided the convergence is uniform for z in compact sets.



June 29, 2017

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Clarke derivative

f 0 (u; z ) := lim sup w !u, t !0 +

∂f (u ) := fξ 2 Rn : hξ, z i

f (w + tz ) t

f (w )

.

f 0 (u; z ), for all z 2 Rn g.

If f is strictly di¤erentiable the Clarke subdi¤erential reduces to a singleton, i.e. its derivative.



June 29, 2017

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PS condition and MPT u is called a critical point of f , if 0 2 ∂f (u ). J : Rn !R satis…es non-smooth P-S condition if every fun g such that fJ (un )g is bounded and J 0 (un ; u

un )

εn ku

un k for all u 2 Rn ,

where εn ! 0+ , admits a convergent subsequence.

Theorem

Let J : Rn !R be a locally Lipschitz and satis…es the P-S condition. If there exist u1 , u2 2 Rn , u1 6= u2 and r 2 (0, ku2 u1 k) such that inf fJ (u ) : ku

u1 k = r g

maxfJ (u1 ), J (u2 )g

Γ the family of continuous paths γ : [0, 1] ! Rn joining u1 and u2 , then c := inf max J (γ(s )) γ2Γs 2[0,1 ]

maxfJ (u1 ), J (u2 )g

is a critical value for Rn and Kc nfu1 , u2 g 6= ∅. M. G. (LUT, Institute of Mathematics)


June 29, 2017

23 / 28

Main di¤erences in the proof

non-smooth Fermat’s rule, i.e. 0 2 f (x )

y ) f 0 (x )

chain rule, g locally Lipschitz, f is C 1 then ∂ (g


f)

∂g

rf


June 29, 2017

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Invertibility of Lipschitz mappings

Lemma If f : D ! Rn satis…es a Lipschitz condition in some neighbourhood of x0 and ∂f (x0 ) Rn is of maximal rank, then there exist neighborhoods U D of x0 and V of f (x0 ) and a Lipschitz function g : V ! Rn such that i ) for every u 2 U, g (f (u )) = u, and ii) for every v 2 V , f (g (v )) = v . Set ∂f (x ) Rn being of maximal rank means that all matrices in ∂f (x ) are nonsingular, so it re‡ects the assumption that det[f 0 (x )] 6= 0 for every x 2D



June 29, 2017

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Global lipeomorphism

Theorem (M.G. M.R. submitted) Assume f : Rn ! Rn is a locally Lipschitz mapping such that (b1) for any y 2 Rn the functional ϕ : Rn ! R de…ned by ϕ (x ) =

1 kf (x ) 2

y k2

satis…es the non-smooth P-S condition; (b2) for any x 2 Rn we have that ∂f (x ) Rn is of maximal rank; Then f is invertible on Rn and f 1 is locally Lipschitz.



June 29, 2017

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Applications

Ax = F (x ) + ξ,

(7)

ξ 2 RN is …xed, A positive de…nite, symmetric; F : RN ! RN is a locally Lipschitz function.

Theorem (i) There exists a constant 0 < a