## On singular perturbation problems for systems of PDEs

Ivar Ekeland and Eric Séré

CEREMADE, Université Paris-Dauphine

Chern Institute, Nankai University, May 14th, 2018

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Chern Institute, Nankai University, May 14th, 2018 1 / 31

## The problem

We want to solve a system of PDEs:

S(*ε,*u,Du) =f
nearS(0,0,0) =0,

the linearized operatorv !^{S}^{u}(*ε,*u,Du)v+S_{Du}(*ε,*u,Du)Dv has a
right-inverseL* _{ε}*(u)for (

*ε,*u,Du) small

k^{L}*ε*(u)k !∞ when *ε*!^{0}

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

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## The linear case

Consider on the 2-dimensional torus the linear di¤erential operator

*∂**ω* =*ω*1

*∂*

*∂θ*1

+*ω*2

*∂*

*∂θ*2

where *ω*:= (*ω*1,*ω*2) satis…es the Diophantine condition:

*ω*1

*ω*2

p q

K

j^{q}j^{2}^{+}^{α}^{,}* ^{α}*>0

Clearly *∂** _{ω}* mapsH

^{k}into H

^{k}

^{1}, that is, it loses one derivative. It also has an inverse:

u =

## ∑

n

u_{n}exp(*ω*1n_{1}*θ*1+*ω*2n_{2}*θ*2)

*∂** _{ω}*u = f () 8

^{n}2

**Z**

^{d}, un =fn(

*ω*1n1+

*ω*2n2)

^{1}Because of the Diophantine condition, j

^{u}

^{n}j

^{K}

^{1}

*2*

^{ω}^{1}k

^{n}k

^{1}

^{+}

*j*

^{α}^{f}

^{n}j

^{. So}

*∂*_{ω}^{1} sendsH^{k} into H^{k} ^{1} * ^{α}*, that is, it loses many derivatives

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

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## The spaces

Typically, for PDEs, both the operator and its inverse lose derivatives, so
the inverse does not send you back into the initial space. For this reason
one must introduce a scale of Banach spaces (X_{s},k k^{s}),s_{0} s < _{∞}

s_{0} s_{1} s_{2} S =)[X_{s}_{2} X_{s}_{1} and k k^{s}1 k k^{s}2]

For instance,X_{s} =H^{s}(Ω)(Sobolev space) andC^{∞} =\^{s}^{H}^{s} ^{with} ^{S} =∞.
So, in the preceding example, where Ωis the torus,

*∂**ω* : H^{s} !^{H}^{s} ^{1}

*∂*_{ω}^{1} : H^{s} !^{H}^{s} ^{1} ^{α}

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## Polynomials

We shall need an additional structure. Let E_{N},N 0 be an increasing
sequence of …nite-dimensional spaces, the union being dense in all the X_{s}

X_{s} =[^{n} ^{0}^{E}^{N}

The projection ΠN :X_{0} !^{E}^{N} is assumed to satisfy:

(Polynomial growth): for all nonnegative numbers s,d satisfying
s+d S, and all x 2^{E}N, there is a constant C_{S} such that:

k^{x}k^{s}^{+}^{d} ^{C}^{S}^{N}^{d}k^{x}k^{s}
k(1 ΠN)xk^{s} ^{C}SN ^{d}k^{x}ks+d

(Interpolation): for 0 t 1:

k^{x}kts1+(1 t)s2 C_{S}k^{x}k^{t}s1k^{x}k^{1}s2 ^{t} .

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

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## Nonlinear operators

Nonlinear operators from C^{∞} into itself, such asS(u,Du) arise from maps
S :R^{d} R^{nd} !^{R}^{m} through some basic operations. The behaviour of the
Sobolev normsH^{s} is ruled by the so-called Hölder estimates:

(Product) Ifu andv are in L^{∞}\^{H}^{s} then there is a constantm_{s}
depending only on s such that:

k^{uv}ks m_{s}(k^{u}kssupj^{v}j+k^{v}kssupj^{u}j)

(Composition) Letf :R!^{R}^{be}^{C}^{∞}^{. If} ^{u} 2^{L}^{∞}\^{H}^{s} ^{then}
f (u)2^{L}^{∞}\^{H}^{s} and there is a constant `s depending on s,f,and
supj^{u}j ^{such that:}

k^{f} (u)ks `s(1+k^{u}ks)

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## Towards a statement

Let (Y_{s}, k k^{0}s),0 s S, be another scale of Hilbert spaces with the
same properties. We consider a map F from the unit ballB_{s}_{0}_{+}_{max}_{f}_{m}_{.}_{`}_{g}
into Y0, wheres0,m,` are given. We normalize by settingF(0) =0 and
we want to solveF (u) =v . Assume F is continuous and

Gâteaux-di¤erentiable.

Let m,m^{0},`,`^{0} be given. We assume that for any S and any s S,
there are constants a_{S} andb_{S} and a linear map L(x):Y !^{X} satisfying,
for all u with k^{u}ks0+maxf^{m,}`g 1

F^{0}(u)h ^{0}_{s} a_{S} k^{u}ks+m+k^{u}ks0k^{h}ks+m

F^{0}(u)L(u)k = k

k^{L}(u)kks bS k^{k}ks+`^{0}0+k^{k}k^{0}s0k^{u}ks+`

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## A local surjection theorem

Theorem Suppose:

s >s_{0}+maxf^{m,}`g
*δ* >s+`^{0}

and S is large enough. Then, there exists some r >0 and some k >0
such that, if k^{v}k*δ* <r , the equation F(u) =v can be solved with

k^{u}ks0+maxf^{m,}`g 1
k^{u}ks kk^{v}k^{0}*δ*

For any *ε*>0, we can solveF(u) =v with:

v 2^{Y}^{δ}^{for any}* ^{δ}* =s

_{0}+maxf

^{m,}`g+`

^{0}+

*ε*u 2

^{X}

^{s}

^{with}

^{s}=

*δ*(`

^{0}+

*ε*)

The linearized operator yields *ε*=0, so the nonlinear operator does almost
as well

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## Comparison with earlier results

There is a long history of results of this type, following the seminal papers of Arnol’d, Nash and Moser, the sharpest results to date being due to Hörmander. Most of them use smoothing operators instead of projections.

We are not aware that any of these papers gets the optimal value for
*δ* (lowest regularity of the RHS)

Every single one of them require an additional tame estimate on the second derivative:

F^{00}(x)u,v ^{0}_{s} c1 k^{u}ks+p1k^{v}kp2 +k^{u}kp1k^{v}ks+p2

+c_{2}k^{x}ks+p3 k^{u}ks+p4 k^{v}kp5 +k^{u}kp4k^{x}ks+p5

They then use second-order expansion to get quadratic convergence:

k^{x}^{n}^{+}^{1} ^{x}^{n}k ^{1}_{2}^{MK}k^{x}^{n} ^{x}^{n} ^{1}k^{2} = ^{1}

2MKk^{x}^{n} ^{x}^{n} ^{1}k k^{x}^{n} ^{x}^{n} ^{1}k
with k^{F}^{00}(x)k ^{K}. This is Newton’s method. However the range of
convergence is extremely small: ^{1}_{2}MKk^{x}^{n} ^{x}^{n} ^{1}k ^{1,}^{so}

k^{x}^{1}k ^{2}(MK) ^{1}.This is the key to the following

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

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## The Inverse Function Theorem in Banach spaces

Theorem (Newton-Kantorovitch)

Assume F is C^{2} with F(0) =0, and fork^{x}k ^{R we have:}

F^{0}(x) ^{1} M and F^{00}(x) K

For every y such that k^{y}k ^{min} M^{1}^{2}K,_{M}^{R} there is unique solution to
F (x) =y in the ball k^{x}k ^{min} MK^{1} ,R

Theorem ((Wazewski 1947, IE 2011))

Assume F is G-di¤erentiable with F (0) =0, and for k^{x}k R there is a
map L(x)such that

F^{0}(x)L(x) =I_{Y} and sup

k^{x}k ^{R}k^{L}(x)k<M

Then for every y with k^{y}k M^{R} there is a solution with k^{x}k ^{R.}

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

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## First approach to singular perturbations

Suppose we want to solveF (*ε,*u) =v with F (*ε,*0) =0 and for *ε* andu
small enough:

DuF (*ε,*u) ^{1} *ε* ^{1}M
D_{uu}^{2} F (*ε,*u) K
Then the two preceding theorems give the extimates:

(NK) k^{v}k ^{ε}

2

M^{2}K, k^{u}k _{KM}* ^{ε}*
(WE) k

^{v}k

^{εR}_{M}

^{,}k

^{u}k

^{R}We gain one order of magnitude, but we lose uniqueness

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## A singular perturbation result.

We have F(*ε,*u)with F (*ε,*0) =0. We assume

k^{DF}*ε*(u)hks^{0} a_{S} k^{u}ks+m+k^{u}ks0k^{h}ks+m

DF* _{ε}*(u)L(u)k = k

k^{L}(u)kks b_{S}*ε* ^{g} k^{k}ks+`^{0}0+k^{k}k^{0}s0k^{u}ks+`

Theorem

Suppose s >s_{0}+maxf^{m,}`g^{,}* ^{δ}* >s+`

^{0}g

^{0}>g , and S is large enough.

Then, for some r >0and k >0, whenever k^{v}k^{0}*δ* <rε^{g}^{0} there exists for
every *ε*>0 some u* _{ε}* such that:

F(*ε,*u* _{ε}*) = v
k

^{u}

*ε*ks0+maxf

^{m}

^{,}`g 1

k^{u}* ^{ε}*ks k

*ε*

^{g}

^{0}k

^{v}k

^{0}

*δ*

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## A singular perturbation problem

Rauch and Métivier (2010) and Texier and Zumbrun (2013) have studied the following sysem of nonlinear coupled Schrödinger equations in

v = (u,u¯)for j,k K andn N

*∂*

*∂t* +i*λ*j∆ v_{j} =

## ∑

N n=1b_{j}^{k}_{n}(v)^{∂v}^{k}

*∂x*_{n} +c_{jn}^{k} (v)^{∂}^{v}^{¯}^{k}

*∂x*_{n}
v_{k} 2 ^{C}^{,} ^{λ}^{j} 2 ^{R}^{,} ^{j} 6=j^{0} =)^{λ}^{j} 6=*λ*j^{0}

The coe¢ cients b_{j}^{k}

n(v) andc_{j}^{k}

n(v) must satisfy a growth condition near the origin:

9 ^{p} ^{2}^{:}8^{a}= (a_{1}, ,a_{N})2^{N}^{d}^{,} ^{∂b}

k jn

*∂v*^{a} + ^{∂c}

k jn

*∂v*^{a} C_{j}_{a}_{j}j^{v}j^{(}^{p} ^{j}^{a}^{j}^{)}^{+}
and we also need some nonresonance conditions:

*λ*j +*λ*j^{0} = 0=)^{c}^{jj}^{0} ^{c}^{j}^{0}^{j} =0

Imbjj = 0

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## The initial condition

This system arises in optics, gives rise to several kinds of solutions, according to the initial condition:

v* _{ε}*(0,x) =

*ε*

*(a*

^{σ}*(x),a¯*

_{ε}*(x)) There are two important classes of solutions:*

_{ε}a* _{ε}*(x) = a

^{0}x

*ε* (concentration)
a* _{ε}*(x) = a

^{0}(x)exp i

*ε*x *ξ*_{0} (oscillation)

Here g will be an (explicit) function of *σ,*p andd.Applying our result, we
can solve the problem for:

*σ*> ^{1} ^{σ}^{a}
p+1 +^{d}

2

p (p+1) (p 1) This is an improvement on earlier results.

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## The results

We seek a solution on [0, T], whereT is independent of *ε.*

For *σ* large enough, namely:

*σ*> ^{1}+^{N}_{2} (oscillating case)
1 (concentrating case)

this is not a singular perturbation problem. Metivier and Rauch, using the standard inverse function theorem in Banach spaces, prove that solutions exist for all d andp.

The case when *σ* is smaller than these bounds has been investigated by
Texier and Zumbrun using a Nash-Moser theorem. We will compare our
results to theirs

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## Comparison: two space dimensions

Concentrating case

TZ requirep 4 and _{2(p+1)}^{9} <*σ*<1.

ES gives anyp 3, and _{(p+1)(p}^{2p} ^{1} _{1)}<*σ*<1
Oscillating case

TZ requirep 3 and _{p}_{+1}^{5} <*σ*<2

ES gives anyp 2 and _{(p+1)(p}^{3p} ^{2} _{1)} <*σ*<2

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## Comparison: three space dimensions

Concentrating case

TZ requiresp 4 and _{p+1}^{4} <*σ*<1.

ES gives anyp 3, and _{2(p}_{+1)(p}^{4p} ^{1} _{1)} <*σ*<1
Oscillating case

TZ gives anyp 2 and _{2(p+1)}^{11} <*σ*< ^{5}_{2}
ES gives anyp 2 and _{2(p+1)(p}^{7p} ^{4} _{1)} <*σ*< ^{5}_{2}

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## Proof of the Theorem

We introduce positive numbers *α*>1,*θ* <1,*σ*<S and *β*with *αβ*<*σ.*

We choose an integerN0 2 and we de…ne two in…nite sequences of
integres Nn andMn byNn = N_{0}^{α}^{n} and Mn = N_{n}* ^{θ}* , so that Nn !

^{∞}very fast andN

_{n}

_{1}M

_{n}N

_{n}with dim(M

_{n})<<dim(N

_{n}). We then construct by induction a sequence u

_{n}2

^{E}

^{N}n such that:

ΠM^{0} nF (un) =_{Π}^{0}_{M}_{n} _{1}v for n 1

8^{n} ^{1,} k^{u}^{n} ^{u}^{n}^{+}^{1}ks0 CNn^{αβ}^{σ}^{+}^{s}^{0}k^{v}k^{0}d

8^{n} ^{1,} k^{u}^{n} ^{u}^{n}^{+}^{1}k*σ* CNn* ^{αβ}*k

^{v}kd

^{0}

Since *β* *σ/α*<0 the sequence xn is Cauchy in Hs0. So it converges to
some u2 ^{H}^{s}0 and by continuityF(u) =v.Using the interpolation
inequality, we see that u_{n} will also converge in all norms s <*σ* *αβ*

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## Initialization

Introduce a norm N^{1} ^{on}^{E}^{N}1 and E_{N}^{0}

1 :

N^{1}(z) =k^{z}k*δ*+N_{1} ^{θ}^{α}^{(}^{σ}^{δ}^{)}k^{z}k*σ*

Set B^{1} ^{:}=f^{z} 2^{E}^{N}1 j N^{1}(z) 1g and de…ne a functionf1 :B^{1} !^{E}M^{0} 1

by:

f_{1}(u) =Π^{0}M1F (u)
Lemma

If (1 *θ*) (*σ* *δ*)>*θm*+maxf`,*θ*`^{0}g+g/η, then Df1(u)has a right
inverse L_{1}(u)and we have:

k^{L}(u)k_{N}_{1} ^{C}^{ε}^{g} ^{M}1^{`}^{0}+N_{1}^{`}

We then apply the local surjection theorem to solvef_{1}(u_{1}) =ΠM0v,
provided:

k^{v}k^{0}*δ* C*ε*^{g} M_{1}^{`}^{0}+N_{1}^{`} ^{1}

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## The induction

Introduce a norm N^{n} ^{on}^{E}^{N}n :

N^{n}(z) =k^{z}ks0 +N_{n}^{σ}_{1}^{+}^{s}k^{z}k*σ*

Set B^{n} ^{:}= ^{n}z 2^{E}^{N}n j N^{n}(z) *ε* ^{g}N_{n}^{αβ}_{1}^{σ}^{+}^{s}^{0}k^{v}k^{0}*δ*

o

and de…ne a
functionf_{n} :B^{n} !^{E}^{N}n by:

fn(z) =ΠM^{0} n(F(un 1+z) F(un 1))

Taking into account the induction assumption ΠN^{0}n 1F (u_{n} _{1}) =_{Π}^{0}_{N}_{n} _{2}v,
this can be rewritten as:

fn(z) =en+hn

hn = _{Π}^{0}_{M}_{n} _{1} _{Π}^{0}_{M}_{n} _{2} v 2^{E}M^{0} n 1

e_{n} = _{Π}^{0}_{M}_{n} _{1} _{Π}^{0}_{M}_{n} F (u_{n} _{1})2^{E}M^{0} n
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## The induction

Lemma Assume:

(1+*α* *θα*) (*σ* s_{0}) > *αβ*+*α*(m+`) +`^{0}+ ^{g}
*αη*
(1 *θ*) (*σ* s0) > m+*θ*`^{0}+ ^{g}

*αη*
Then Dfn(z)has a right-inverse Ln(z)on the ball
N^{n}(z) *ε* ^{g}N_{n}^{αβ}_{1}^{σ}^{+}^{s}^{0}k^{v}k^{0}*δ* and :

k^{L}^{n}(z)k_{N}n *ε* ^{g}

Nn^{β}^{+`}N_{n i}^{σ}^{+}^{s}^{0}^{+`}^{0}
M_{1}^{`}^{0}+N_{1}^{`} +M_{n}^{`}^{0}

!

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## The induction

We have to solve fn(z) =en+hn with N^{n}(z) *ε* ^{g}N_{n}^{αβ}_{1}^{σ}^{+}^{s}^{0}k^{v}k^{0}*δ*.By
the local surjection theorem, this is possible provided

k^{L}^{n}(z)k_{N}n k^{e}^{n}k_{N}_{n}+k^{h}^{n}k_{N}_{n} ^{ε}^{g}^{N}n* ^{αβ}*1

^{σ}^{+}

^{s}

^{0}k

^{v}k

^{0}

*δ*.

Lemma Assume:

*δ* >s_{0}+ ^{α}

*ϑ*(*σ* s_{0} *αβ*+`")
(*α* 1)*β*>(1 *ϑ*) (*σ* s0) +*ϑm*+`"+^{g}

*η*

`"=max (*α* 1)`+`^{0},*ϑ*`^{0}

Then the local surjection theorem obtains, and un can be found

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## Showing that the values of the parameters are compatible

*α,β,ϑ* and*σ* must satisfy, with `"=maxf(*α* 1)`+`^{0},*ϑ*`^{0}g
1

*α* <*ϑ*<1

(1 *ϑ*) (*σ* *δ*)>*ϑm*+max `,*ϑ*`^{0} +^{g}
*η*
*β*>max `,*ϑ*`^{0} + ^{ϑ}

*α*(*σ* *δ*)
*σ* >*αβ*+s

(1+*α* *ϑα*) (*σ* s0)>*αβ*+*α*(m+`) +`^{0}+ ^{g}
*η*
(1 *ϑ*) (*σ* s0)>m+*ϑ*`^{0}+ ^{g}

*αη*
*δ* >s_{0}+ ^{α}

*ϑ*(*σ* s_{0} *αβ*+`")
(*α* 1)*β*>(1 *ϑ*) (*σ* s_{0}) +*ϑm*+`"+^{g}

*η*

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## Continuous sections

Theorem (Eric Séré)

Let X and Y be Banach spaces, B X the unit ball, and F :B !^{Y}
with F (0) =0.Suppose F is continuous and Hadamard-di¤erentiable on
X . Assume there are a map L:B ! L(X,Y)and a constant a<1 such
that, for every (x,v)2^{B} Y there exists some *ε*>0 with:

DF x^{0} L(x)v v ak^{v}k

Assume moreover that supfk^{L}(x)k j^{x} 2^{B}g<M.Denote by B^{0} Y
the ball of radius (1 a)rM ^{1}. Then there exists a continuous map
G :B^{0} !B such that F G =I_{Y}

So, there is no uniqueness in solving F(*ε,*x) =y, but we can ask for
continuous dependence on the right-hand side. Our plan is to extend this
to the singular perturbation problem with loss of derivative and show
continuous dependence on v and*ε.*

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THE INVERSE FUNCTION THEOREM IN BANACH SPACES

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## A variational principle

Theorem (Ekeland, 1972)

Let (X,d) be a complete metric space, and f :X !**R**[ f+_{∞}g ^{be a}
lower semi-continuous map, bounded from below:

f(x,a) j^{a} ^{f} (x)g is closed in X **R**
inff > ∞

Suppose f (x_{0})<∞. Then for every R >0, there exists somex such that:¯
f (x¯) f (x0)

d(x,¯ x0) R

f (x) f (x¯) ^{f} (x_{0}) inff

R d(x,x¯) 8^{x}

The proof relies on the Baire theorem: if Fn,n2N, is a sequence of
closed bounded subsets of a complete metric space, with F_{n}+1 F_{n} and
diam F_{n} !0, then their intersection is a singleton.

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## First-order version

De…nition

We shall say that f is Gâteaux-di¤erentiableat x if there exists a
continous linear map Df (x):X !^{X} ^{such that}

8* ^{ξ}* 2

^{X}

^{,}

^{lim}

_{t}

^{1}

_{t}[f (x+tξ) f (x)] =h

^{Df}(x),

*ξ*i In other words, the restriction of f to every line is di¤erentiable (for instance, partial derivatives exist). Note that a G-di¤erentiable function need not be continuous.

Theorem

Let (X,d) be a complete metric space, and f :X !** ^{R}**[ f+∞g

^{be a}lower semi-continuous map, bounded from below. Then for every x

_{0}2

^{X}and R >0, there exists some x such that:¯

f (x¯) f (x0),
k^{x}^{¯} ^{x}^{0}k ^{R}

k^{Df} (x¯)k ^{f} (x_{0}) inff
R
Corollary

There is a sequence xn such that:

f (x_{n})!^{inf}^{f}
k^{Df} (x_{n})k !^{0}
Proof.

Just take x_{0} such that f (x_{0}) inff *ε* and R=p

*ε* and let *ε*= ^{1}_{n}

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## Proof

For simplicity, take x0 =0 and inff =0. Apply EVP to x =x¯ +tu and
let u !^{0. We get:}

f (x¯ +tu) f (x¯) ^{f} (0)

R tk^{u}k 8(t,u)

t!lim+0

1

t (f (x¯ +tu) f (x¯)) ^{f} (0)

R k^{u}k 8^{u}
h^{Df} (x),ui ^{f} (0)

R k^{u}k 8^{u}^{, or} k^{Df} (x)k ^{f} (0)
R

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## A non-smooth inverse function theorem

Theorem

Let X and Y be Banach spaces. Let F :X !Y be continuous and Gâteaux-di¤erentiable, with F (0) =0. Assume that the derivative DF(x) has a right-inverse L(x), uniformly bounded in a neighbourhood of0:

8^{v} 2^{Y}^{,} ^{DF}(x)L(x)v =v
supfk^{L}(x)k j k^{x}k ^{R}g<M
Then, for every y such that¯

k^{y}^{¯}k _{M}^{R}
there is some x such that:¯

k^{x}^{¯}k ^{M}k^{y}^{¯}k
F (x¯) =y¯

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 29 / 31

## Proof

Take any y¯ with k^{y}^{¯}k M^{R}. Consider the functionf :X !^{R} ^{de…ned by:}

f (x) =k^{F} (x) y¯k

It is continuous and bounded from below, so that we can apply EVP. We can …nd x¯ with:

f (x¯) f (0) =k^{y}^{¯}k _{M}^{R}
k^{x}^{¯}k ^{M}k^{y}^{¯}k ^{R}
8^{x,} ^{f} (x) f (x¯) ^{f} (0)

R k^{x} ^{x}^{¯}k=f (x¯) ^{1}

M k^{x} ^{x}^{¯}k
I claim F(x¯) =y.¯

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 30 / 31

## Proof (ct’d)

Assume F(x¯)6=y¯. The last equation can be rewritten:

8^{t} ^{0,} 8^{u} 2^{X}^{,} ^{f} (x¯ +tu) f (x¯)
t

1
M k^{u}k
Simplify matters by assuming X is Hilbert. Then:

F(x¯) y¯

k^{F}(x¯) y¯k^{,}^{DF}(x¯)u =h^{Df} (x¯),ui _{M}^{1} k^{u}k

We now take u = L(x¯) (F(x¯) y¯), so thatDF(x¯)u = (F (x¯) y¯). We get a contradiction:

k^{F} (x¯) y¯k k^{L}(x¯)k

M k^{F} (x¯) y¯k<k^{F}(x¯) y¯k

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 31 / 31