Inverse function theorems: soft and hard

Inverse function theorems: soft and hard

Ivar Ekeland,

CEREMADE, Université Paris-Dauphine

June 2013, DacorognaFest, Lausanne

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 f (x) 0, 8^x

Suppose f (0)<∞. Then for every A>0, there exists some x such that:¯ f (x¯) f (0)

d(x,¯ 0) A

f (x) f (x¯) ^f (0)

A d(x,x¯) 8^x

This is a Baire-type result: relies on completeness, no compactnes needed

Gâteaux-di¤erentiability

De…nition

Let X and Y be Banach spaces.We shall say that F :X !^{Y is} Gâteaux-di¤erentiable atx if there exists a continous linear map DF (x):X !^{Y such that}

8^ξ 2 ^{X, lim}_t ¹_t [F (x+tξ) F (x)] =DF(x)ξ in Y Example

Let Ω Rⁿ be a bounded domain. De…ne F :L¹(Ω)!^L1(Ω)by F (u):=R

Ωf (u(x))dx, wheref is C¹ and f⁰ is bounded: j^f0(u)j ^a.

Then F is G-di¤erentiable, with:

[DF(u)v] (x) =f⁰(u(x))v(x)

but it is NOT C¹ (unlessf (u) =au+b). If f⁰(u) c >0, then DF(u) is invertible, and the inverse L(u)is uniformly bounded

[L(u)v] (x) =f⁰(u(x)) ¹v(x)

Note that a G-di¤erentiable function need not be continuous.

Ivar Ekeland, (CEREMADE, Université Paris-Dauphine)Inverse function theorems: soft and hard

June 2013, DacorognaFest, Lausanne 4 / 31

First-order version

SupposeX is a Banach space, and d(x₁,x₂) =k^{x1 x2}k. Apply EVP to x =x¯ +tu and letu !^{0. We get:}

f (x¯ +tu) f (x¯) ^f (0)

A tk^uk 8(t,u)

t!lim+0

1

t (f (x¯ +tu) f (x¯)) ^f (0)

A k^uk 8^u h^Df (x),ui ^f (0)

A k^uk 8^{u, or} k^Df (x)k ^f (0) A Corollary

Suppose F is everywhere …nite and Gâteaux-di¤erentiable. Then there is a sequence xn such that:

f (x_n)!^inff k^Df (x_n)k !⁰

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^xk ^Rg<m Then, for every y such that¯

k^y¯k _m^R there is some x such that:¯

k^x¯k ^mk^y¯k F (x¯) =y¯

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Proof

Consider the function f :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 with A=mk^y¯k. We can …nd x¯ with:

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

mk^y¯kk^{x x¯}k=f (x¯) ¹

mk^{x x¯}k I claim F(x¯) =y.¯

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 mk^uk Simplify matters by assuming X is Hilbert. Then:

F (x¯) y¯

k^F (x¯) y¯k^,DF(x¯)u =h^Df (x¯),ui _m¹ k^uk

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

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THE INVERSE FUNCTION THEOREM IN FRÉCHETSPACES

Fréchet spaces.

A Fréchet spaceX is gradedif its topology is de…ned by an increasing sequence of norms:

8^x 2^X, k^xkk k^xkk+1, k 0

A point x2^{X is controlled}if there is a constantc₀(x)such that:

k^xkk c₀(x)^k De…nitions

A graded Fréchet space is standard if, for everyx 2^X, there is a constant c3(x)and a sequence xn of controlled vectors such that:

8^k _n^lim

!∞k^{xn x}kk =0

8^n, k^xnkk c₃(x)k^xkk

The graded Fréchet spaces C^∞ Ω¯,R^d =\^Ck Ω^¯,R^d and C^∞ Ω¯,R^d =\^{Hk Ω,Rd} are both standard.

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Normal maps

We are given two Fréchet spacesX andY, and a neighbourhood of zero B = x j k^xkk0 R inX

De…nition

A mapF :X !^{Y is normaloverB} if there are two integersd₁, d₂ and two non-decreasing sequences m_k >0, m⁰_k >0 such that:

1 F (0) =0 and F is continuous on B

2 F is Gâteaux-di¤erentiable onB and for allx 2^B 8^k 2N, k^DF(x)ukk mkk^ukk+d1

There exists a linear map L(x):Y !^{X such that:}

8^v 2^{Y, DF} (x)L(x)v =v 8^k 2^{N, sup}

x2^Bk^L(x)vkk <m_k⁰ k^vkk+d2

An inverse function theorem

Theorem

Suppose Y is standard, and F :X !Y is normal over B = x j k^xkk0 R . Then, for every y with

k^ykk0+d2

R m⁰_k

0

there is some x 2B such that:

k^xkk0 m_k⁰

0k^ykk0+d2 and F(x) =y

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Corollary (Lipschitz inverse)

For every y₁,y₂ with k^yikk0+d2 m_k^{0 1}

0 R and every x₁ 2^{B with} F (x₁) =y₁, there is some x₂ with:

k^{x2 x1}kk0 m⁰_k0k^{y2 y1}kk0+d2 and F(x₂) =y₂

Corollary (Finite regularity)

Suppose F extends to a continuous map F¯ :X_k0 !^Yk0 d1. Take some y 2 ^Yk0+d2 with k^ykk0+d2 <Rm_k^{0 1}

0 . Then there is some x 2^Xk0 such that k^xkk0 <R and F¯ (x) =y.

Proof: step 1

Let y¯ be given, with k^y¯kk0+d2

R m⁰_k

0

. Letβk 0 be a sequence with unbounded support satisfying:

∑

βkm_km_k⁰_+d1n^k < ∞, 8ⁿ2 ^N, 1

βk0+d2

∑

βkk^y¯kk

R m_k⁰

0

Set αk :=m_k^{0 1}

0 βk+d2 and de…ne:

k^xkα:=

∑

αkk^xkk, X_α =f^x2 ^X j k^xkα< ∞g

Then X_α X is a linear subspace, X_α is a Banach space and the identity map X_α !^X is continuous: So the restriction F :X_α !^Y is continuous.

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Step 1 (ct’d)

Now consider the function f :X_α !^R[ f+∞g ^{(the value} +∞ is allowed) de…ned by:

f (x) =

∑

βkk^F(x) y¯kk

f is lower semi-continuous, and 0 inff f (0) =_∑^∞_k=0βkk^y¯kk <∞. By the EVP there is a point x¯ 2 ^Xα such that:

f (x¯) f (0) =

∑

βkk^y¯kk, k^x¯kα αk0R

f (x) f (x¯) ^f (0)

αk0R k^{x x¯}kα, 8^x 2^Xα It follows that:

∑

αkk^x¯kk αk0R,so k^x¯kk0 R

∑

∑

∑

Proof: step 2

Assume thenF (x¯)6=y. If¯ u 2^Xα, we can set x=x¯ +tu, replace f by its value and divide by t.

tlim!⁰ t>0

1 t

"

∑

βkk^{y¯ F} (x¯ +tu)kk

∑

βkk^{y¯ F}(x¯)kk

#

A

∑

k 0

αkk^ukk

with A=_∑βkk^y¯kk(αk0R) ¹<1. We would like to go one step further:

∑

βk

F(x¯) y¯ k^F(x¯) y¯_kkk

,DF (x¯)u

k

A

∑

k 0

αkk^ukk

This program can be carried through (by repeated use of Lebesgue’s dominated convergence theorem) if we takeu =un, whereun =L(x¯)vn

and:

vn !^F (x¯) y,¯ vn controlled, k^vnkk c₃(F(x¯) y¯)k^F(x¯) y¯kk

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Proof: step 3

Plugging inu_n =L(x¯)v_n, we get:

∑

βk

F (x¯) y¯ k^F (x¯) y¯_kkk

,DF(x¯)L(x¯)v_n

k

A

∑

k 0

αkk^L(x¯)v_nkk

∑

βk

F(x¯) y¯ k^F(x¯) y¯_kkk

,vn k

A

∑

k 0

αkk^L(x¯)vnkk

Letting n !^{∞, so vn} !^F (x¯) y, this becomes:¯

∑

βkk^F (x¯) y¯_kkk A

∑

k 0

αkm_kk^F (x¯) y¯_kkk+d2

=A

∑

k d2

βkk^F (x¯) y¯_kkk

which is a contradiction sinceA<1

THE HARDINVERSE FUNCTION THEOREM:

NASH-MOSER FOR NON-SMOOTH FUNCTION

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Let (Xs, k k^s), 0 s S, be a scale of Banach spaces:

0 s₁ s₂ S =)(X_s2 X_s1 and k k^s1 k k^s2)

We shall assume that there exists a sequence of projectors ΠN : X₀ !^EN (smoothing operators) whereE_N ^T_{s 0}X_s is the range ofΠN, with Π0 =0,E_N E_N+1 and^S_{N 1}E_N is dense in each space Xs for the norm k k^s. We assume that:

k^ΠNuk^{s+d CNd}k^uk^s k(1 ΠN)uk^{s CN d}k^uk^s+d

Note that these properties imply some interpolation inequalities, for 0 t 1 and 0 s₁,s₂ A

k^xkts1+(1 t)s2 C₂^Ak^xk^ts1k^xk¹s2 ^t .

Let (Ys, k k⁰s)s 0 be another regular scale of Banach spaces, with smoothing operators Π⁰_N :Y₀ !^EN⁰

T

s 0Y_s

In the following,R >0 and S >0 are prescribed, with possibly S =_∞ De…nition

We shall say that F : B₀(R)!^{Y0 is} roughly tamewith loss of regularity µif:

(a) F is continuous and Gâteaux-di¤erentiable fromB₀(R)\^Xs toY_s for anys 2 [0,S).

(b) There is a constantK such that, for all s S and x2^B0(R):

8^h2^X, k^DF(x)hks⁰ K(k^hk^s+k^xk^sk^hk⁰) (c) Forx 2^B0(R)\^EN, the linear

mapsΠ⁰_NDF (x)j^EN : E_N !^EN⁰ have a right-inverse, denoted byL_N(x). There are constants µ>0 and γ>0, such that, for all s S andx 2^B0(R)we have:

8^k 2^EN⁰, k^LN(x)kk^{s 1}

γN^µ(k^kk⁰s+k^xk^sk^kk⁰0)

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A hard inverse function theorem.

Note the nonlinear estimates inH^s: ifk^uk_∞ ^andk^vk_∞ are …nite, then:

k^f (u,v)kH^s C(k^ukH⁰k^vkH^s +k^ukH^sk^vkH⁰) The loss of derivatives for F andDF is normalized to 0.

The loss of derivatives for DF ¹ is µ, and we cannot expect any better forF ¹

De…ne a real function ϕon[4, ∞[by:

ϕ(x) = 8<

:

1 2x 1

q

1 ⁴_x +1 if 4<x ³p^+p5 5 1 x²

4 1

q 1 ⁴_x

2

if x ³p^+p5 5 1

4 5 6 7 8 1.0

1.5 2.0 2.5 3.0

S/mu delta/mu

The authorized region is above the graph

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Theorem (IE, Eric Séré)

Assume F(0) =0 and F :B₀(R)\^Xs !^{Ys,0 s} <S, is roughly tame with loss of regularity µ.Ifδ/µis in the authorized region, then, for any α with

α

µ <min δ

µ ϕ S µ , S

µ ϕ ¹ δ µ

one can solve F (x) =y with y 2^Yδ and x 2^Xα. More precisely, we can

…nd ρ >0and C >0 such that, whenever k^yk⁰δ ρ , there is some x 2^Xα with:

F(x) =y k^xk0 1 k^xk_α ^Ck^yk_δ

Comments

S >4µ(you need some room upstairs)

3µ>δ α>µ(the loss of derivatives for F is larger than the one for DF)

ifS/µ!∞ (lots of room upstairs), δ α!^µ(lowest possible value)

IfS/µ!4 (little room upstairs), δ α!^3µ(large loss of regularity)

Corollary

Assume F(x) sends X_s into Y_s and is roughly tame atx with loss of¯ regularity µ. Suppose x¯ 2^XS,y¯ 2^{YS and F}(x¯) =y . Then one can solve¯ F(x) =y,with k^{x x¯}k^{α C}k^{y y¯}k⁰δ.

Proof.

Consider the map Φ(x,y):=F (x) y+y¯ fromX_s Y_s into Y_s. It is roughly tame, with F(0,0) =0, and we can apply the preceding

Theorem June 2013, DacorognaFest, Lausanne 24

The proof: constructing approximate solutions

Theorem

Consider an integer N₀ 2and de…ne N_n '(N₀)^κn, where

κ =κ(δ,µ)>1 is appropriately chosen. Then one can …ndρ>0 and c >0such that, for any y with k^ykδ <ρ , there is a sequence(xn)n 1

with k^xnk0 1 satisfying:

(case 1) Π⁰NnF(xn) =_Π⁰_{Nn 1}y andxn 2^ENn

(case 2) Π⁰NnF(xn) =_Π⁰_Nny andxn 2^ENn

and in both cases, for appropriate σ and β with κβ< σ<S : k^x1kσ cN₁^βk^yk⁰δ andk^{xn+1 xn}kσ c N_n^κβk^yk⁰δ

k^x1k^{0 cN}1^µk^yk⁰δ andk^{xn+1 xn}k^{0 c Nnκβ σ}k^yk⁰δ

Auxiliary constants

(S,δ,µ)are given.

We …rst choose κ:

1<κ<2 and κ² κ 1 < ^S

µ and minf^{κ2, κ}+1g<µδ µ This gives two possibilities: 1<κ ¹⁺

p5

2 and ¹⁺₂^p5 κ <2. Then we choose σ and β:

κ² κ 1 < ^κ

µβ< ¹ µσ< ^S

µ κβ> σ+κµ δ

κ

for 1<κ 1+p 5 2 β> µ+σ δ for 1+p

5

2 κ<2

June 2013, DacorognaFest, Lausanne 26

Passing to the limit

The sequence (x_n)has a limit x in X₀, with k^xk^{0 C}k^yk⁰δ,for

C :=c(N₁^µ+_{∑n 1}N_n^{κβ σ}).Then F(x_n)converges toF(x)in Y₀,by the continuity of F :X0 !^Y0^.. On the other hand:

F(x_n) = (1 Π⁰_Nn)F(x_n) +Π⁰_{Nn 1}y. One proves by induction an estimate of the form

k(1 Π⁰N)F(xn)k⁰0 C Nn^{β σ}k^yk⁰δ !⁰

because the exponent(β σ)is negative. So(1 Π⁰_N)F(x_n)converges to zero in Y0. On the other hand, by the de…nition of a smoothing operator,

k(1 Π⁰Nn 1)yk⁰0 CN_n^δ₁k^yk⁰δ !⁰ Passing to the limit we get F(x) =y.

Checking the induction.

Here we assume that we have chosen ρ andc, and found x₁, ,x_n. We are going to construct xn+1 =xn+u. Using the induction hypothesis Π_N⁰_nF(x_n) =Π⁰_{Nn 1}y, the equation to be solved by ∆x_n may be written in the following form:

fn(u) =en+_∆yn 1

f_n(u):=_ΠNn+1 (F(x_n+u) F(x_n))2^EN⁰n+1

e_k :=ΠNk+1(ΠNk 1)F(x_k)

∆yk := _ΠN_k+1(1 ΠN_k)y

The function f_n is continuous and Gâteaux-di¤erentiable withf (0) =0.

So this is an inverse problem for u

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Choosing the right norms

We choose the norm

Nⁿ(u) =k^uk⁰+N_n ^σk^ukσ on E_Nn+1 Nn⁰(v) =k^vk⁰0+N_n^σk^vk⁰σ on E_N⁰_n+1

Set R_n :=cN_n^{κβ σ}k^ykδ. For Nⁿ(u) R_n we have:

[Df_n(u)] ¹k ₀ 2N_n^µ₊₁ γ k^kk⁰ [Df_n(u)] ¹k

σ

N_n^µ₊₁

γ (k^kkσ+B⁽²⁾R_nN_n^σk^kk⁰) Hence:

Nⁿ([Df(u)] ¹k) ^B

(2)+2

γ N_n^µ₊₁Nn⁰(k)

Applying the soft IVT

The IVT gives the existence of u¯_n 2 Bⁿ(R_n)such that f_n(u¯_n) =e_n+∆y_n provided:

Nn⁰(e_n+_∆y_{n 1})< ^Rn

B⁽²⁾+2 γ ¹N_n^µ₊₁ . The condition on en+_∆yn 1 is ful…lled provided

Nn⁰(en) +Nn⁰(_∆yn 1)< ^2c

B⁽²⁾+2N_n+^µ₁Nn^{κβ σ}k^ykδ . This is satis…ed if:

2B⁽³⁾c N_n^β+C⁽¹⁾ 0

@g^δκN_n^{σ δκ} + ^g Nn

(δ σ)+

κ

N_n^{(σ δ)+} 1

A< ^{2c g}

µ

B⁽²⁾+2N_n^{κ(β µ)} holds. We …nd that the exponents of Nn in the left-hand side of are

strictly smaller than the one in the right-hand side. So, for N₀ chosen large enough, is satis…ed for all n

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Bibliography

I.Ekeland, "Nonconvex minimization problems", Bull. AMS 1 (New Series) (1979) p. 443-47

I. Ekeland, "Some lemmas on dynamical systems", Math. Scan. 52 (1983), p.262-268

I.Ekeland, "An inverse function theorem in Fréchet spaces" 28 (2011), Annales de l’Institut Henri Poincaré, Analyse Non Linéaire, p. 91–105 I. Ekeland and E. Séré, work in progress

All papers can be found on my website:

http://www.ceremade.dauphine.fr/~ekeland/