A local surjection theorem with continuous inverse in Banach spaces

. In this paper we prove a local surjection theorem with continuous right-inverse for maps between Banach spaces, and we apply it to a class of inversion problems with loss of derivatives.


Introduction
In the recent work [4] we introduced a new algorithm for solving nonlinear functional equations admitting a right-invertible linearization, but with an inverse losing derivatives.These equations are of the form F (u) = v with F (0) = 0, v small and given, u small and unknown.The main difference with the classical Nash-Moser algorithm (see e.g.[7,12]) was that, instead of using a regularized Newton scheme, we constructed a sequence (u n ) n of solutions to Galerkin approximations of the "hard" problem and proved the convergence of (u n ) n to a solution u of the exact equation.Each u n was obtained thanks to a topological theorem on the surjectivity of maps between Banach spaces, due to one of us in [3].However, this theorem does not provide the continuous dependence of u n as a function of v.As a consequence, nothing was said in [4] on the existence of a continuous selection of solutions u(v).Theorem 8 of the present work overcomes this limitation thanks to a variant of the topological argument, stated in Theorem 2.
In the sequel, L(X, Y ) is the space of bounded linear operators between Banach spaces X and Y ; the operator norm on this space is denoted by • X,Y .We first restate the result of [3] below for the reader's convenience: Theorem 1. [3] Let X and Y be Banach spaces.Denote by B the open ball of radius R > 0 around the origin in X.Let f : B → Y be continuous and Gâteaux-differentiable, with f (0) = 0. Assume that the derivative Df (x) has a right-inverse L (x), uniformly bounded on the ball B R : Then, for every y ∈ Y with y Y < Rm −1 there is some x ∈ B satisfying: We recall that in the standard local inversion theorem, one assumes that f is of class C 1 , with Df (0) invertible and y small.An explicit bound on y Y is provided by the classical Newton-Kantorovich invertibility condition (see [1]) when f is of class C 2 .The bound y Y < Rm −1 of Theorem 1 is much less restrictive than the Newton-Kantorovich condition, at the price of losing uniqueness, even in the case when L(x) is also a left inverse of Df (x).To illustrate this, we consider a finite-dimensional example.
Example A. We take X = Y = C viewed as a 2-dimensional real vector space and f (z) = (2+z) n −2 n , for any complex number z in the open disc of center 0 and radius R = 1 (here n is a positive integer).In that case Df (z) is the multiplication by n(2 + z) n−1 and L(z) is the multiplication by n −1 (2 + z) 1−n , so f satisfies the assumptions of Theorem 1 for R = 1 and any real number m > n −1 .Thus, Theorem 1 tells us that the equation f (z) = Z has a solution of modulus less than or equal to m|Z|, provided Z has modulus less than m −1 .However uniqueness does not hold: the solutions of the algebraic equation (2 + z) n − 2 n = Z are of the form z k = 4ie i kπ n sin kπ n + O(2 −n ), and for |Z| > 4π the three solutions z 0 , z 1 , z −1 lie in the closed disc {|z| ≤ n −1 |Z|} when n is large enough.Yet, there is a unique continuous function g such that g(0) = 0 and f • g(Z) = Z for all complex numbers Z of modulus less than 1/m.This continuous selection is This example raises the following question: in the general case, can we select a solution x depending continuously on y, even in infinite dimension and when Df (x) does not have a left inverse?The following theorem gives a positive response, under mild additional assumptions: (ii) There are a function L : B → L (Y, X) , a constant a < 1 and, for any (iii) There is some m < ∞ such that: Denote by B ′ ⊂ Y the open ball of radius R ′ := (1 − a) Rm −1 and center 0. Then there is a continuous map g : B ′ → B such that: If, in addition, one has: (iv) f is Fréchet differentiable on B, Df (x) has a left-inverse for all x ∈ B and there is a non-decreasing function ε : (0, ∞) → (0, ∞) with lim t→0 ε(t) = 0, such that for all then g is the unique continuous right-inverse of f defined on B ′ and mapping 0 Y to 0 X .Remark 3. If a function f satisfies the assumptions (i), (ii) and (iii) then, for every x 0 ∈ B, taking the radius R x 0 = R − x 0 X , one can apply Theorem 1 to the function z ∈ B X (0, R x 0 ) → f (x 0 + z) − f (x 0 ), and one concludes that the restriction of f to B X (x 0 , R x 0 ) has a continuous right-inverse g x 0 defined on B Y (f (x 0 ), (1 − a)R x 0 m −1 ) and such that g x 0 (y) Conversely, Assumption (ii) is satisfied, for instance, if (i) and (ii') hold true, with: (ii') For each x ∈ B, Df (x) has a right-inverse L(x) ∈ L(Y, X) .Moreover, the map x → Df (x) is continuous for the strong topology of B and the strong operator topology of L(X, Y ) : in other words, if The function f of Example A satisfies the assumptions (i), (ii ′ ) and (iii).In that finite-dimensional case, f is of course differentiable in the classical sense of Fréchet.Let us give an example for which Fréchet differentiability does not hold.[9,10].However, Φ satisfies conditions (i), (ii ′ ) and (iii) for any r > 0 and m > (inf R ϕ ′ ) −1 .Therefore, Theorem 2 applies to Φ, but the inverse Ψ is easily found without the help of this theorem, as a Nemitskii operator: It turns out that any function f satisfying (i) has the Hadamard differentiability property which is stronger than the Gâteaux differentiability and that we recall below: This notion is weaker than Fréchet differentiability but in finite dimension Hadamard and Fréchet differentiability are equivalent.On the other hand, Hadamard differentiability is stronger than Gâteaux differentiability, but if a map f is Gâteaux differentiable and Lipschitz, then it is Hadamard differentiable (see, e.g., [6]).In particular, the functions f of Theorems 1, 2 are Hadamard differentiable.
Note that the chain rule holds true for Hadamard differentiable functions, while this is not the case with Gâteaux differentiability (see [6]).Hadamard differentiable functions are encountered for instance in statistics [14,13,6] and in the bifurcation theory of nonlinear elliptic partial differential equations [5].
The paper is organized as follows.In Section 2 we prove Theorem 2. In Section 3 we state the hard surjection theorem with continuous right-inverse (Theorem 8) that can be proved using Theorem 2 and proceeding as in [4].Finally, under additional assumptions we state and prove the uniqueness of the continuous rightinverse (Theorem 9).

Proof of Theorem 2
In [3], Theorem 1 was proved by applying Ekeland's variational principle in the Banach space X, to the map x → f (x) − y Y .This principle provided the existence of an approximate minimiser x.Assuming that f (x) − y Y > 0 and considering the direction of descent L(x) (y − f (x)), a contradiction was found.Thus, f (x) − y was necessarily equal to zero and x was the desired solution of the equation f (x) = y.However, there was no continuous dependence of x as a function of y.In order to obtain such a continuous dependence, it is more convenient to solve all the equations f (x) = y for all possible values of y ∈ B ′ simultaneously, by applying the variational principle in a functional space of continuous maps from B ′ to X.The drawback is that it is more difficult to construct a direction of descent, as this direction should be a continuous function of y.In order to do so, we use an argument inspired of the classical pseudo-gradient construction for C 1 functionals in Banach spaces [8], which makes use of the paracompactness property of metric spaces.
Consider the space C of continuous maps g : B ′ → X such that y −1 g(y) is bounded on Ḃ′ , with the notation Ḃ′ := B ′ \ {0}.Endowed with the norm C is a Banach space.Consider the function: The function ϕ is lower semi-continuous on C and its restriction to the closed ball 1−a } is finite-valued.In addition, we have: Choose some m 0 with: By Ekeland's variational principle [2], there exists some g 0 ∈ C such that: R, and the last equation can be rewritten: If ϕ (g 0 ) = 0, then f (g 0 (y)) − y = 0 for all y ∈ B ′ and the existence proof is over.If not, then ϕ (g 0 ) > 0 and we shall derive a contradiction.In order to do so, we are going to build a deformation g t of g 0 which contradicts the optimality property We define a continuous map w : B ′ → Y by the formula By continuity of w, the set Now, Df is bounded since f is Lipschitz continuous, and L is bounded on B by Assumption (iii).Therefore, combining these bounds with the continuity of w, we see that for each (x, y) ∈ B × ( Ḃ′ \ V) , there exists a positive radius β(x, y) such that, if Let γ(x, y) := min (α(x, w(y)); β(x, y)) where α(x, w) is the radius introduced in Assumption (ii).Then this assumption combined with (2.5) implies that for each (x, y) ∈ B × ( Ḃ′ \ V) and all (x ′ , y ′ ) ∈ B X (x, γ(x, y)) × B Y (y, β(x, y)) .
We are now ready to define One easily checks that L is locally Lipschitz on B × Ḃ′ .Moreover, it satisfies the same uniform estimate as L: and due to (2.6), it is an approximate right-inverse of Df "in the direction w(y)".
More precisely, for all (x, y) in B × Ḃ′ , one has: Now, to each y ∈ Ḃ′ we associate the vector field on B X y (x) := L(x, y)w(y) and we consider the Cauchy problem x(0) = g 0 (y).
The vector field X y is locally Lipschitz in the variable x ∈ B and from (2.7) we have the uniform estimate Thus, recalling that g 0 (y) X < Rm 0 m −1 , we see that our Cauchy problem has a unique solution x(t) = g t (y) ∈ B on the time interval [0, τ ] with τ = m−m 0 (1−a ′ )m 0 .In addition, we take g t (0) = 0.This gives us a one-parameter family of functions g t : B ′ → B. For 0 < t ≤ τ , g t satisfies the estimate (2.9) sup y −1 Since g 0 C ≤ m 0 1−a and ϕ(g 0 ) ≤ 1, the inequality (2.9) implies that We recall that L(•, •) is locally Lipschitz on B × Ḃ′ and g 0 , w are continuous.Thus, by Gronwall's inequality, for each t ∈ [0, τ ] the function g t is continuous on Ḃ′ .We can conclude that g t ∈ C , and (2.10) implies that ϕ(g t ) < ∞ .Now, to each (t, y) ∈ [0, τ ] × Ḃ′ we associate s t (y) := t 0 θ(g u (y), y)du and we consider the function Since f is Lipschitzian, its Gâteaux differential Df (x) at any x ∈ B is also a Hadamard differential, as mentioned in the introduction.This implies that for any function γ : (−1, 1) → B differentiable at 0 and such that γ(0) = x, the function f •γ is differentiable at 0 and the chain rule holds true: (f •γ) ′ (0) = Df (γ(0))γ ′ (0) .Therefore, using (2.8), we get: In addition h(0, y) = 0, so by the mean value theorem, By the triangle inequality, this implies that (2.12) We are now ready to get a contradiction.The estimate (2.9) may be written as follows: As a first consequence of (2.12), if then one has w(y) Y ≥ ϕ(g 0 ) 2 y Y , hence s t (y) = t.Thus, the estimate (2.12) implies that for all (t, y) ∈ [0, τ ] × Ḃ′ ,

Moreover we always have y
for all 0 ≤ t ≤ τ ′ and y ∈ Ḃ′ .This means that Combining (2.13) with (2.14), we find the following, for 0 < t ≤ τ ′ : which contradicts (2.4).This ends the proof of the existence statement in Theorem 2.
The uniqueness statement is proved by more standard arguments: if g 1 and g 2 are two continuous right-inverses of f such that g 1 (0) = g 2 (0) = 0, then the set is nonempty and closed.On the other hand, if Df (x) is left and right invertible, it is an isomorphism.By Remark 4 its inverse L(x) is bounded independently of x.We fix an arbitrary y 0 in Z and we consider a small radius ρ > 0 (to be chosen later) such that B Y (y 0 , ρ) ⊂ B ′ .By continuity of g 1 − g 2 at y 0 , there is η(ρ) > 0 such that lim ρ→0 η(ρ) = 0 and, for each y in the ball B Y (y 0 , ρ) , Moreover, we have f (g 2 (y)) − f (g 1 (y)) = y − y = 0. Thus, using (iv), we find that Then, multiplying Df (g 1 (y))(g 2 (y) − g 1 (y)) on the left by L(g 1 (y)) and using the uniform bound on L, we get a bound of the form with lim ρ→0 ξ(ρ) = 0.As a consequence, for ρ small enough one has g 2 (y) − g 1 (y) = 0, so y ∈ Z.This proves that Z is open.By connectedness of B ′ we conclude that Z = B ′ , so g 1 and g 2 are equal.This ends the proof of Theorem 2.

A hard surjection theorem with continuous right-inverse
In this section we state our hard surjection theorem with continuous right-inverse and we shortly explain its proof which is a variant of the arguments of [4] in which Theorem 1 is replaced by Theorem 2.
Let (V s , • s ) 0≤s≤S be a scale of Banach spaces, namely: We shall assume that to each Λ ∈ [1, ∞) is associated a continuous linear projection Π(Λ) on V 0 , with range E(Λ) ⊂ V S .We shall also assume that the spaces E(Λ) form a nondecreasing family of sets indexed by [1, ∞), while the spaces Ker Π(Λ) form a nonincreasing family.In other words: Finally, we assume that the projections Π(Λ) are "smoothing operators" satisfying the following estimates: Polynomial growth and approximation: There are constants A 1 , A 2 ≥ 1 such that, for all numbers 0 ≤ s ≤ S, all Λ ∈ [1, ∞) and all u ∈ V s , we have: When the above properties are met, we shall say that (V s , • s ) 0≤s≤S endowed with the family of projectors { Π(Λ) , Λ ∈ [1, ∞) } , is a tame Banach scale.Let (W s , • ′ s ) 0≤s≤S be another tame scale of Banach spaces.We shall denote by Π ′ (Λ) the corresponding projections defined on W 0 with ranges E ′ (Λ) ⊂ W S , and by A ′ i (i = 1, 2, 3) the corresponding constants in (3.1), (3.2).We also denote by B s the unit ball in V s and by B ′ s (0, r) the ball of center 0 and positive radius r in W s : In the sequel we fix nonnegative constants s 0 , m, ℓ and ℓ ′ .We will assume that S is large enough.
We first recall the definition of Gâteaux-differentiability, in a form adapted to our framework: Definition 6.We shall say that a function F : B s 0 +m → W s 0 is Gâteauxdifferentiable (henceforth G-differentiable) if for every u ∈ B s 0 +m , there exists a linear map DF (u) : V s 0 +m → W s 0 such that for every s ∈ [s 0 , S − m], if u ∈ B s 0 +m ∩ V s+m , then DF (u) maps continuously V s+m into W s , and Note that, even in finite dimension, a G-differentiable map need not be C 1 , or even continuous.However, if DF : B s 0 +m ∩ V s+m → L(V s+m , W s ) is locally bounded, then F : B s 0 +m ∩ V s+m → W s is locally Lipschitz, hence continuous.In the present paper, we are in such a situation.
We now define the notion of S-tame differentiability: Definition 7.
• We shall say that the map F : B s 0 +m → W s 0 is S-tame differentiable if it is G-differentiable in the sense of Definition 6, and, for some positive constant a and all s ∈ [s 0 , S − m] , if u ∈ B s 0 +m ∩ V s+m and h ∈ V s+m , then DF (u) h ∈ W s with the tame direct estimate • Then we shall say that DF is tame right-invertible if there are b > 0 and ℓ, ℓ ′ ≥ 0 such that for all u ∈ B s 0 +max{m,ℓ} , there is a linear map L (u) : W s 0 +ℓ ′ → V s 0 satisfying and for all s 0 ≤ s ≤ S − max {ℓ, ℓ ′ }, if u ∈ B s 0 +max{m,ℓ} ∩ V s+ℓ and k ∈ W s+ℓ ′ , then L (u) k ∈ V s , with the tame inverse estimate In the above definition, the numbers m, ℓ, ℓ ′ represent the loss of derivatives for DF and its right-inverse.
The main result of this section is the following theorem.
Theorem 8. Assume that the map F : B s 0 +m → W s 0 is S-tame differentiable between the tame scales (V s ) 0≤s≤S and (W s ) 0≤s≤S with F (0) = 0 and that DF is tame right-invertible.Let s 0 , m, ℓ, ℓ ′ be the associated parameters.
Assume in addition that for each Λ, Λ ′ ∈ [1, S] the map )) is continuous for the norms • s 0 and • ′ s 0 .Let s 1 ≥ s 0 + max{m, ℓ} and δ > s 1 + ℓ ′ .Then, for S large enough, there exist a radius r > 0 and a continuous map G : B ′ δ (0, r) → B s 1 such that: As mentioned in the introduction, compared with the results of [4] the novelty in Theorem 8 is the continuity of G. To prove this theorem, one repeats with some modifications the arguments of [4] in the case ε = 1 (in that paper a singularly perturbed problem depending on a parameter ε was dealt with, but for simplicity we do not consider such a dependence here).With the notation of that paper, let us explain briefly the necessary changes.
We recall that in [4] a vector v was given in B ′ δ (0, r) and the goal was to solve the equation F (u) = v.The solution u was the limit of a sequence u n of approximate solutions constructed inductively.Each u n was a solution of the projected equation Π . The existence of z n was proved by applying Theorem 1 to the function f n in a ball B Nn (0, R n ) (see [4] Instead, in order to prove Theorem 8 we construct inductively a sequence of continuous functions G where g n is a continuous right-inverse of f n such that g n (0) = 0, obtained thanks to Theorem 2.
Under the same conditions on the parameters as in [4], we find that the sequence of continuous functions (G n ) n converges uniformly on B ′ δ (0, r) for the norm • s 1 and this implies the continuity of their limit G : B ′ δ (0, r) → B s 1 .This limit is the desired continuous right inverse of F .We insist on the fact that the conditions on r are exactly the same as in [4].Indeed, in order to apply Theorem 2 to f n we just have to check assumptions (i), (ii ′ ) and (iii).This is done with exactly the same constraints on the parameters as in [4].
We end the paper with a uniqueness result, which requires additional conditions.Theorem 9. Suppose that we are under the assumptions of Theorem 8, and that the following two additional conditions hold true: • For each s ∈ [s 0 , S − m] and each c > 0 there exists a non-decreasing function ε s,c : (0, ∞) → (0, ∞) such that lim t→0 ε s,c (t) = 0 and, for all u 1 , u 2 in B s 0 +m ∩ E s+m with u 1 s+m ≤ c : (3.7) F (u 2 ) − F (u 1 ) − DF (u 1 )(u 2 − u 1 ) s ≤ ε s,c ( u 2 − u 1 s+m ) u 2 − u 1 s 0 +m .
In order to prove Theorem 9, we assume that G 1 , G 2 both satisfy (3.8) and we introduce the set Z := {v ∈ B ′ δ (0, r) | G 1 (v) = G 2 (v)}.This set is nonempty since it contains 0, and it is closed in B ′ δ (0, r) for the norm • ′ δ by continuity of G 1 − G 2 .It remains to prove that it is open.

Theorem 2 .
Let X, Y be two Banach spaces.Denote by B the open ball of radius R > 0 around the origin in X.Consider a map f : B → Y with f (0) = 0. We assume the following:(i) f is Lipschitz continuous and Gâteaux-differentiable on B.