Local Inversion for Differentiable Functions and Darboux Property

Jean Saint Raymond

Equipe d’Analyse – Boîte 186 Université Paris VI 4, place Jussieu F - 75252 Paris cedex 05, France E-mail : jsr@ccr.jussieu.fr

ABSTRACT : The main goal of this paper is to prove that the classical theorem of local inversion for �^{1} functions extends in finite dimension to everywhere differentiable functions. As usual a theorem of implicite functions can be deduced of this “Local Inversion Theorem”. The deepest part of the local inversion theorem consists to show that a differentiable function with non-vanishing Jacobian determinant is locally one-to-one. And in turn this fact allows us to extend the Darboux property of derivative functions on ℝ ( the range of the derivative is an interval) to the Jacobian function D_{f} of a differentiable function, under the condition that this Jacobian function does not vanish. We also prove that these results are no longer true in infinite dimension. These results should be known in whole or part, but I was not able to find references to a complete proof.

The main result of this paper is the following theorem which answers a question raised about lectures on Differential Calculus : “What about the classical Theorem of Local Inversion if the function is not assumed to be �^{1} but only differentiable ?”

I am grateful to D. Preiss who pointed out to me that the theorem of local inversion was already contained in papers about light and open maps by M. Cristea (see [3] and [4]), even with exceptional sets. Unfortunately these papers refer to the deep theory of discrete and open mappings, see also [5] and [6], and in order to employ this theory some preliminary steps are needed. I was not able to find a complete analytic proof of the result in the litterature, even in the simple case where the function is everywhere differentiable with non-zero Jacobian determinant. It is why I give the following proof, which is quite elementary. It could be useful to people who need such a proof to get a complete new proof of this theorem and a counterexample for the infinite-dimensional case.

It is an old and well-known result that a derivable real valued function on a real interval possesses the Darboux property (its range over any subinterval is an interval) and even has a connected graph in the plane. We prove here that this extends for the Jacobian determinant of a differentiable function from a connected open subset of ℝ^{n} to ℝ^{n} under the additional assumption that its Jacobian determinant does not vanish, and show by a counterexample that this extra hypothesis is necessary even for n = 2.

Whenever E and F are Banach spaces and T is a linear mapping from E to F we will denote by T h the image of an element h of E under T. In particular, if f : E → F is differentiable at some point x ∈ E we will use the notation f′(x)h for the image of h ∈ E under the differential of f at x.

Theorem 1. Let Ω be an open subset of ℝ^{n} and f a differentiable function from Ω to ℝ^{n}. Assume that the Jacobian determinant of f does not vanish on Ω. Then for every point x of Ω there are an open neighborhood V of x and an open neighborhood W of f(x) such that f be a homeomorphism from V onto W and that f^{−1} be differentiable on W.

We equip the space ℝ^{n} with the euclidean norm. In all the sequel we denote for r > 0, B(a, r) (resp. B̃(a, r), S(a, r)) the open ball (resp. the closed ball, the sphere) of center a and radius r. In the following lemmas and corollaries 2 – 11 we assume that the mapping f is as in Theorem 1.

Lemma 2. For every x in Ω, one can define a ρ(x) > 0 and a δ(x) > 0 such that : i) B̃(x, ρ(x)) ⊂ Ω.

ii) for all x′ ∈ B̃(x, ρ(x)), one has ∥f(x′) − f(x)∥ ≥ δ(x) ∥x′ − x∥

Let δ(x) = \frac{1}{2 ∥f′(x)^{−1}∥} and ε > 0 such that ε < δ(x). Since f is differentiable at x

there exists a ρ > 0 such that B̃(x, ρ) ⊂ Ω and that

∥x′ − x∥ ≤ ρ ⟹ ∥f(x′) − f(x) − f′(x)(x′ − x)∥ < ε ∥x′ − x∥

Then if ∥x′ − x∥ < ρ, one has

∥f(x′) − f(x)∥ ≥ ∥f′(x)(x′ − x)∥ − ∥f(x′) − f(x) − f′(x)(x′ − x)∥

≥ \frac{1}{∥f′(x)^{−1}∥} ∥x′ − x∥ − ε ∥x′ − x∥

≥ δ(x) ∥x′ − x∥

and this completes the proof. ■

The previous inequality allows to prove easily that f is open using for 0 ≤ r ≤ ρ(x) the index of the mapping H_{r} defined from S_{n−1} to S_{n−1} by :

H_{r}(u) = \{ \begin{matrix}\frac{f(x + ru) − f(x)}{∥f(x + ru) − f(x)∥} & if r > 0 \\ \frac{f′(x)u}{∥f′(x)u∥} & if r = 0 \end{matrix}

But in order to show later that f is locally one-to-one, we give an other proof which uses the differentiability of f.

Lemma 3. Let U be a bounded open set whose closure is contained in Ω and V a connected open subset of ℝ^{n}. Assume that V ∩ f(U) ≠ ∅ and that V ∩ f(∂U) = ∅. Then V ⊂ f(U).

Since V is connected, it is enough to show that the non-empty set f(U) ∩ V is clopen in V to prove that f(U) ∩ V = V, that means V ⊂ f(U). And for this we shall show that the boundary of V ∩ f(U) in V is empty.

Assume for contradiction that b^{∗} ∈ V ∩ ∂(f(U)) ; since b^{∗} ∈ V, there exists some η > 0 such that B(b^{∗}, 2η) ⊂ V, hence that B(b^{∗}, 2η) ∩ f(∂U) = ∅, and there exists since b^{∗} ∈ \overline{f(U)} some x ∈ U such that f(x) ∈ B(b^{∗}, η). Then if b ∈ B(b^{∗}, η) we put for z ∈ Ω :

φ_{b}(z) = ∥f(z) − b∥^{2}

Since the mapping u ↦ ∥u∥^{2} is �^{1}, it is clear that φ_{b} is differentiable. Moreover the differential of φ_{b} at z is the mapping :

h ↦ 2⟨f(z) − b, f′(z)h⟩

and φ′_{b}(z) is a zero mapping only if f(z) − b is orthogonal to the range of f′(z), thus if f(z) = b since one assumes that f′(z) is an isomorphism for all z ∈ Ω.

Since \overline{U} is compact in Ω, the continuous function φ_{b} attains its minimum over \overline{U} at some point z^{∗}, which cannot belong to ∂U, since, for all z of ∂U,

φ_{b}(z^{∗}) ≤ φ_{b}(x) ≤ (∥f(x) − b^{∗}∥ + ∥b − b^{∗}∥)^{2} < (2η)^{2} ≤ φ_{b}(z)

It results that z^{∗} belongs to U, thus that z^{∗} is a critical point of φ_{b}, that means f(z^{∗}) = b, hence b ∈ f(U).

We conclude that B(b^{∗}, η) ⊂ f(U) ∩ V, and this shows that b^{∗} does not belong to the boundary of f(U) ∩ V, in contradiction with the choice of b^{∗}. ■

Lemma 4. The function f is open.

Let x ∈ Ω. We shall show that if r ≤ ρ(x), we have f(B(x, r)) ⊃ B(f(x), δ(x).r). Let U = B(x, r) and V = B(f(x), δ(x).r). Since r ≤ ρ(x), \overline{U} is compact in Ω. We have f(x) ∈ f(U)∩V and following Lemma 2, f(∂U)∩B(f(x), δ(x).r) = ∅. It results by Lemma 3 that f(U) ⊃ V. ■

Lemma 5. Let x ∈ Ω and ε > 0. There exists r ≤ ρ(x) such that if ∥b − f(x)∥ < r, we have

B(x, ρ(x)) ∩ f^{−1}(b) ⊂ B(x + f′(x)^{−1}(b − f(x)), ε ∥b − f(x)∥)
Let z ∈ B(x, ρ(x)) such that b = f(z). We have
b = f(x) + f′(x)(z − x) + η(z)

where η(z) = o(z − x). Thus

f′(x)^{−1}(b − f(x)) = z − x + f′(x)^{−1}η(z)

it is

∥z − (x + f′(x)^{−1}(b − f(x)))∥ ≤ ∥f′(x)^{−1}∥ ∥η(z)∥ = \frac{∥η(x)∥}{2δ(x)}

There exists r_{1} > 0 such that, if ∥z − x∥ < r_{1} we have

∥η(z)∥ < 2δ(x)^{2}. ∥z − x∥

and if r = δ(x).r_{1}, we have ∥z − x∥ < r_{1} whenever z ∈ B(x, ρ(x)) ∩ f^{−1}(b)) and ∥b − f(x)∥ < r, thus

∥η(z)∥ < 2εδ(x)^{2}. ∥z − x∥ < 2εδ(x) ∥b − f(x)∥

and

∥z − (x + f′(x)^{−1}(b − f(x)))∥ < ε ∥b − f(x)∥

and this shows that every point of f^{−1}(b) ∩ B(x, ρ(x)) belongs to the ball of center x + f′(x)^{−1}(b − f(x)) and radius ε ∥b − f(x)∥ when ∥b − f(x)∥ < r. ■

Corollary 6. If f is one-to-one on the open set U ⊂ Ω, f is a homeomorphism from U onto the open set f(U), and, for all x in U, f^{−1} is differentiable at f(x), with differential f′(x)^{−1}.

Lemma 7. Let a ∈ Ω and b ∈ ℝ^{n}. If ∥f(a) − b∥ = η > 0, there exist h ∈ ℝ^{n}, λ > 0, t_{0} > 0 and a connected component U of f^{−1}(B(b, η)) such that

∥u − h∥ ≤ λ and 0 < t < t_{0} ⟹ a − tu ∈ U

In particular a ∈ \overline{U}.

Since f′(a) is invertible, there exists h such that f′(a)h = f(a) − b. Then if φ_{b} denotes the function : z ↦ ∥f(z) − b∥^{2}, we have φ′(a)h = 2⟨f(a) − b, f′(a)h⟩ = 2η^{2}. There exists λ > 0 such that ∥u − h∥ ≤ λ ⟹ φ′_{b}(a)u ≥ η^{2} and β > 0 such that

∥z − a∥ < β ⟹ |φ_{b}(z) − φ_{b}(a) − φ′_{b}(a)(z − a)| < \frac{η^{2}}{2(∥h∥ + λ)} ∥z − a∥

Then if t_{0} = \frac{β}{∥h∥ + λ}, we have, for 0 < t < t_{0} and ∥u − h∥ ≤ λ, ∥tu∥ ≤ t(∥h∥ + λ) < β and

\begin{aligned}φ_{b}(a − tu) & < φ_{b}(a) + tφ′_{b}(a)u + \frac{η^{2}}{2} \frac{t ∥u∥}{∥h∥ + λ} ≤ η^{2} + t(φ′_{b}(a)h + φ′_{b}(a).(u − h) + \frac{η^{2}}{2} ) \\ & ≤ η^{2}(1 − \frac{t}{2}) < η^{2} \end{aligned}

which shows that f(a − tu) ∈ B(b, η). And since f^{−1}(B(b, η)) contains the connected set C = {a − tu : 0 < t < t_{0} and ∥u − h∥ ≤ λ} , there exists a connected component U of f^{−1}(B(b, η)) which contains C. Then clearly a is in the closure of C, hence of U. ■

Lemma 8. Let U a connected component of f^{−1}(B(b, η)). Assume \overline{U} ⊂ Ω, a lies in the boundary of U and W is a neighborhood of a. Then f(U ∩ W) is a neighborhood of f(a) in B(b, η).

Let r < ρ(a) such that B(a, r) ⊂ W and put V = B(b, η)∩B(f(a), δ(a).r). The open set V is convex, hence connected. Since a ∈ \overline{U}, there exist points of U in f^{−1}(B(f(a), δ(a).r)), which shows that f(B(a, r) ∩ U) ∩ V ≠ ∅. Finally if z lies in the boundary of B(a, r) ∩ U, we have

- either ∥z − a∥ = r whence ∥f(z) − f(a)∥ ≥ δ(a).r and f(z) ∉ V

- or z ∈ ∂U, whence f(z) ∈ S(b, η) and f(z) ∉ V.

We deduce that V ∩ f(∂(B(a, r) ∩ U)) = ∅. Then it follows from Lemma 3 that V ⊂ f(B(a, r) ∩ U) ⊂ f(W ∩ U). ■

Lemma 9. For every b, the set f^{−1}(b) ∩ B̃(x, ρ(x)) is finite.

This set is compact since f^{−1}(b) is closed. Moreover Lemma 2 shows that every point z of f^{−1}(b) is isolated, since f^{−1}(b) ∩ B(z, ρ(z)) = {z}. ■

Lemma 10. Assume that 0 < ρ^{∗} < ρ, that ∥f(z) − w∥ ≥ δρ if ∥z − x∥ = ρ and that

f(B̃(x, ρ^{∗})) ⊂ B(w, \frac{δρ}{3}). Then f is one-to-one on B(x, ρ^{∗}).

Assume for contradiction that there exist two distinct points z_{1} and z_{2} of B(x, ρ^{∗}) such that b = f(z_{1}) = f(z_{2}). Denote by � the set of connected compact subsets of B̃(x, ρ^{∗}) containing both z_{1} and z_{2}. It is compact when equipped with Hausdorff metric, and nonempty since B̃(x, ρ^{∗}) ∈ � . The non-negative function Φ defined on � by

Φ(T) = sup_{z∈T} ∥f(z) − b∥^{2}

is continuous and cannot vanish because a compact connected set containing several points cannot be finite, hence cannot be contained in f^{−1}(b). Thus

η^{2} = inf_{T ∈ �} Φ(T) > 0

and this minimum is attained. Let K ∈ � such that Φ(K) = η^{2}. Then f(K) ⊂ B̃(b, η).

Let U be a connected component of f^{−1}(B(b, η)) which meets B(x, ρ). Then \overline{U} ⊂ Ω. Indeed if z ∈ B̃(x, ρ^{∗}), one has

∥f(z) − b∥ = ∥f(z) − f(z_{1})∥ ≤ ∥f(z) − w∥ + ∥f(z_{1}) − w∥ ≤ \frac{2}{3}δρ

thus

η^{2} ≤ Φ(B̃(x, ρ^{∗})) ≤ (\frac{2}{3}δρ)^{2}

If z ∈ U, one has

\begin{aligned}∥f(z) − w∥ & ≤ ∥f(z) − f(z_{1})∥ + ∥f(z_{1}) − w∥ ≤ η + ∥f(z_{1}) − w∥ \\ & ≤ η + \frac{1}{3}δρ < δρ \end{aligned}

from which we deduce that U ∩ S(x, ρ) = ∅ since ∥f(z) − w∥ ≥ δρ for all z of S(x, ρ). And since U is connected, one has U ⊂ B(x, ρ) thus \overline{U} ⊂ B̃(x, ρ) ⊂ Ω.

Since U ≠ ∅, we have f(U) ∩ B(b, η) ≠ ∅. Moreover f(∂U) ⊂ S(b, η). Then Lemma 3 applied to V = B(b, η) shows that b ∈ V ⊂ f(U), hence that U ∩ f^{−1}(b) ≠ ∅.

Since each connected component of f^{−1}(B(b, η)) meeting B(x, ρ) meets also f^{−1}(b), there are finitely many such connected components. If � denotes the set of these connected components, which are all open, it follows from Lemma 7 that every point of K belongs to the closure of some element of �. And since the connected compact set K is finitely covered by the closed sets (K ∩ \overline{U})_{U∈�} , those sets cannot be all disjoint. Hence there exists a ∈ K such that �_{1} = \{U ∈ � : a ∈ K ∩ \overline{U}\} contains several elements. If a did belong to some U ∈ �_{1}, it could not belong to the closure of another U′ ∈ � since if not U ∩ U′ would be non-empty and we would get U = U′. Hence ∥f(a) − b∥ = η.

It follows from Lemma 7 that exist h, λ, t_{0} and U_{0} ∈ �_{1} such that U_{0} contains C = {a − tu : 0 < t < t_{0} and ∥u − h∥ ≤ λ}. But since �_{1} has several elements, there exists some U_{1} ≠ U_{0} in �_{1}.

We deduce from Lemma 8 that for t > 0 small enough, the point y_{t} = f(a) + t(b − f(a)) has a preimage z_{t} in U_{1} ∩ B(a, ρ(a)), and from Lemma 5 that

∥z_{t} − (a + th)∥ = o(t)

Thus if t > 0 is chosen small enough in order to satisfy ∥z_{t} − (a + th)∥ < λt, we get

v = \frac{z_{t} − a}{t} ∈ B(h, λ)

hence z_{t} ∈ C ⊂ U_{0}, which shows that U_{0} ∩ U_{1} ≠ ∅, in contradiction with U_{0} ≠ U_{1} since they are connected components of f^{−1}(B(b, η)).

This contradiction completes the proof. ■

Corollary 11. The function f is one-to-one on a neighborhood of x.

It is enough to take w = f(x), ρ = ρ(x) and δ = δ(x) in the previous lemma, and one

can find some ρ^{∗} > 0 such that f(B̃(x, ρ^{∗})) ⊂ B (f(x), \frac{1}{3}δ(x)ρ(x)). ■

Proof of Theorem 1

The above corollary shows that every point x of Ω has a neighborhood V on which f is one-to-one. Corollary 6 shows that f is a homeomorphism from V onto the open set W = f(V) and that f^{−1} is differentiable on W. ■

It is also possible with the methods used above to prove a theorem of implicit functions.

Theorem 12. Let E be a Banach space, Ω an open subset of E ×ℝ^{n} and F a differentiable function from Ω to ℝ^{n}. Assume that F vanishes at the point (x_{0}, y_{0}) of Ω and that \frac{∂F}{∂y} (x, y) is invertible for all (x, y) of Ω. Then there exist a neighborhood W of y_{0} in ℝ^{n}, a neighborhood U of x_{0} in E and a differentiable function ψ from W to U such that U × W ⊂ Ω and that

F (x, y) = 0 ⟺ x = ψ(y)

holds for (x, y) ∈ U × W.

The standard method consists to look at the mapping : (x, y) ↦ (x, F (x, y)) from Ω to E × ℝ^{n} and apply to it the local inversion theorem. It works fine if E is finite-dimensional, but if we need to extend the result to the infinite-dimensional case, we have to be more careful.

It follows from Lemma 3 that if g is a differentiable function with values in ℝ^{n}, defined on a neighborhood of B̃(y_{0}, ρ), satisfying ∥g(y_{0})∥ < δρ and ∥g(y)∥ ≥ δρ for every y ∈ S(y_{0}, ρ), and if the differential of g is everywhere invertible, one has g(B(y_{0}, ρ)) ⊃ B(0, δρ).

Moreover, Lemma 10 shows that if in addition g(B̃(y_{0}, ρ^{∗})) is contained in B(0, \frac{δρ}{3}),

the function g is one-to-one on B̃(y_{0}, ρ^{∗}).

Since F is differentiable at (x_{0}, y_{0}) and since T = \frac{∂F}{∂y} (x_{0}, y_{0}) is a linear isomorphism of

ℝ^{n}, we can define S = \frac{∂F}{∂x} (x_{0}, y_{0}), δ = \frac{1}{3 ∥T^{−1}∥}, and find some ρ > 0 such that ∥x − x_{0}∥ ≤ ρ

and ∥y − y_{0}∥ ≤ ρ imply (x, y) ∈ Ω and :

∥F (x, y) − F (x_{0}, y_{0}) − S.(x − x_{0}) − T (y − y_{0})∥ ≤ δ(∥x − x_{0}∥ + ∥y − y_{0}∥)

It follows that if ∥x − x_{0}∥ ≤ ρ we get, by letting g_{x}(y) = F (x, y) :

\begin{aligned}∥g_{x}(y)∥ & = ∥F (x, y) − F (x_{0}, y_{0})∥ ≥ 3δ ∥y − y_{0}∥ − ∥S∥ ∥x − x_{0}∥ − δ(∥x − x_{0}∥ + ∥y − y_{0}∥) \\ & ≥ 2δ ∥y − y_{0}∥ − ∥x − x_{0}∥ (∥S∥ + δ) \end{aligned}

and that

∥g_{x}(y_{0})∥ = ∥F (x, y_{0}) − F (x_{0}, y_{0})∥ ≤ ∥S∥ ∥x − x_{0}∥ + δ ∥x − x_{0}∥

hence that if ∥x − x_{0}∥ < ρ_{1} = \frac{δρ}{∥S∥ + δ}, we have ∥g_{x}(y_{0})∥ < δρ and ∥g_{x}(y)∥ ≥ δρ for

y ∈ S(y_{0}, ρ). We deduce that g_{x}(B(y_{0}, ρ)) ⊃ B(0, δρ) thus that the equation g_{x}(y) = 0 possesses a solution in B(y_{0}, ρ) for all x in B(x_{0}, ρ_{1}). One sees in the same way that the equation g_{x}(y) = 0 possesses then a solution in W = B(y_{0}, ρ^{∗}) for all x in U = B(x_{0}, ρ_{1}^{∗})

where ρ_{1}^{∗} = \frac{δρ^{∗}}{∥S∥ + δ}. This solution ψ(x) is then unique.

Let ε ∈]0, δ[. There exists r_{0} < ρ^{∗} such that
∥F (x, y) − F (x_{0}, y_{0}) − S.(x − x_{0}) − T.(y − y_{0})∥ ≤ ε(∥x − x_{0}∥ + ∥y − y_{0}∥)

holds whenever ∥x − x_{0}∥ ≤ r_{0} and ∥y − y_{0}∥ ≤ r_{0}.

Then putting y_{1} = y_{0} − T^{−1}S(x − x_{0}) for x ∈ B(x_{0}, r_{0}), we get

∥F (x, y) − T (y − y_{1})∥ ≤ ε(∥x − x_{0}∥ + ∥y − y_{0}∥)

thus

\begin{aligned}∥g_{x}(y)∥ & ≥ 3δ ∥y − y_{1}∥ − ε(∥x − x_{0}∥ + ∥y − y_{1}∥ + ∥T^{−1}S∥ ∥x − x_{0}∥) \\ & ≥ 2δ ∥y − y_{1}∥ − ε(1 + ∥T^{−1}S∥) ∥x − x_{0}∥ \end{aligned}

and

∥g_{x}(y_{1})∥ < ε(∥x − x_{0}∥ + ∥y_{1} − y_{0}∥) ≤ ε(1 + ∥T^{−1}S∥) ∥x − x_{0}∥

We conclude that for all y such that ∥y − y_{1}∥ = r = \frac{ε}{δ}(1 + ∥T^{−1}S∥) ∥x − x_{0}∥ one has

∥g_{x}(y)∥ ≥ δr > ∥g_{x}(y_{1})∥

whence, by Lemma 3, 0 ∈ B(0, δr) ⊂ g_{x}(B(y_{1}, r)), that means ψ(x) ∈ B(y_{1}, r), and also

∥ψ(x) − ψ(x_{0}) + T^{−1}S(x − x_{0})∥ ≤ \frac{ε}{δ}(1 + ∥T^{−1}S∥) ∥x − x_{0}∥

which shows that ψ is differentiable at x_{0}, with differential −T^{−1}S, since ε is arbitrary. The same proof shows that if x_{1} ∈ U and y_{1} = ψ(x_{1}), there exists on a neighborhood de x_{1} an implicit function solving F (x, y) = 0, hence agreeing locally with ψ, differentiable at x_{1} and

that its differental at x_{1} is −\frac{∂F}{∂y} (x_{1}, y_{1})^{−1} \frac{∂F}{∂x} (x_{1}, y_{1}). Thus ψ is differentiable on U. ■

We can also deduce from the proof of Lemma 10 a simple proof of the following result, which extends in some way the Darboux property for derivable functions on ℝ :

Corollary 13. Let Ω be a connected open subset of ℝ^{n} and f be a differentiable mapping from Ω to ℝ^{n} whose Jacobian determinant D_{f} does not vanish on Ω. Then the sign of D_{f} is constant on Ω.

This result is already known and published with an other proof, but I can remember neither where I saw it, neither the title of the paper, nor the name of its author.

It is enough to show that the sign of D_{f} is constant in a neighborhood of each point a ofe Ω. And it follows from Lemma 10 that we can assume that Ω is the ball B(a, 2ρ) and that f is one-to-one. If ∥b − a∥ ≤ ρ, we let for t ∈ [0, 1],

\{ \begin{matrix}x(t) & = a + (3t^{2} − 2t^{3})(b − a) \\ r(t) & = t(1 − t)ρ \end{matrix}

It is easily checked that x(t) belongs to the segment [a, b], that \frac{dx}{dt}(0) = \frac{dx}{dt}(1) = 0,

and that r(t) ≤ \frac{ρ}{4}, hence that if ∥z − x(t)∥ = r(t) one has z ∈ B(a, 2ρ). Finally for S_{n−1} = {u : ∥u∥ = 1}, we define a mapping H : S_{n−1} × [0, 1] → S_{n−1} by :

H(u, t) = \{ \begin{matrix}\frac{f(x(t) + r(t).u) − f(x(t))}{∥f(x(t) + r(t).u) − f(x(t))∥} & if 0 < t < 1 \\ \frac{f′(a)u}{∥f′(a)u∥} & if t = 0 \\ \frac{f′(b)u}{∥f′(b)u∥} & if t = 1 \end{matrix}

The mapping H is clearly continuous on S_{n−1}×]0, 1[ since f is one-to-one and r(t) > 0. Moreover, when t tends to 0, one has x(t) = a + o(t) and r(t) = t.ρ + o(t), thus :

f(x(t) + r(t).u) − f(x(t)) = (f(x(t) + r(t).u) − f(a)) − (f(x(t)) − f(a))

= (f′(a)(r(t).u + x(t) − a) + o(t)) − (f′(a)(x(t) − a) + o(t))

= tρ(f′(a)u + o(1))

uniformly in u ∈ S_{n−1}, from which we deduce that H is continuous at each point of S_{n−1} × {0}. It is shown in the same way that H is continuous at each point of S_{n−1} × {1}.

It follows that the mappings from S_{n−1} to S_{n−1} : u ↦ \frac{f′(a)u}{∥f′(a)u∥} and u ↦ \frac{f′(b)u}{∥f′(b)u∥} are

homotopic, and this is possible only if the determinants of the linear mappings f′(a) and f′(b) have the same sign, it is only if D_{f}(a) and D_{f}(b) have the same sign. ■

Nevertheless, without hypotheses on the Jacobian determinant D_{f}, it is not true that the set D_{f}(Ω) be connected if the dimension exceeds 1, as shown by the following example.

Example 14. There are a connected open subset Ω of ℝ^{2} and a differentiable function F : U → ℝ^{2} such that the Jacobian determinant D_{F} takes only two values.

Define U = ℝ×]0, +∞[, F (0, y) = 0 for y > 0 and, for (x, y) ∈ ℝ^{∗}×]0, +∞[,

F (x, y) = ( \begin{matrix}x^{2}\sqrt{y} cos \frac{1}{x^{3}} \\ x^{2}\sqrt{y} sin \frac{1}{x^{3}} \end{matrix} )

It is easily seen that F is differentiable and that the Jacobian matrix of F at (x, y) is 0 if x = 0 and if not

J_{F} (x, y) = ( \begin{matrix}2x\sqrt{y} cos \frac{1}{x^{3}} + 3\frac{\sqrt{y}}{x^{2}} sin \frac{1}{x^{3}} & \frac{x^{2}}{2\sqrt{y}} cos \frac{1}{x^{3}} \\ 2x\sqrt{y} sin \frac{1}{x^{3}} − 3\frac{\sqrt{y}}{x^{2}} cos \frac{1}{x^{3}} & \frac{x^{2}}{2\sqrt{y}} sin \frac{1}{x^{3}} \end{matrix} )

and we get D_{F} (x, y) = \frac{3}{2} if x ≠ 0 but D_{F} (0, y) = 0. ■

We now show that the previous phenomenon disappears if the Jacobian determinant does not vanish, and that the Jacobian determinant satisfies then the “intermediate values property”.

For any Borel subset X of ℝ^{n} we will denote by vol(X) its Lebesgue measure. The following lemmas should be well known.

Lemma 15. Let U be an open subset of ℝ^{n} and f a differentiable one-to-one function : U → ℝ^{n} with positive Jacobian determinant. If A is a bounded convex open neighborhood of 0 in ℝ^{n}, x ∈ U, (x_{p}) a sequence converging to x and (t_{p}) a sequence of positive real numbers converging to 0 such that ∥x − x_{p}∥ = O(t_{p}), then

lim_{p→∞} \frac{vol(f(x_{p} + t_{p}A))}{vol(x_{p} + t_{p}A)} = lim_{p→∞} t_{p}^{−n} \frac{vol(f(x_{p} + t_{p}A))}{vol(A)} = D_{f}(x)

Let ε > 0. Since f′(x)(A) := {f′(x)h : h ∈ A } is a bounded open convex set there are M and a positive η such that ∥u∥ ≤ M for all u ∈ A and

z ∈ f′(x).(∂A) and ∥z′ − z∥ < η ⟹ z′ ∈ ((1 + ε)f′(x)(A)) \ ((1 − ε)f′(x)(A))

The number M can also be chosen in such a way that ∥x − x_{p}∥ ≤ Mt_{p}. Since f is differentiable at x, there is a r > 0 such that B̃(x, r) ⊂ U and

∥f(y) − f(x) − f′(x)(y − x)∥ ≤ \frac{η}{2M} ∥y − x∥

for ∥y − x∥ < r. Since A is bounded it is easily seen that for n large enough the set x_{p} + t_{p}A is contained in B(x, r). Then if y ∈ x_{p} + t_{p}A we have y = x_{p} + t_{p}u with u ∈ A and

∥y − x∥ ≤ ∥x − x_{p}∥ + t_{p} ∥u∥ ≤ 2Mt_{p}

∥f(y) − (f(x) + f′(x)(x − x_{p})) − t_{p}f′(x)u∥ ≤ \frac{η}{2M} ∥y − x∥ ≤ ηt_{p}

Thus, with b_{p} = f(x) + f′(x)(x − x_{p}), we have

∥\frac{f(y) − b_{p}}{t_{p}} − f′(x)u∥ ≤ η

thus f(y) ∈ b_{p} + t_{p}(1 + ε)f′(x)(A), and

vol(f(x_{p} + t_{p}A)) ≤ t_{p}^{n}(1 + ε)^{n} vol(f′(x)(A)) = t_{p}^{n}(1 + ε)^{n}D_{f}(x) vol(A)

Simillarly by Lemma 3 it is seen that f(x_{p} + t_{p}A) ⊃ b_{p} + t_{p}(1 − ε)f′(x).A and thus that

vol(f(x_{p} + t_{p}A)) ≥ t_{p}^{n}(1 − ε)^{n} vol(f′(x)(A)) = t_{p}^{n}(1 − ε)^{n}D_{f}(x) vol(A)

We conclude that

(1 − ε)^{n}D_{f}(x) ≤ \frac{vol(f(x_{p} + t_{p}A))}{vol(x_{p} + t_{p}A)} ≤ (1 + ε)^{n}D_{f}(x)

and this completes the proof since ε is arbitrary. ■

Lemma 16. Let U be an open subset of ℝ^{n}, f a differentiable one-to-one function : U → ℝ^{n} with positive Jacobian determinant and Q a compact cube. Then if we denote by U_{Q} the open set {x ∈ ℝ^{n} : x + Q ⊂ U }, the function m defined on U_{Q} by m(x) = vol(f(x + Q)) is continuous.

Let μ the Borel measure on U defined by μ(A) = vol(f(A)). The classical theorem of differentiation for Borel measures on ℝ^{n} (see [7]) asserts that μ is absolutely continuous since denoting by ν the singular part of μ, the derivative of μ with respect the Lebesgue measure should be +∞ ν-a.e. and that this derivative at x is equal to D_{f}(x) < +∞ following the previous lemma.

Since the boundary of x+Q is μ-null, Lebesgue’s dominated convergence theorem shows that for every sequence x_{p} converging to x ∈ U_{Q} we have

m(x_{p}) = ∫_{U} 1_{x_{p}+Q} dμ(x) → ∫_{U} 1_{x+Q} dμ(x) = m(x)

which proves the continuity of m. ■

Lemma 17. Let U be an open subset of ℝ^{n}, f a differentiable one-to-one function : U → ℝ^{n} with positive Jacobian determinant and Q a compact cube contained in U. Then there is a point x^{∗} ∈ Q such that

D_{f}(x^{∗}) = \frac{vol(f(Q))}{vol(Q)}

We construct inductively a decreasing sequence (Q_{j}) of cubes such that Q_{0} = Q, diam(Q_{j}) = 2^{−j} diam(Q) and

\frac{vol(f(Q_{j}))}{vol(Q_{j})} = \frac{vol(f(Q))}{vol(Q)}

Suppose Q_{j} = ∏_{i=1}^{n}[a_{i}, a_{i} + 2h] is defined and consider, for α ∈ {0, 1}^{n} the cubes

Q^{(α)} = ∏_{i=1}^{n}[a_{i} + α_{i}h, a_{i} + (1 + α_{i})h]

Then these cubes have the same volume h^{n}, cover Q_{j} and are almost disjoint. It follows that

vol(f(Q_{j})) = ∑_{α∈{0,1}^{n}} vol(f(Q^{(α)})

and that

\frac{vol(f(Q_{j}))}{vol(Q_{j})} = \frac{1}{2^{n}} ∑_{α∈\{0,1\}^{n}} \frac{vol(f(Q^{(α)}))}{vol(Q^{(α)})}

Hence

inf_{α∈\{0,1\}^{n}} \frac{vol(f(Q^{(α)}))}{vol(Q^{(α)})} ≤ \frac{vol(f(Q_{j}))}{vol(Q_{j})} ≤ sup_{α∈\{0,1\}^{n}} \frac{vol(f(Q^{(α)}))}{vol(Q^{(α)})}

and the continuity of the function m : x ↦ vol(x + Q′), for Q′ = ∏_{i=1}^{n}[0, h], shows that there is a point x_{j} ∈ ∏_{i=1}^{n}[a_{i}, a_{i} + h] such that m(x_{j}) = \frac{vol(f(Q_{j}))}{vol(Q_{j})}. Then Q_{j+1} = x_{j} + Q′ satisfies the requirement.

The decreasing sequence of compact cubes (Q_{j}) has intersection {x^{∗}}, for some x^{∗} ∈ Q. And Lemma 15 shows that \frac{vol(f(Q_{j}))}{vol(Q_{j})} → D_{f}(x^{∗}). But since \frac{vol(f(Q_{j}))}{vol(Q_{j})} = \frac{vol(f(Q))}{vol(Q)} for all j we are done. ■ Lemma 18. Let U be a connected open subset of ℝ^{n}, f a differentiable one-to-one function : U → ℝ^{n} with non-vanishing Jacobian determinant, and B̃(a, r) a closed ball contained in U. Then the set {D_{f}(x) : x ∈ B̃(a, r) } is an interval.

By Theorem 13, the sign of D_{f} is constant on U, and up to composing with a symmetry, we can assume it is positive. It is enough to prove that if x ∈ B̃(a, r) and θ is strictly between D_{f}(a) and D_{f}(x), there is some x^{∗} ∈ B(a, r) such that D_{f}(x^{∗}) = θ.

Let Q = [−1, 1]^{n} the unit cube in ℝ^{n}. There is a λ > 0 such that λQ ⊂ B(0, 1). Then, for x ∈ B(a, r) and 0 < t < 1, we have x + t((a − x) + λrQ) ⊂ B(a, r). Thus with

t_{p} = \frac{λr}{p + 1} and x_{p} = x + t_{p} \frac{a − x}{λr}, we have x_{p} → x, x_{p} + t_{p}Q ⊂ B(a, r), ∥x_{p} − x∥ = O(t_{p})

and a + t_{p}Q ⊂ B(a, r). Then by Lemma 15

lim_{p→∞} \frac{vol(f(x_{p} + t_{p}Q))}{vol(x_{p} + t_{p}Q)} = D_{f}(x) and lim_{p→∞} \frac{vol(f(a + t_{p}Q))}{vol(a + t_{p}Q)} = D_{f}(a)

And there is a p such that θ is strictly between \frac{vol(f(x_{p} + t_{p}Q))}{vol(x_{p} + t_{p}Q)} and \frac{vol(f(a + t_{p}Q))}{vol(a + t_{p}Q)}. By

continuity of the mapping x ↦ \frac{vol(f(x + t_{p}Q))}{vol(x + t_{p}Q)}, one can find a x′ on the segment [a, x_{p}]

such that θ = \frac{vol(f(x′ + t_{p}Q))}{vol(x′ + t_{p}Q)}. The convexity of B(a, r) implies that x′ + t_{p}Q ⊂ B(a, r), and then Lemma 17 proves there exists a x^{∗} ∈ x′ + t_{p}Q ⊂ B(a, r) such that D_{f}(x^{∗}) = θ. ■

We now are able to give the following extension of the Darboux property for the Jacobian function.

Theorem 19. Let Ω be a connected open subset of ℝ^{n} and f a differentiable function : Ω → ℝ^{n} with non-vanishing Jacobian determinant. Then the set D_{f}(Ω) = {D_{f}(x) : x ∈ Ω } is an interval.

It is enough to show that if x_{1} and x_{2} are points of Ω and D_{f}(x_{1}) < θ < D_{f}(x_{2}) there is some x ∈ Ω such that D_{f}(x) = θ. Let � the collection of all balls B(y, r) whose closure is contained in Ω and on which f is one-to-one. By Lemma 10, � is an open cover of Ω, and it is easily checked that the set

X = {x ∈ Ω : ∃(B_{j})_{0≤j≤ℓ} in � such that x_{1} ∈ B_{0}, x ∈ B_{ℓ}, B_{j} ∩ B_{j+1} ≠ ∅}

is clopen and contains x_{1}. Thus X = Ω and x_{2} ∈ X. Then J_{j} = D_{f}(B_{j}) is an interval and J_{j} ∩ J_{j+1} ≠ ∅. We conclude that J = ⋃_{0≤j≤ℓ} D_{f}(B_{j}) is an interval containing D_{f}(x_{1}) and D_{f}(x_{2}), hence θ.

We can even prove the following extension of a one-dimensional well-known result.

Theorem 20. Let Ω be a connected open subset of ℝ^{n} and f a differentiable function : Ω → ℝ^{n} with non-vanishing Jacobian determinant. Then the graph Γ of D_{f} in Ω × ℝ is connected.

Assume by contradiction that Γ is not connected. There would exist a clopen subset X of Γ with X ≠ ∅ and X ≠ Γ. Denote by π the projection (x, t) ↦ x from Γ onto Ω and by Y the set π(X). Since Ω is connected, the boundary ∂Y of Y is a closed non-empty subset of Ω. Since the mapping x ↦ D_{f}(x) is of first Baire class, the sets Y = {x ∈ Ω : (x, D_{f}(x)) ∈ X } and Z = {x ∈ Ω : (x, D_{f}(x)) ∉ X } are F_{σ}-sets. There are two sequences (A_{p}) and (B_{p}) of closed subsets of Ω such that Y = ⋃_{p} A_{p} and Z = ⋃_{p} B_{p}. We then have

∂Y = ⋃_{p}(A_{p} ∩ ∂Y) ∪ ⋃_{p}(B_{p} ∩ ∂Y)

And by Baire’s Category Lemma, there exists an open set W such that W ∩ ∂Y is nonempty and contained in some of these sets. Up to exchanging X and Γ \ X we can assume W ∩ ∂Y ⊂ A_{p} ∩ ∂Y.

Let a ∈ ∂Y ∩ W ⊂ A = π(X). There are η > 0 and a ball B(a, r) contained in W, on which f is one-to-one (by Lemma 10). Since a ∈ ∂Y, there is some b ∈ Z ∩ B(a, \frac{r}{3}).

Then δ = d(b, ∂Y) ≤ ∥b − a∥ < \frac{r}{3}. Since B(b, δ) is connected and disjoint from ∂Y, we have B(b, δ) ⊂ Z, while S(b, δ) ∩ ∂Y ≠ ∅. For each a′ ∈ S(b, δ) ∩ ∂Y, we have

∥a′ − a∥ ≤ ∥b − a∥ + ∥a′ − b∥ < 2\frac{δ}{3}, hence a′ ∈ W ∩ ∂Y ⊂ Y. Since (a′, D_{f}(a′)) ∈ X

there are ρ ≤ δ and η > 0 such that

Γ ∩ (B(a′, 2ρ)×]D_{f}(a′) − η, D_{f}(a′) + η[) ⊂ X

Applying Lemma 18 to B̃(b′, ρ) where b′ = a′ + \frac{ρ}{δ}(b − a′) since B(b′, ρ) ⊂ B(b(, δ) ⊂ Z and

a′ ∈ B̃(b′, ρ), we get that D_{f}(B(b′, ρ)) contains the interval [D_{f}(b′), D_{f}(a′)[ ; thus there is a point x ∈ B(b′, ρ) such that D_{f}(x) ∈]D_{f}(a′) − η, D_{f}(a′) + η[. And then

(x, D_{f}(x)) ∈ Γ ∩ (B(a′, 2ρ)×]D_{f}(a′) − η, D_{f}(a′) + η[) ⊂ X

although x ∈ B(b′, ρ) ⊂ Z. This contradiction completes the proof. ■

The proofs above use in a heavy way the local compactness of the space E. In fact this is needed since the result of Theorem 1 cancels if E has infinite dimension as shown by the following example.

Example 21. There exists on the Hilbert space ℓ^{2} a Fréchet-differentiable function F whose differential is everywhere invertible but such that F is neither open nor locally one-to-one.

We shall prove several lemmas. We will denote in the sequel by P the orthogonal projection from ℓ^{2} on the hyperplane {x ∈ ℓ^{2} : x_{0} = 0 } : P x = (0, x_{1}, x_{2}, …) for x = (x_{0}, x_{1}, x_{2}, …) and P̌ = Id − P. We shall alse denote by Σ the set

\{x = (x_{n}) ∈ ℓ^{2} : x_{0} = 0 and ∥x∥ ≤ \frac{1}{2} \}

Lemma 22. There exists a �^{1}-diffeomorphism Φ from ℓ^{2} onto ℓ^{2} \ Σ such that Φ(x) = x if ∥x∥ ≥ 1.

A classical theorem (see [2], [1] and the references inside this last) asserts that there exists a �^{1}-diffeomorphism Ψ from ℓ^{2} onto ℓ^{2} \ {0} such that Ψ(x) = x if ∥x∥ ≥ 1. We will prove the existence of a �^{1}-diffeomorphism Ψ_{1} from ℓ^{2} \Σ onto ℓ^{2} \{0} wich agrees with the identity outside the unit ball. Then Φ = Ψ_{1}^{−1}∘Ψ will satisfy the condition of the statement.

Let ρ be the function defined on ℝ by

ρ(t) = \{ \begin{matrix}0 & if t ≤ 0 \\ 1 & if t ≥ 1 \\ 3t^{2} − 2t^{3} & if 0 ≤ t ≤ 1 \end{matrix}

Since ρ′(t) = 6t(1 − t) for 0 < t < 1, ρ is clearly non-decreasing, positive, bounded by 1 and �^{1}. Define further, for s and t real

ψ(s, t) = ρ(2s) + (1 − ρ(2s))ρ(4t − 1)

Since 1 − ψ(s, t) = (1 − ρ(2s))(1 − ρ(4t − 1)), it is easily checked that ψ is �^{1} and positive,

that ψ(s, t) > 0 if s > 0 or if t > \frac{1}{4} and also that ψ(s, t) = 1 if s > \frac{1}{2} or if t > \frac{1}{2}.

Then put, for x ∈ ℓ^{2},
Ψ_{1}(x) = P̌x + ψ(∥P̌x∥^{2} , ∥P x∥^{2}).P x

It is clear that Ψ_{1} is �^{1} and that Ψ_{1}(x) = x if ∥P̌x∥^{2} ≥ \frac{1}{2} or ∥P x∥^{2} ≥ \frac{1}{2}, thus whenever

∥x∥ ≥ 1. Furthermore if x ∉ Σ, one has ψ(∥P̌x∥^{2} , ∥P x∥^{2}) > 0, thus Ψ_{1}(x) ≠ 0.

We show that Ψ_{1} is a bijection from ℓ^{2} \ Σ onto ℓ^{2} \ {0}. Let y ≠ 0 and consider the equation Ψ_{1}(x) = y. Then we need P̌x = P̌y and

\{ \begin{matrix}P x & = \frac{1}{ψ(s, t)}.P y \\ t & = ∥P x∥^{2} \end{matrix}

with s = ∥P̌y∥^{2} and t = ∥P x∥^{2}.

That means that by letting H(t, y) = tψ(∥P̌y∥^{2} , t)^{2} − ∥P y∥^{2}, we need H(t, y) = 0. Since

\frac{∂H}{∂t} (t, y) = ψ(s, t)^{2} + 2tψ(s, t)(1 − ρ(2s)).ρ′(4t − 1) ≥ ψ(s, t)^{2}

the function t ↦ H(t, y) is increasing on [0, +∞[ if P̌y ≠ 0 and on [\frac{1}{2}, +∞[ if P̌y = 0. If

P̌y = 0 we have P y = y ≠ 0, hence H(\frac{1}{2}, y) < 0. And if P̌y ≠ 0 we have H(0, y) ≤ 0. Since lim_{t→+∞} H(t, y) = +∞, there exists a unique t = τ(y) ≥ 0 such that H(t, y) = 0, and one

has τ(y) > \frac{1}{2} if P̌y = 0. This shows that Ψ_{1} is a bijection from ℓ^{2} \ Σ onto ℓ^{2} \ \{0\}. We

always have ψ(∥P̌y∥^{2} , τ(y)) > 0, hence

\frac{∂H}{∂t} (∥P̌y∥^{2} , τ(y)) ≥ ψ(∥P̌y∥^{2} , τ(y)) > 0

and we can then deduce from the Implicit Functions Theorem that y ↦ τ(y) is �^{1} on ℓ^{2} \ {0}. Finally since

Ψ_{1}^{−1}(y) = P̌y + \frac{1}{ψ(∥P̌y∥^{2} , τ(y))}.P y

we conclude that Ψ_{1}^{−1}is �^{1} and that Ψ_{1} is a �^{1}-diffeomorphism. ■

Lemma 23. There exists a �^{1} mapping Ψ_{2} from ℓ^{2} \ Σ into ℓ^{2}, whose differential is everywhere invertible, which agrees with the identity outside the unit ball, and which is neither one-to-one nor onto.

Denote by h the �^{1} non-negative function from ℝ to ℝ defined by

h(w) = \{ \begin{matrix}0 & if w ≥ \frac{1}{4} \\ (1 − 4w)^{2} & if w ≤ \frac{1}{4} \end{matrix}

and by φ the function defined for λ and t real by

φ(λ, t) = \{ \begin{matrix}0 & if t = 0 \\ t − λ ∫_{t}^{1/4} h(w) dw & if t > 0 \\ −φ(λ, −t) & if t < 0 \end{matrix}

For |t| ≥ \frac{1}{4} and any λ, we have φ(λ, t) = t. And for λ = 0, we have φ(0, t) = t for all t.

Moreover φ is �^{1} on {(λ, t) : t ≠ 0 }.

Finally for x = (x_{n}) we define Ψ_{2}(x) = y = (y_{n}) by

\{ \begin{matrix}y_{0} & = φ(h(∥P x∥^{2}), x_{0}) \\ P y & = P x \end{matrix}

It is clear that Ψ_{2} is �^{1} on \{x ∈ ℓ^{2} : x_{0} ≠ 0 \}, and also on \{x ∈ ℓ^{2} : ∥P x∥ > \frac{1}{2} \} because, on

these sets, it agrees with the identity. So Ψ_{2} is �^{1} on ℓ^{2} \ Σ.

Furthermore if |x_{0}| ≥ \frac{1}{4}, we have φ(h(∥P x∥^{2}), x_{0}) = x_{0}, thus Ψ_{2}(x) = x. This shows

that Ψ_{2} agrees with the identity outside the set \{x : |x_{0}| ≤ \frac{1}{4} and ∥P x∥ ≤ \frac{1}{2} \}, hence in particular, outside the unit ball.

If y satisfies y_{0} = 0 and ∥y∥ = \frac{1}{2}, there cannot be any x ∈ ℓ^{2} \ Σ such that Ψ_{2}(x) = y :

such a x should satisfy P x = y, hence 0 = y_{0} = φ(h(∥P x∥^{2}), x_{0}) = x_{0}, it is x = y ∈ Σ. Thus Ψ_{2}(ℓ^{2} \ Σ) ≠ ℓ^{2}.

Finally

lim_{t→0,t>0} φ(1, t) = −\frac{1}{12} < 0 < φ(1, \frac{1}{4}) = \frac{1}{4}

Thus there exists a γ ∈]0, \frac{1}{4}[ such that φ(1, γ) = 0. And, for a = (γ, 0, 0, …), we have a ≠ −a and Ψ_{2}(a) = Ψ_{2}(−a) = 0, which shows that Ψ_{2} is not one-to-one.

It remains to show that the differential of Ψ_{2} is invertible, it is bijective, at each point of ℓ^{2} \ Σ ; we have thus to prove that the equation Ψ′_{2}(x)u = v has a unique solution u for

all v ∈ ℓ^{2} and all x ∈ ℓ^{2} \ Σ. Since Ψ′_{2} agrees with the identity if |x_{0}| > \frac{1}{4} or if ∥P x∥ > \frac{1}{2}

and that Ψ′_{2}(x) = Ψ′_{2}(−x), we are led to the case where ∥P x∥ ≤ \frac{1}{2} and 0 < x_{0} ≤ \frac{1}{4}. Then if u = (u_{n}) and v = Ψ′_{2}(x)u = (v_{n}), we have P v = P u and

v_{0} = 2\frac{∂φ}{∂λ}(h(∥P x∥^{2}), x_{0})h′(∥P x∥^{2})⟨P x, u⟩ + \frac{∂φ}{∂t}(h(∥P x∥^{2}), x_{0})u_{0}

and we get

u_{0} = \frac{1}{1 + (1 − 4 ∥P x∥^{2})(1 − 4x_{0})} (v_{0} + \frac{(1 − 4x_{0})^{3}(1 − 4 ∥P x∥^{2})}{6}⟨P x, v⟩)

wich proves the existence and the uniqueness of u, and completes the proof of the lemma. ■

Corollary 24. There exists a �^{1} mapping f from ℓ^{2} into itself, which agrees with the identity outside the unit ball, whose differential is everywhere invertible, and wich is neither one-to-one nor onto.

It is enough to put f = Ψ_{2}∘Φ, where the functions Φ and Ψ_{2} have the properties as in Lemmas 22 and 23.

Construction of Example 21

Let A be a closed non-empty and nowhere dense subset of ℓ^{2} (e.g. {0}). There exists then a sequence (a_{n}) of points of A of which each point of A is a cluster point. And one can find a sequence (b_{n}) of mutually distinct points of ℓ^{2} \ A such that ∥b_{n} − a_{n}∥ < 2^{−n}, then a sequence (r_{n}) of positive real numbers such that :

i) r_{n} < 2^{−n}d(b_{n}, A)

ii) B(b_{n}, r_{n}) ∩ B(b_{p}, r_{p}) = ∅ for n ≠ p.

Then we define the function F : ℓ^{2} → ℓ^{2} by

F (x) = \{ \begin{matrix}b_{n} + r_{n}f(\frac{x − b_{n}}{r_{n}}) & if ∥x − b_{n}∥ < r_{n} \\ x & if x ∈ ℓ^{2} \ ⋃_{n} B(b_{n}, r_{n}) \end{matrix}

Since the family of balls B(b_{n}, r_{n}) is locally finite outside A, it is clear that F is �^{1} on ℓ^{2} \A,

that Ψ′(x) is the identity whenever x ∈ ℓ^{2} \ (A ∪ ⋃_{n} B(b_{n}, r_{n})) and that Ψ′(x) = f′(\frac{x − b_{n}}{r_{n}})

whenever ∥x − b_{n}∥ < r_{n}. Finally, if a ∈ A, we have :

F (x) − F (a) − (x − a) = F (x) − x = \{ \begin{matrix}r_{n} (f(\frac{x − b_{n}}{r_{n}}) − \frac{x − b_{n}}{r_{n}} ) & if x ∈ B(b_{n}, r_{n}) \\ 0 & if x ∉ ⋃_{n} B(b_{n}, r_{n}) \end{matrix}

When x ∈ B(b_{n}, r_{n}), one has

2^{n}r_{n} ≤ d(b_{n}, A) ≤ ∥b_{n} − a∥ ≤ ∥x − a∥ + r_{n}

thus

r_{n} < \frac{1}{2^{n} − 1} ∥x − a∥

We then have : ∥\frac{x − b_{n}}{r_{n}} ∥ < 1 and ∥f(\frac{x − b_{n}}{r_{n}})∥ < 1, hence

∥F (x) − F (a) − (x − a)∥ ≤ 2r_{n} ≤ \frac{2}{2^{n} − 1} ∥x − a∥

For all ε > 0, there exists an integer n_{0} such that \frac{2}{2^{n_{0}} − 1} < ε.

Then W = ℓ^{2} \ ⋃_{n<n_{0}} B̃(b_{n}, r_{n}) is a neighborhood of a and we have
∥F (x) − F (a) − (x − a)∥ ≤ ε ∥x − a∥

on W, which shows that F is differentiable at a, with F′(a) = Id. This completes the proof that F is differentiable and that F′(x) is invertible for all x in ℓ^{2}.

If a is a point of A and V a neighborhood of a, there is some n such that B(b_{n}, r_{n}) ⊂ V. and since f is not one-to-one on the unit ball, there are two distinct points x′ and x′′ such that f(x′) = f(x′′) ; then the points b_{n} +r_{n}x′ and b_{n} +r_{n}x′′ are two distinct points of V and one has F (b_{n} + r_{n}x′) = F (b_{n} + r_{n}x′′). One deduces from this that F is locally one-to-one at the neighborhood of no point of A.

Finally, if a ∈ A, one has a = F (a) ∈ F (ℓ^{2}). Since ψ is not onto, there exists some y′ such that y′ ∉ ψ(ℓ^{2}) ; and since f agrees with the identity outside the unit ball, we have ∥y′∥ < 1. There exists then a sequence (n_{k}) such that a = lim_{k→∞} b_{n_{k}}. Notice that if F (x) ∈ B(b_{n}, r_{n}), we necessarily have x ∈ B(b_{n}, r_{n}). If the point b_{n_{k}} + r_{n_{k}}y′ did not belong to F (ℓ^{2}), it would exist x′_{k} in the unit ball such that b_{n_{k}} + r_{n_{k}}y′ = b_{n_{k}} + r_{n_{k}}f(x′_{k}), hence that y′ = f(x′_{k}) ∈ f(ℓ^{2}). We then see that the points b_{n_{k}} + r_{n_{k}}y′ converge to a and do not belong to F (ℓ^{2}) ; thus a is not an interior point of F (ℓ^{2}) and F is not an open mapping. ■

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