OPEN DIFFERENTIABLE MAPPINGS

JEAN SAINT RAYMOND

Abstract. It is a well-known result that a C^{1}-mapping defined on an open subset Ω of a Banach space E with values in a Banach space F whose differential is everywhere onto is an open mapping from Ω onto an open subset of F. We prove that this result still holds if E is finite-dimensional and both the critical set and the set of critical values have small topological dimensions. The restriction on the set of critical values can be removed if E and F have same dimension.

Let Ω be an open subset of the Banach space E, F be a Banach space and L(E, F) denote the Banach space of continuous linear mappings from E to F. Let f : Ω → F be a C^{1} function. A point a ∈ Ω is said to be critical if the differential f′(a) ∈ L(E, F) is not onto. We call critical set of f the set Γ of critical points of f, and a point b ∈ F is said to be a critical value of f if there is some critical point a ∈ Ω such that f(a) = b. It is a well-known fact that the set {u ∈ L(E, F) : u(E) = F } of continuous onto linear mappings is open in L(E, F) ; so the set Γ is closed in Ω.

If V is a connected open set in ℝ^{n} we say that a closed set Z ⊂ ℝ^{n} cuts V if the open set V \ Z is not connected.

Theorem 1. Let Ω be an open subset of ℝ^{n} and φ : Ω → ℝ^{p} (2 ≤ p ≤ n) a C^{1} function. Assume that the set Γ of critical points of φ is 0-dimensional and that the set φ(Γ) of critical values of φ has dimension ≤ p − 2. Then φ is an open mapping.

Remark first that such a result cannot hold when n = p = 1 without the hypothesis on the set of critical values. The C^{∞} function q : x ↦ x^{2} has a unique critical point 0, and a unique critical value 0, but is not open : indeed q(ℝ) = ℝ^{+} is not a neighborhood of q(0) = 0 in ℝ. Nevertheless for p = 1 if the topological dimension of φ(Γ) is −1, the set φ(Γ) has to be empty, so φ has no critical point and the above statement reduces to the standard theorem.

Up to replacing Ω by an open subset of Ω, it is enough to prove that φ(Ω) is open in ℝ^{p}. The classical “Local Inversion Theorem” (or a variant of it) shows that the set φ(Ω \ Γ) of regular values is open in ℝ^{p}. So we have only to show that φ(Ω) is a neighborhood of every point b ∈ φ(Ω) such that Ω ∩ φ^{−1}(b) ⊂ Γ.

Lemma 2. Let a ∈ Γ ∩ Ω, b = φ(a) and r > 0. Assume that Ω ∩ φ^{−1}(b) ⊂ Γ. Then there exists a connected open neighborhood V of a, contained in the ball B(a, r) and some ρ > 0 such that ∂V ∩ Γ = ∅ and ∥φ(z) − b∥ ≥ 3ρ for all z in the boundary ∂V of V.

Proof. Without loss of generality we can and do assume that r is small enough for the closed ball B̃(a, r) being contained in Ω. Since Γ is a closed 0-dimensional subset of Ω there is a compact open neighborhood K of a in Γ ∩ B(a, r). Then K and H = (Γ\K)∪(ℝ^{n} \B(a, r)) are two disjoint closed subsets of ℝ^{n}. The open set V_{0} = {x ∈ ℝ^{n} : d(x, K) < d(x, H)} is a neighborhood of K in ℝ^{n} contained in B(a, r) and for every x ∈ ∂V_{0} we have d(x, K) = d(x, H), hence d(x, K) = d(x, H) > 0 since K ∩ H = ∅. Replacing V_{0} by the connected component V of a in V_{0}, we get a connected open set V whose boundary is contained in ∂V_{0}. In particular ∂V ∩Γ ⊂ ∂V_{0} ∩Γ = ∅, and b ∉ φ(∂V). Since ∂V is compact in Ω, φ(∂V) is compact

in ℝ^{p} and because b ∉ φ(∂V), we have ρ = \frac{1}{3} inf_{w∈φ(∂V)} ∥w − b∥ > 0. □

Lemma 3. Let Γ_{0} = \{y ∈ φ(Γ ∩ \overline{V}) : φ^{−1}(y) ∩ V ⊂ Γ\}. Then Γ_{0} is a closed set which does not cut the open ball B(b, ρ).

Proof. Since \overline{V} is compact in Ω and Γ is closed, φ(Γ ∩ \overline{V}) is compact. Let (y_{m}) be a sequence in Γ_{0} which converges to some y^{∗} ∉ Γ_{0} : there exists x^{∗} ∈ V \ Γ such that φ(x^{∗}) = y^{∗}, and y^{∗} ∈ φ(Γ ∩ \overline{V}) since this last set is compact.

Since x^{∗} ∉ Γ, φ is open on a neighborhood W of x^{∗} and φ(V ∩ W \ Γ) is a neighborhood of y^{∗}. Thus there exists at least one index k such that y_{k} ∈ φ(V ∩ W \ Γ) and then some x_{k} ∈ V ∩ W \ Γ such that φ(x_{k}) = y_{k}, in contradiction with y_{k} ∈ Γ_{0}. This shows that Γ_{0} is closed. And since Γ ∩ ∂V = ∅ we also have Γ_{0} = \{y ∈ φ(Γ ∩ \overline{V}) : φ^{−1}(y) ∩ \overline{V} ⊂ Γ\}.

Denote by U_{0} the ball B(b, ρ). By hypothesis φ(Γ) has dimension ≤ p − 2, and Γ_{0} ⊂ φ(Γ). Then Γ_{0} ∩ U_{0} is a closed subset of U_{0} contained in φ(Γ), hence a closed subset of U_{0} of dimension ≤ p − 2.

A closed set X of dimension ≤ p − 2 cannot cut a ball B of ℝ^{p} : if X cuts such a ball, by homeomorphism between B and ℝ^{p} there would be some closed set X′ in ℝ^{p} having dimension ≤ p − 2 and cutting ℝ^{p}, then by inversion with respect to some point in ℝ^{p} \ X′ we could find a bounded open set in ℝ^{p} (hence also a basis of the topology) whose boundary has dimension ≤ p−2. So ℝ^{p} would have topological dimension ≤ p − 1. Thus the open set U_{0} \ Γ_{0} is connected. □

Lemma 4. Let a be a critical point of φ, b = φ(a) and r > 0 such that the closed ball B̃(a, r) is contained in Ω. Assume that b ∉ φ(B(a, r) \ Γ). Then there exists ρ > 0 such that φ(B(a, r)) ⊃ B(b, ρ).

Proof. By Lemma 2 we can find some neighborhood V of a contained in B(a, r), hence relatively compact in Ω, and ρ > 0 such that ∥φ(z) − b∥ ≥ 3ρ for all z ∈ ∂V. If w satisfies ∥w − b∥ < ρ we then have ∥φ(z) − w∥ > 2ρ > ∥b − w∥ for all z ∈ ∂V. The C^{1} function ψ_{w} defined by ψ_{w}(z) = ∥φ(z) − w∥^{2} attains its minimum over the compact set \overline{V} at a point z_{0} which cannot belong to ∂V since ψ_{w}(z_{0}) ≤ ψ_{w}(a) < ρ^{2} ≤ inf_{z∈∂V} φ_{w}(z). It follows that z_{0} ∈ V and that the differential of ψ_{w} vanishes at z_{0}. Then ⟨ψ′_{w}(z_{0}), h⟩ = 2⟨φ(z_{0}) − w, φ′(z_{0}).h⟩ for all h ∈ ℝ^{n}. And if moreover φ′(z_{0}) is onto, there must exist h^{∗} ∈ ℝ^{n} such that φ′(z_{0}).h^{∗} = φ(z_{0}) − w, hence 0 = ⟨ψ′_{w}(z_{0}), h^{∗}⟩ = 2 ∥φ(z_{0}) − w∥^{2} and φ(z_{0}) = w.

This implies that either φ(z_{0}) = w or z_{0} ∈ Γ. It follows from Lemma 3 that the open set U = B(b, ρ)\Γ_{0} is connected. By compactness of \overline{V}, the set F = φ(\overline{V})∩U is closed in U. The open set V_{1} = V ∩φ^{−1}(B(b, ρ)) is a neighborhood of a, and since Γ has empty interior there exists some z^{∗} ∈ V_{1} \Γ, hence φ(z^{∗}) ∈ B(b, ρ)∩φ(\overline{V})\Γ_{0} = F. This shows that F is a nonempty closed subset of the connected open set U.

We now prove that F is open in U and this will imply that F = U hence that U ⊂ φ(\overline{V}), whence B(b, ρ) ⊂ \overline{U} ⊂ φ(\overline{V}) ⊂ φ(B(a, r)), since φ(\overline{V}) is compact. So let c ∈ F ⊂ B(b, ρ) \ Γ_{0}. Then d(c, φ(∂V)) > ρ and c ∉ φ(∂V). Since c ∈ U and U is open, there exists ε ∈]0, ρ[ such that d(c, ℝ^{p}\U) > 2ε. Then if w ∈ B(c, ε), and if ψ_{w} attains its minimum over \overline{V} at z_{0}, we have d(w, ℝ^{p} \U) ≥ d(c, ℝ^{p} \U)−∥c − w∥ > ε and

∥w − φ(z_{0})∥^{2} = ψ_{w}(z_{0}) ≤ ψ_{w}(c) < ε^{2} < d(w, ℝ^{p} \ U)^{2}

which shows that φ(z_{0}) ∈ U = B(b, ρ) \ Γ_{0}. Thus there exists z_{1} ∈ V \ Γ such that φ(z_{1}) = φ(z_{0}), hence ψ_{w}(z_{1}) = ψ_{w}(z_{0}), and φ_{w} attains its minimum over \overline{V} also at z_{1}. As above φ′_{w}(z_{1}) is onto and we deduce that w = φ(z_{1}), that w ∈ U ⊂ φ(\overline{V}) \ φ(∂V), hence that z_{1} ∈ V and that w ∈ F. Thus F ⊃ B(c, ε), and F is open in U. □

Remark 5. It appears quite clearly from the above proof that what is really used is not that the set φ(Γ) of critical values has dimension ≤ p − 2 but that the (smaller) set of “hypercritical values” of φ has dimension ≤ p − 2, where hypercritical points of φ are defined as points a ∈ Ω for which exists a neighborhood V_{a} such that φ^{−1}(φ(a)) ∩ V_{a} ⊂ Γ and hypercritical values as images under φ of hypercritical points.

Proof of Theorem 1.

Proof. We have to prove that φ(Ω) is a neighborhood of each point b ∈ φ(Ω) such that Ω ∩ φ^{−1}(b) ⊂ Γ, what is the result of Lemma 4. □

Corollary 6. Let Ω be an open subset of ℝ^{n} and f : Ω → ℝ^{p} (2 ≤ p ≤ n) be a C^{1} function. If the critical set of f is countable, then f is open.

Proof. If Γ is countable it has dimension 0. And the set f(Γ) of critical values is countable hence 0-dimensional too. □

We now show that when n = p, the hypothesis on the dimension of the set of critical values can be removed. A bounded open set V will be said to have a piecewise-C^{1} boundary if its boundary is the union of finitely many piecewise-C^{1} hypersurfaces. For example a finite union of cubes in ℝ^{n} has a piecewise-C^{1} boundary.

Lemma 7. Let Ω be an open subset of ℝ^{n}, φ : z ↦ (X_{1}(z), X_{2}(z), …, X_{n}(z)) be a C^{1} mapping from Ω to ℝ^{n}, J_{φ} its jacobian determinant and V be a relatively compact open subset of Ω with piecewise-C^{1} boundary. Then

∫_{∂V} X_{1} dX_{2} ∧ dX_{3} ∧ ⋯ ∧ dX_{n}(z) = ∫_{V} J_{φ}(z) dz

Proof. Let ρ : ℝ^{n} → ℝ be a non-negative C^{∞} function with compact support such

that ∫∫ ρ(x, y) dx dy = 1 and φ_{k} the “regularized by convolution” function φ ∗ ρ_{k}

where ρ_{k}(z) = k^{n}ρ(k.z) for z ∈ ℝ^{n}. Then the C^{∞} functions (φ_{k}) are defined on a neighborhood of \overline{V} for k large enough and this sequence converges in C^{1}-norm to φ.

For proving the above equality for the function φ it is enough to prove it for the functions φ_{k} since both members depend continuously on φ in C^{1} topology. So we can assume φ to be C^{∞} and apply Stokes formula to the (n − 1)-form ω = X_{1} dX_{2} ∧ dX_{3} ∧ ⋯ ∧ dX_{n} :

∫_{V} dω = ∫_{∂V} ω

And since dω = dX_{1} ∧ dX_{2} ∧ dX_{3} ∧ ⋯ ∧ dX_{n} = J_{φ}.dz_{1} ∧ dz_{2} ∧ ⋯ ∧ dz_{n}, the result follows. □

Theorem 8. Let Ω be an open subset of ℝ^{n}, φ be a C^{1} mapping from Ω to ℝ^{n}, θ a linear functional on ℝ^{n} and V a connected relatively compact open subset of Ω with piecewise-C^{1} boundary. Assume that θ ∘ φ(z) < 0 for all z ∈ ∂V and that θ ∘ φ(a) = 0 for some a ∈ V. Assume moreover that the critical set of φ is nowhere dense in V. Then the jacobian determinant J_{φ} of φ takes on V both positive and negative values.

Proof. Since θ ≠ 0, we can find linear functionals η_{1}, η_{2}, …, η_{n} on ℝ^{n} such that η_{1} = θ and (η_{1}, η_{2}, …, η_{n}) is a basis of the dual space. Then define α : ℝ^{n} → ℝ^{n} by

α(z) = (⟨η_{1}, z − φ(a)⟩, ⟨η_{2}, z⟩, …, ⟨η_{n}, z⟩)

Then α is an affine isomorphism of ℝ^{n} to itself and the function φ̃ = α ∘ φ satisfies

sup_{x∈∂V} X_{1}(x) < X_{1}(a) = 0 where X_{1}(x) = ⟨θ, φ(x) − φ(a)⟩ and X_{k}(x) = ⟨η_{k}, φ(x)⟩

(2 ≤ k ≤ n). So J_{α} is constant, and we can replace φ by φ̃ and assume φ(z) = (X_{1}(z), X_{2}(z), …, X_{n}(z)).

Assume towards a contradiction that J_{φ} takes only non-negative or only nonpositive values. Up to replacing X_{n} by −X_{n} if necessary we can and do assume that J_{φ}(z) ≥ 0 for all z ∈ V. Fix λ > 0. Since ∂V is compact and X_{1} < 0 on ∂V there is some δ > 0 such that X_{1}(z) ≤ −δ for all z ∈ ∂V. Then the set

W = \{z ∈ V : X_{1}(z) > −\frac{δ}{2}\} is an open set containing a. Applying lemma 7 to the

function ψ_{λ} : z ↦ (e^{λX_{1}(z)}, X_{2}(z), …, X_{n}(z)) we get J_{ψ_{λ}}(z) = e^{λX_{1}(z)}.J_{φ}(z) and

∫_{∂V} e^{λX_{1}} dX_{2} ∧ … dX_{n} = ∫_{V} e^{λX_{1}}.J_{φ}(z) dz ≥ e^{−λδ/2} ∫_{W} J_{φ}(z) dz

Since X_{k} and ∇X_{k} ( 2 ≤ k ≤ n) are bounded on ∂V, there is some M ∈ ℝ^{+} such

that |∫_{∂V} u(z) dX_{2} ∧ … dX_{n}| ≤ M sup_{z∈∂V} |u(z)| for any continuous function u on ∂V.

So we get |∫_{∂V} e^{λX_{1}} dX_{2} ∧ … dX_{n}| ≤ M.e^{−λδ} and

0 ≤ ∫_{W} J_{φ}(z) dz ≤ inf_{λ>0} e^{λδ/2}.Me^{−λδ} = inf_{λ>0} M.e^{−λδ/2} = 0

This implies that the continuous non-negative function J_{φ} vanishes everywhere on W, so that is every point of W is critical, in contradiction with the hypothesis since W ≠ ∅. □

Theorem 9. Let Ω be an open subset of ℝ^{n}, φ be a C^{1} mapping from Ω to ℝ^{n} and V a connected relatively compact open subset of Ω with piecewise-C^{1} boundary ∂V. Assume that d(b, φ(∂V)) > d(b, φ(\overline{V})) > 0 for some b. Assume moreover that the critical set of φ is nowhere dense in V. Then the jacobian determinant J_{φ} of φ takes on V both positive and negative values.

Proof. The continuous function x ↦ ∥b − φ(x)∥ attains its minimum over \overline{V} at a point a which cannot belong to ∂V since ∥b − φ(a)∥ = d(b, φ(\overline{V})) < d(b, φ(∂V)). Morover, from the assumption d(b, φ(\overline{V})) > 0, one has r_{∗} = ∥b − φ(a)∥ > 0. Consider the inversion h with pole b which transforms φ(a) into itself : h(y) =

b + r_{∗}^{2} · \frac{y − b}{∥y − b∥^{2}} . Then h is a C^{1}-diffeomorphism from ℝ^{n} \ \{b\} onto itself and

it is easily checked that J_{h}(y) = −r_{∗}^{2n}. ∥y − b∥^{−2n} < 0 for all y ≠ b in ℝ^{n}, and ψ = h ∘ φ satisfies J_{ψ}(x) = J_{h}(φ(x)).J_{φ}(x), ψ(\overline{V}) ⊂ B̃(b, r_{∗}) and for X : x ↦ ⟨φ(a) − b, ψ(x) − b⟩ − r_{∗}^{2} we have X(a) = 0 > sup_{x∈∂V} X(x) since for x ∈ ∂V we have

∥b − ψ(x)∥ = \frac{r_{∗}^{2}}{∥b − φ(x)∥} ≤ \frac{r_{∗}^{2}}{d(b, φ(∂V))} < \frac{r_{∗}^{2}}{d(b, φ(\overline{V}))}

hence X(x) ≤ ∥φ(a) − b∥ . ∥ψ(x) − b∥ − r_{∗}^{2} ≤ r_{∗}^{2} \frac{∥φ(a) − b∥ − d(b, φ(∂V))}{d(b, φ(∂V))} < 0 and

ψ(x) = h ∘ φ(x) ≠ h ∘ φ(a) = φ(a). Then it follows from the previous theorem that J_{ψ} takes on V both positive and negative values. And so does J_{φ} since J_{h} < 0. □

Theorem 10. Let Ω be an open subset of ℝ^{n} and φ : Ω → ℝ^{n} a C^{1} function. Assume that the set Γ of critical points of φ is 0-dimensional. Then φ is an open mapping.

We can assume that Ω is connected. First remark that Γ cannot cut Ω : hence the jacobian determinant J_{φ} has constant sign, and up to composing φ with a symmetry we can assume that J_{φ} ≥ 0 everywhere on Ω.

Lemma 11. Let a be a critical point of φ, b = φ(a) and r > 0 such that the closed ball B̃(a, r) is contained in Ω. Assume that b ∉ φ(B(a, r) \ Γ). Then there exists ρ > 0 such that φ(B(a, r)) ⊃ B(b, ρ).

Proof. By Lemma 2 we can find some connected neighborhood V of a contained in B(a, r), hence relatively compact in Ω, and ρ > 0 such that ∂V ∩ Γ = ∅ and

∥φ(z) − b∥ ≥ 3ρ for all z ∈ ∂V. Since ∂V ∩ Γ = ∅, there exists an integer q > \frac{\sqrt{n}}{ρ}

such that inf_{z∈∂V} d(z, Γ) > \frac{\sqrt{n}}{q}. For all ℓ = (ℓ_{1}, ℓ_{2}, …, ℓ_{n}) ∈ ℤ^{n} define Q_{ℓ} as the

cube ∏_{j=1}^{n} [\frac{ℓ_{j}}{q} , \frac{ℓ_{j} + 1}{q}] ⊂ ℝ^{n}.

Then V′ = ⋃{Q_{ℓ} : Q_{ℓ} ∩ V ≠ ∅} contains V and its boundary is disjoint from Γ : indeed if z ∈ ∂V′ there is ℓ ∈ ℤ^{n} such that z ∈ ∂Q_{ℓ} and Q_{ℓ} ⊄ V, hence Q_{ℓ} ∩∂V ≠ ∅

and d(z, ∂V) ≤ diam(Q_{ℓ}) = \frac{\sqrt{n}}{q} < inf_{z∈∂V} d(z, Γ).

Moreover ∂V′ is the union of a subfamily of the faces of the Q_{ℓ}’s hence a finite union of piecewise-C^{1} hypersurfaces. So replacing V by the interior of V′ we can

assume that V has a piecewise-C^{1} boundary and that ∥φ(z) − b∥ ≥ 3ρ − \frac{\sqrt{n}}{q} >

3ρ − ρ = 2ρ for all z ∈ ∂V.

If w satisfies ∥w − b∥ < ρ and w ∉ φ(\overline{V}) we then have ∥φ(z) − w∥ > 2ρ − ρ > ∥b − w∥ for all z ∈ ∂V. The C^{1} function ψ_{w} defined by ψ_{w}(z) = ∥φ(z) − w∥^{2} attains its minimum over the compact set \overline{V} at a point z_{0} ∉ ∂V since ψ_{w}(z_{0}) ≤ ψ_{w}(a) < ρ^{2} ≤ inf_{z∈∂V} φ_{w}(z). Moreover φ_{w}(z_{0}) > 0. Since the closed set Γ has dimension 0, it cannot cut V and is nowhere dense in V. By Theorem 9 we conclude that J_{φ} has to take negative values in contradiction with the fact that J_{φ} ≥ 0 on Ω. This contradiction implies that w ∈ φ(\overline{V}) and since w ∉ φ(∂V) that w ∈ φ(V). So φ(B(a, r)) ⊃ φ(V) ⊃ B(b, ρ). □

Proof of Theorem 10.

Proof. As for the proof of Theorem 1, notice that φ(Ω) is a neighborhood of φ(Ω\Γ) by the Local Inversion Theorem and a neighborhood of every hypercritical value in virtue of lemma 11. □

We now show by an example that the hypothesis on the critical set of φ cannot be weakened, even if n = p.

Example 12. Let H be a Hilbert space of dimension ≥ 1. Then there exists a nonopen C^{1} mapping Φ from H × ℝ to itself such that the critical set has dimension 1 and that there exists a unique critical value.

Consider the “vertical” segment J = {0}×[−1, 1] of the Hilbert space E = H ⊕ℝ. For (x, z) ∉ J, there is a unique ρ > 0 such that

\frac{∥x∥^{2}}{ρ} + \frac{z^{2}}{1 + ρ} = 1 .

Indeed the function f : t ↦ \frac{∥x∥^{2}}{t} + \frac{z^{2}}{1 + t} is decreasing on ]0, +∞[, satisfies

lim_{t→+∞} f(t) = 0 < 1, lim_{t→0^{+}} f(t) = +∞ > 1 if x ≠ 0, and lim_{t→0^{+}} f(t) = z^{2} > 1 if x = 0.

More precisely we have (1 + ρ) ∥x∥^{2} + ρz^{2} = ρ(1 + ρ), hence

ρ^{2} − ρ(∥x∥^{2} + z^{2} − 1) − ∥x∥^{2} = 0 ,

and ρ is the largest root of this equation. It follows that if (x, z) ∉ J, we have ρ > 0 and there are some u ∈ H and ψ ∈ [−π/2, π/2] such that ∥u∥ = 1 and

(x, z) = (\sqrt{ρ}.u. cos ψ, \sqrt{1 + ρ}. sin ψ)

We then put ω = (0, 0) ∈ E,

Φ(x, z) = \begin{cases}ω & if (x, z) ∈ J \\ (ρ^{2}.u. cos ψ, ρ^{2} sin ψ) & if not \end{cases}

so Φ(x, z) ∈ E and ∥Φ(x, z)∥ = ρ^{2}. We will show that Φ is C^{1}, that its critical set is J, which has dimension 1, and that the only critical value is ω, since Φ(J) = {ω}.

Looking at ρ as a function from E to ℝ_{+}, we then have Φ(x, z) = (X, Z) with

\{ \begin{matrix}X & = x.ρ^{3/2}(x, z) \\ Z & = z.\frac{ρ^{2}(x, z)}{\sqrt{1 + ρ(x, z)}} \end{matrix}

Since

ρ(x, z) = \frac{1}{2}(∥x∥^{2} + z^{2} − 1 + \sqrt{(∥x∥^{2} + z^{2} − 1)^{2} + 4 ∥x∥^{2}})

the function ρ is continuous on E and vanishes on J. We have to show that Φ is C^{1} on E \ J and that ∥∇Φ(x, z)∥ tends to 0 as (x, z) tends to a point (0, t) of J.

From the above formula it is clear that ρ is C^{1} at each q = (x, z) where ∥x∥^{2} > 0 or |z| > 1, it is outside J. For the study of Φ around J, we can restrict ourself to the case where q ∈ Q = {(x, z) : ρ(x, z) ≤ 1}, which is a neighborhood of J.

Lemma 13. If q ∈ Q and δ = d(q, J) we have ρ ≤ 6δ.

OPEN DIFFERENTIABLE MAPPINGS 7

Proof. Since \frac{∥x∥^{2}}{ρ} + \frac{z^{2}}{1 + ρ} = 1, we have ∥x∥^{2} ≤ ρ and z^{2} ≤ 1 + ρ, from what we get ∥x∥ ≤ 1 and z^{2} ≤ 2, hence |z| ≤ \sqrt{2}, for q = (x, z) ∈ Q. Moreover

∥x∥ ≤ δ = d(q, J) ≤ ∥x∥ + (|z| − 1)_{+} ≤ 1 + (\sqrt{2} − 1) = \sqrt{2} .

If q ∈ Q and ρ ≥ 5δ, we have ∥x∥^{2} ≤ δ^{2} and |z| ≤ 1 + δ, hence

\frac{(1 + δ)^{2}}{1 + ρ} ≥ \frac{z^{2}}{1 + ρ} = 1 − \frac{∥x∥^{2}}{ρ} ≥ 1 − \frac{δ^{2}}{5δ} = 1 − \frac{δ}{5}

whence

ρ ≤ \frac{(1 + δ)^{2}}{1 − δ/5} − 1 = \frac{1 + 2δ + δ^{2} − 1 + δ/5}{1 − δ/5} ≤ \frac{2δ + δ^{2} + δ/5}{1 − δ/5} ≤ δ.\frac{2 + \sqrt{2} + 1/5}{1 − \sqrt{2}/5} ≤ 6δ

It follows that ρ < 5δ or ρ ≤ 6δ, hence that ρ ≤ 6δ for each q ∈ Q. □

Lemma 14. We have \frac{∥x∥^{2}}{ρ^{2}} + \frac{z^{2}}{(1 + ρ)^{2}} = \frac{∥x∥^{2} + ρ^{2}}{ρ^{2}(1 + ρ)} .

Proof. Indeed

\frac{∥x∥^{2}}{ρ^{2}} + \frac{z^{2}}{(1 + ρ)^{2}} = \frac{∥x∥^{2}}{ρ^{2}} + \frac{1}{1 + ρ} · (1 − \frac{∥x∥^{2}}{ρ} )

= \frac{(1 + ρ) ∥x∥^{2} + ρ^{2} − ρ ∥x∥^{2}}{ρ^{2}(1 + ρ)} = \frac{∥x∥^{2} + ρ^{2}}{ρ^{2}(1 + ρ)} ,

which is the announced equality. □ Lemma 15. We have

⟨\frac{∂ρ}{∂x}, h⟩ = \frac{2ρ(1 + ρ)}{∥x∥^{2} + ρ^{2}} · ⟨x, h⟩ ; \frac{∂ρ}{∂z} = \frac{2ρ^{2}z}{∥x∥^{2} + ρ^{2}}

and ∥∇ρ(q)∥ ≤ 2\sqrt{6}.ρ(q)^{−1/2} whenever q ∈ Q \ J. Proof. Since θ(x, z) := \frac{∥x∥^{2}}{ρ} + \frac{z^{2}}{1 + ρ} = 1, differentiating with respect to x, we get for h ∈ H :

0 = \frac{∂θ}{∂x}.h = \frac{⟨2x, h⟩}{ρ}−\frac{∥x∥^{2}}{ρ^{2}} ·⟨\frac{∂ρ}{∂x}, h⟩−\frac{z^{2}}{(1 + ρ)^{2}} ·⟨\frac{∂ρ}{∂x}, h⟩ = \frac{⟨2x, h⟩}{ρ}−⟨\frac{∂ρ}{∂x}, h⟩·\frac{∥x∥^{2} + ρ^{2}}{ρ^{2}(1 + ρ)}

whence ⟨\frac{∂ρ}{∂x}, h⟩ = \frac{2ρ(1 + ρ)}{∥x∥^{2} + ρ^{2}} .⟨x, h⟩. Analogously,

0 = \frac{∂θ}{∂z} = −\frac{∥x∥^{2}}{ρ^{2}} · \frac{∂ρ}{∂z} + \frac{2z}{1 + ρ} − \frac{z^{2}}{(1 + ρ)^{2}} · \frac{∂ρ}{∂z} = \frac{2z}{1 + ρ} − \frac{∂ρ}{∂z} · \frac{∥x∥^{2} + ρ^{2}}{ρ^{2}(1 + ρ)}

whence \frac{∂ρ}{∂z} = \frac{2ρ^{2}z}{∥x∥^{2} + ρ^{2}} .

Since ∥x∥^{2} ≤ ρ ≤ 1 and z^{2} ≤ 1 + ρ ≤ 2, we deduce :

∥∇ρ(q)∥^{2} = 4ρ^{2} · \frac{(1 + ρ)^{2} ∥x∥^{2} + ρ^{2}z^{2}}{(∥x∥^{2} + ρ^{2})^{2}} ≤ 4ρ^{2} · \frac{ρ(1 + ρ)^{2} + ρ^{2}(1 + ρ)}{ρ^{4}}

= 4 · \frac{(1 + ρ)(2 + ρ)}{ρ} ≤ 4 · \frac{(2 × 3)}{ρ} = \frac{24}{ρ}

whence ∥∇ρ(q)∥ ≤ 2\sqrt{6}.ρ(q)^{−1/2}. □

Lemma 16. For q = (x, z) ∈ Q \ J, we have ∥∇Φ(q)∥ ≤ 41.δ^{1/2}(q).

Proof. For (h, s) ∈ E, we have X′(q).(h, s) = ρ^{3/2}.h + \frac{3}{2}x.ρ^{1/2}(q).⟨∇ρ(q), (h, s)⟩,

hence

∥∇X(q)∥ ≤ ρ^{3/2}(q) + \frac{3}{2} ∥x∥ .ρ^{1/2}(q).2\sqrt{6}.ρ^{−1/2}(q) ≤ ρ(q) + 3\sqrt{6}. ∥x∥
≤ 6δ(q) + 3\sqrt{6}.δ(q) ≤ 14.δ(q) ≤ 14\sqrt[4]{2}.δ^{1/2}(q) ≤ 17.δ^{1/2}(q)

Finally ⟨Z′(u), (h, s)⟩ = s.\frac{ρ^{2}}{\sqrt{1 + ρ}} + z.\frac{2ρ(1 + ρ) − ρ^{2}/2}{(1 + ρ)^{3/2}}.⟨∇ρ(q), (h, s)⟩, whence

since |z| ≤ \sqrt{2}, ρ ≤ 1 and 0 ≤ 2ρ(1 + ρ) − \frac{ρ^{2}}{2} ≤ 2ρ(1 + ρ) :

∥∇Z(q)∥ ≤ \frac{ρ^{2}}{\sqrt{1 + ρ}} + \sqrt{2}.(2\sqrt{6}.ρ^{−1/2}(q)).\frac{2ρ}{\sqrt{1 + ρ}}
≤ ρ^{2}(q) + 8\sqrt{3}.ρ^{1/2}(q) ≤ ρ^{1/2}(q).(ρ^{3/2}(q) + 8\sqrt{3})
≤ \sqrt{6}.δ^{1/2}(q).(1 + 8\sqrt{3}) ≤ 37.δ^{1/2}(q) .

Hence ∥∇Φ(q)∥ ≤ (∥∇X(q)∥^{2} + ∥∇Z(q)∥^{2})^{1/2} ≤ \sqrt{(17^{2} + 37^{2}).δ(q)} ≤ 41.δ^{1/2}(q). □

Lemma 17. The function Φ is C^{1} on E, and its critical points are the points of J.

Proof. We have previously seen that Φ is continuous on E, C^{1} on E \ J and that ∥∇Φ(q)∥ → 0 as δ(q) = d(q, J) → 0. It follows that Φ is differentiable at each point q of J with Φ′(q) = 0 and that Φ′ is continuous on E. Since Φ′(q) = 0 for every q ∈ J, J is contained in the critical set of Φ, and since Φ vanishes on J, there is a unique critical value ω = (0, 0) corresponding to these critical points. It remains only to prove that Φ is regular at each point of E \ J. It is clear that Φ(E \ J) ⊂ E \ {ω}.

We will show that Ψ = Φ^{−1} is differentiable on E \ {ω}, and it will follow that, for q ∈ E \ J and v = Φ(q), Φ′(q) ∘ Ψ′(v) = Ψ′(v) ∘ Φ′(q) = Id_{E}, hence that Φ′(q) is invertible. Indeed for v = Φ(q) = (X, Z) ≠ ω in E, we must have ρ(q) = ∥v∥^{1/2}. Hence for r = ∥v∥^{1/2} = (∥X∥^{2} + Z^{2})^{1/4} the point q has to belong to the ellipsoid

E = \{(x, z) ∈ E : \frac{∥x∥^{2}}{r} + \frac{z^{2}}{1 + r} = 1\} and we must have for q = (x, z) = Ψ(X, Z),

\{ \begin{matrix}x & = X.r^{−3/2} \\ z & = Z.\frac{\sqrt{1 + r}}{r^{2}} \end{matrix}

Since on E \ {ω} the mapping : v = (X, Z) ↦ r = (∥X∥^{2} + Z^{2})^{1/4} is C^{1} and does not vanish, this shows that Ψ is C^{1} on E \ {ω}, hence that Φ is regular at every point of E \ J. □

Lemma 18. The mapping Φ is not open.

Proof. It is enough to notice that the image C under Φ of the open unit ball B of E is not a neighborhood of Φ(ω) = ω, and even that C ∩ ({0} × ℝ) = {ω}. Indeed, if

Φ(q) = (0, t) for some t ≠ 0 we must have r = |t|^{1/2}, hence q = (0, t.\frac{\sqrt{1 + |t|^{1/2}}}{|t|} ),

and ∥q∥ = \sqrt{1 + |t|^{1/2}} > 1, it is q ∉ B. □

It follows from what precedes that if H has dimension ≥ 2 the critical set of Φ has a topological dimension 1 ≤ dim(E) − 2 and that Φ is not open.

Questions Let us complete this note by asking some questions.

In the statement of Theorem 1 an hypothesis is done on the topological dimension of the set of critical values (or more exactly on the dimension of the set of hypercritical values). It follows from Theorem 10 that if n = p this hypothesis is useless. It is then quite natural to ask whether this hypothesis could be removed also when n > p, since Example 12 shows that in turn the hypothesis on the dimension of critical points cannot be weakened.

• Does there exist a C^{1} non-open function from Ω ⊂ ℝ^{3} to ℝ^{2} whose critical set has dimension 0?

The methods used in the above proofs rely heavily on the local compactness of Ω. So it can be guessed that these results hold only for functions defined on open subsets of a finite-dimensional space. For example the following question can be raised.

• Let H be a infinite-dimensional Hilbert space. If f : H → H is C^{1} with a unique critical point, must f be open ?

Sorbonne Universités, Université Pierre et Marie Curie (Univ Paris 06), CNRS Institut de Mathématique de Jussieu - Paris Rive Gauche Université Pierre et Marie Curie (Paris 6) Boîte 186 - 4 place Jussieu, F- 75252 Paris Cedex 05, FRANCE

   E-mail address: jean.saint-raymond@imj-prg.fr