The classical inverse function theorem reads as follows:

Theorem 1 (${C^1}$ inverse function theorem) Let ${\Omega \subset {\bf R}^n}$ be an open set, and let ${f: \Omega \rightarrow {\bf R}^n}$ be an continuously differentiable function, such that for every ${x_0 \in \Omega}$, the derivative map ${Df(x_0): {\bf R}^n \rightarrow {\bf R}^n}$ is invertible. Then ${f}$ is a local homeomorphism; thus, for every ${x_0 \in \Omega}$, there exists an open neighbourhood ${U}$ of ${x_0}$ and an open neighbourhood ${V}$ of ${f(x_0)}$ such that ${f}$ is a homeomorphism from ${U}$ to ${V}$.

It is also not difficult to show by inverting the Taylor expansion

$\displaystyle f(x) = f(x_0) + Df(x_0)(x-x_0) + o(\|x-x_0\|)$

that at each ${x_0}$, the local inverses ${f^{-1}: V \rightarrow U}$ are also differentiable at ${f(x_0)}$ with derivative

$\displaystyle Df^{-1}(f(x_0)) = Df(x_0)^{-1}. \ \ \ \ \ (1)$

The textbook proof of the inverse function theorem proceeds by an application of the contraction mapping theorem. Indeed, one may normalise ${x_0=f(x_0)=0}$ and ${Df(0)}$ to be the identity map; continuity of ${Df}$ then shows that ${Df(x)}$ is close to the identity for small ${x}$, which may be used (in conjunction with the fundamental theorem of calculus) to make ${x \mapsto x-f(x)+y}$ a contraction on a small ball around the origin for small ${y}$, at which point the contraction mapping theorem readily finishes off the problem.

I recently learned (after I asked this question on Math Overflow) that the hypothesis of continuous differentiability may be relaxed to just everywhere differentiability:

Theorem 2 (Everywhere differentiable inverse function theorem) Let ${\Omega \subset {\bf R}^n}$ be an open set, and let ${f: \Omega \rightarrow {\bf R}^n}$ be an everywhere differentiable function, such that for every ${x_0 \in \Omega}$, the derivative map ${Df(x_0): {\bf R}^n \rightarrow {\bf R}^n}$ is invertible. Then ${f}$ is a local homeomorphism; thus, for every ${x_0 \in \Omega}$, there exists an open neighbourhood ${U}$ of ${x_0}$ and an open neighbourhood ${V}$ of ${f(x_0)}$ such that ${f}$ is a homeomorphism from ${U}$ to ${V}$.

As before, one can recover the differentiability of the local inverses, with the derivative of the inverse given by the usual formula (1).

This result implicitly follows from the more general results of Cernavskii about the structure of finite-to-one open and closed maps, however the arguments there are somewhat complicated (and subsequent proofs of those results, such as the one by Vaisala, use some powerful tools from algebraic geometry, such as dimension theory). There is however a more elementary proof of Saint Raymond that was pointed out to me by Julien Melleray. It only uses basic point-set topology (for instance, the concept of a connected component) and the basic topological and geometric structure of Euclidean space (in particular relying primarily on local compactness, local connectedness, and local convexity). I decided to present (an arrangement of) Saint Raymond’s proof here.

To obtain a local homeomorphism near ${x_0}$, there are basically two things to show: local surjectivity near ${x_0}$ (thus, for ${y}$ near ${f(x_0)}$, one can solve ${f(x)=y}$ for some ${x}$ near ${x_0}$) and local injectivity near ${x_0}$ (thus, for distinct ${x_1, x_2}$ near ${f(x_0)}$, ${f(x_1)}$ is not equal to ${f(x_2)}$). Local surjectivity is relatively easy; basically, the standard proof of the inverse function theorem works here, after replacing the contraction mapping theorem (which is no longer available due to the possibly discontinuous nature of ${Df}$) with the Brouwer fixed point theorem instead (or one could also use degree theory, which is more or less an equivalent approach). The difficulty is local injectivity – one needs to preclude the existence of nearby points ${x_1, x_2}$ with ${f(x_1) = f(x_2) = y}$; note that in contrast to the contraction mapping theorem that provides both existence and uniqueness of fixed points, the Brouwer fixed point theorem only gives existence and not uniqueness.

In one dimension ${n=1}$ one can proceed by using Rolle’s theorem. Indeed, as one traverses the interval from ${x_1}$ to ${x_2}$, one must encounter some intermediate point ${x_*}$ which maximises the quantity ${|f(x_*)-y|}$, and which is thus instantaneously non-increasing both to the left and to the right of ${x_*}$. But, by hypothesis, ${f'(x_*)}$ is non-zero, and this easily leads to a contradiction.

Saint Raymond’s argument for the higher dimensional case proceeds in a broadly similar way. Starting with two nearby points ${x_1, x_2}$ with ${f(x_1)=f(x_2)=y}$, one finds a point ${x_*}$ which “locally extremises” ${\|f(x_*)-y\|}$ in the following sense: ${\|f(x_*)-y\|}$ is equal to some ${r_*>0}$, but ${x_*}$ is adherent to at least two distinct connected components ${U_1, U_2}$ of the set ${U = \{ x: \|f(x)-y\| < r_* \}}$. (This is an oversimplification, as one has to restrict the available points ${x}$ in ${U}$ to a suitably small compact set, but let us ignore this technicality for now.) Note from the non-degenerate nature of ${Df(x_*)}$ that ${x_*}$ was already adherent to ${U}$; the point is that ${x_*}$ “disconnects” ${U}$ in some sense. Very roughly speaking, the way such a critical point ${x_*}$ is found is to look at the sets ${\{ x: \|f(x)-y\| \leq r \}}$ as ${r}$ shrinks from a large initial value down to zero, and one finds the first value of ${r_*}$ below which this set disconnects ${x_1}$ from ${x_2}$. (Morally, one is performing some sort of Morse theory here on the function ${x \mapsto \|f(x)-y\|}$, though this function does not have anywhere near enough regularity for classical Morse theory to apply.)

The point ${x_*}$ is mapped to a point ${f(x_*)}$ on the boundary ${\partial B(y,r_*)}$ of the ball ${B(y,r_*)}$, while the components ${U_1, U_2}$ are mapped to the interior of this ball. By using a continuity argument, one can show (again very roughly speaking) that ${f(U_1)}$ must contain a “hemispherical” neighbourhood ${\{ z \in B(y,r_*): \|z-f(x_*)\| < \kappa \}}$ of ${f(x_*)}$ inside ${B(y,r_*)}$, and similarly for ${f(U_2)}$. But then from differentiability of ${f}$ at ${x_*}$, one can then show that ${U_1}$ and ${U_2}$ overlap near ${x_*}$, giving a contradiction.

The rigorous details of the proof are provided below the fold.

— 1. Proof —

Fix ${x_0 \in \Omega}$. By a translation, we may assume ${x_0=f(x_0)=0}$; by a further linear change of variables, we may also assume ${Df(0)}$ (which by hypothesis is non-singular) to be the identity map. By differentiability, we have

$\displaystyle f(x) = x + o(\|x\|)$

as ${x \rightarrow 0}$. In particular, there exists a ball ${B(0,r_0)}$ in ${\Omega}$ such that

$\displaystyle \| f(x)-x\| < \frac{1}{2} \|x\|$

for all ${x\in B(0,r_0)}$; by rescaling we may take ${r_0=1}$, thus

$\displaystyle \| f(x)-x\| < \frac{1}{2} \|x\| \hbox{ whenever } \|x\| \leq 1. \ \ \ \ \ (2)$

Among other things, this gives a uniform lower bound

$\displaystyle \| f(x) \| > \frac{1}{2} \ \ \ \ \ (3)$

for all ${x \in \partial B(0, 1)}$, and a uniform upper bound

$\displaystyle \| f(x) \| < \frac{1}{10} \ \ \ \ \ (4)$

for all ${x \in \partial B(0, \frac{1}{20})}$; thus ${f}$ maps ${B(0,\frac{1}{20})}$ to ${B(0,\frac{1}{10} )}$.

Proposition 3 (Local surjectivity) For any ${0 < r < 1}$, ${f(B(0,r))}$ contains ${B(0,r/2)}$.

Proof: Let ${y \in B(0,r/2)}$. From (2), we see that the map ${f: \partial B(0,r) \rightarrow f(\partial B(0,r))}$ avoids ${y}$, and has degree ${1}$ around ${y}$; contracting ${\partial B(0,r)}$ to a point, we conclude that ${f(x)=y}$ for some ${x \in B(0,r)}$, yielding the claim.

Alternatively, one may proceed by invoking the Brouwer fixed point theorem, noting that the map ${x \mapsto x - f(x) + y}$ is continuous and maps the closed ball ${\overline{B(0,r)}}$ to the open ball ${B(0,r)}$ by (2), and has a fixed point precisely when ${f(x)=y}$.

A third argument (avoiding the use of degree theory or the Brouwer fixed point theorem, but requiring one to replace ${B(0,r/2)}$ with the slightly smaller ball ${B(0,r/3)}$) is as follows: let ${x \in \overline{B(0,r)}}$ minimise ${\|f(x)-y\|}$. From (2) and the hypothesis ${y \in B(0,r/3)}$ we see that ${x}$ lies in the interior ${B(0,r)}$. If the minimum is zero, then we have found a solution to ${f(x)=y}$ as required; if not, then we have a stationary point of ${x \mapsto \|f(x)-y\|}$, which implies that ${Df(x)}$ is degenerate, a contradiction. (One can recover the full ball ${B(0,r/2)}$ by tweaking the expression ${\|f(x)-y\|}$ to be minimised in a suitable fashion; we leave this as an exercise for the interested reader.) $\Box$

Corollary 4 ${f}$ is an open map: the image of any open set is open.

Proof: It suffices to show that for every ${x \in \Omega}$, the image of any open neighbourhood of ${x}$ is an open neighbourhood of ${f(x)}$. Proposition 3 handles the case ${x=0}$; the general case follows by renormalising. $\Box$

Suppose we could show that ${f}$ is injective on ${B(0,\frac{1}{20})}$. By Corollary 4, the inverse map ${f^{-1}: f(B(0,\frac{1}{20})) \rightarrow B(0,\frac{1}{20})}$ is also continuous. Thus ${f}$ is a homeomorphism from ${B(0,\frac{1}{20})}$ to ${f(B(0,\frac{1}{20}))}$, which are both neighbourhoods of ${0}$ by Proposition 3; giving the claim.

It remains to establish injectivity. Suppose for sake of contradiction that this was not the case. Then there exists ${x_1, x_2 \in B(0,\frac{1}{20} )}$ and ${y \in B(0,\frac{1}{10} )}$ such that

$\displaystyle y = f(x_1) = f(x_2).$

For every radius ${r \geq 0}$, the set

$\displaystyle K_r := \{ x \in \Omega: \| f(x)-y\| \leq r \}$

is closed and contains both ${x_1}$ and ${x_2}$. Let ${K_r^1}$ denote the connected component of ${K_r}$ that contains ${x_1}$. Since ${K_r}$ is non-decreasing in ${r}$, ${K_r^1}$ is non-decreasing also.

Now let us study the behaviour of ${K_r^1}$ as ${r}$ ranges from ${0}$ to ${\frac{4}{10} }$. The two extreme cases are easy to analyse:

Lemma 5 ${K_0^1 = \{x_1\}}$.

Proof: Since ${Df(x_1)}$ is non-singular, we see from differentiability that ${f(x) \neq f(x_1)}$ for all ${x \neq x_1}$ sufficiently close to ${x_1}$. Thus ${x_1}$ is an isolated point of ${K_0}$, and the claim follows. $\Box$

Lemma 6 We have ${B(0,\frac{1}{20} ) \subset K_r^1 \subset B(0,1)}$ for all ${\frac{2}{10} \leq r \leq \frac{4}{10}}$. In particular, ${K_r^1}$ is compact for all ${0 \leq r \leq \frac{4}{10} }$, and contains ${x_2}$ for ${\frac{2}{10} \leq r \leq \frac{4}{10} }$.

Proof: Since ${f(B(0,\frac{1}{20} )) \subset B(f(0),\frac{1}{10} ) \subset \overline{B(y,r)}}$, we see that ${B(0,\frac{1}{20} ) \subset K_r}$; since ${B(0,\frac{1}{20} )}$ is connected and contains ${x_1}$, we conclude that ${B(0,\frac{1}{20} ) \subset K^1_r}$.

Next, if ${x \in \partial B(0,1)}$, then by (3) we have ${f(x) \not \in B(0,\frac{1}{2})}$, and hence ${f(x) \not \in \overline{B(y, r)}}$. Thus ${K_r}$ is disjoint from the sphere ${\partial B(0,1)}$. Since ${x_1}$ lies in the interior of this sphere we thus have ${K_r^1 \subset B(0,1)}$ as required. $\Box$

Next, we show that the ${K_r^1}$ increase continuously in ${r}$:

Lemma 7 If ${0 \leq r < \frac{1}{20}}$ and ${\epsilon > 0}$, then for ${r < r' < \frac{1}{20}}$ sufficiently close to ${r}$, ${K_{r'}^1}$ is contained in an ${\epsilon}$-neighbourhood of ${K_r^1}$.

Proof: By the finite intersection property, it suffices to show that ${\bigcap_{r'>r} K_{r'}^1 = K_r^1}$. Suppose for contradiction that there is a point ${x}$ outside of ${K_r^1}$ that lies in ${K_{r'}^1}$ for all ${r'>r}$. Then ${x}$ lies in ${K_{r'}}$ for all ${r'>r}$, and hence lies in ${K_r \cap B(0,1)}$. As ${x}$ and ${x_1}$ lie in different connected components of the compact set ${K_r \cap \overline{B(0,1)}}$ (recall that ${K_r}$ is disjoint from ${\partial B(0,1)}$), there must be a partition of ${K_r \cap \overline{B(0,1)}}$ into two disjoint closed sets ${F, G}$ that separate ${x}$ from ${x_1}$ (for otherwise the only clopen sets in ${K_r \cap \overline{B(0,1)}}$ that contain ${x_1}$ would also contain ${x}$, and their intersection would then be a connected subset of ${K_r \cap \overline{B(0,1)}}$ that contains both ${x_1}$ and ${x}$, contradicting the fact that ${x}$ lies outside ${K_r^1}$). By normality, we may find open neighbourhoods ${U, V}$ of ${F, G}$ that are disjoint. For all ${x}$ on the boundary ${\partial U}$, one has ${\| f(x)-y\| > r}$ for all ${x \in \partial U}$. As ${\partial U}$ is compact and ${f}$ is continuous, we thus have ${\| f(x)-y\| > r'}$ for all ${x \in \partial U}$ if ${r'}$ is sufficiently close to ${r}$. This makes ${U \cap K_{r'}}$ clopen in ${K_{r'}}$, and so ${x}$ cannot lie in ${K_{r'}^1}$, giving the desired contradiction. $\Box$

Observe that ${K_r^1}$ contains ${x_2}$ for ${r \geq \frac{2}{10} }$, but does not contain ${x_2}$ for ${r=0}$. By the monotonicity of the ${K_r^1}$ and least upper bound principle, there must therefore exist a critical ${0 \leq r_* \leq \frac{2}{10} }$ such that ${K_r^1}$ contains ${x_2}$ for all ${r > r_*}$, but does not contain ${x_2}$ for ${r < r_*}$. From Lemma 7 we see that ${K_{r_*}^1}$ must also contain ${x_2}$. In particular, by Lemma 5, ${r_* > 0}$.

We now analyse the critical set ${K_{r_*}^1}$. By construction, this set is connected, compact, contains both ${x_1}$ and ${x_2}$, contained in ${B(0,1)}$, and one has ${\|f(x)-y\| \leq r_*}$ for all ${x \in K_{r_*}^1}$.

Lemma 8 The set ${U := \{ x \in K_{r_*}^1: \|f(x)-y\| < r_* \}}$ is open and disconnected.

Proof: The openness is clear from the continuity of ${f}$ (and the local connectedness of ${{\bf R}^n}$). Now we show disconnectedness. Being an open subset of ${{\bf R}^n}$, connectedness is equivalent to path connectedness, and ${x_1}$ and ${x_2}$ both lie in ${U}$, so it suffices to show that ${x_1}$ and ${x_2}$ cannot be joined by a path ${\gamma}$ in ${U}$. But if such a path ${\gamma}$ existed, then by compactness of ${\gamma}$ and continuity of ${f}$, one would have ${\gamma \subset K_r}$ for some ${r < r_*}$. This would imply that ${x_2 \in K_r^1}$, contradicting the minimal nature of ${r_*}$, and the claim follows.

Lemma 9 ${U}$ has at most finitely many connected components.

Proof: Let ${U_1}$ be a connected component of ${U}$; then ${f(U_1)}$ is non-empty and contained in ${B(y,r_*)}$. As ${U}$ is open, ${U_1}$ is also open, and thus by Corollary 4, ${f(U_1)}$ is open also.

We claim that ${f(U_1)}$ is in fact all of ${B(y,r_*)}$. Suppose this were not the case. As ${B(y,r_*)}$ is connected, this would imply that ${f(U_1)}$ is not closed in ${B(y,r_*)}$; thus there is an element ${z}$ of ${B(y,r_*)}$ which is adherent to ${f(U_1)}$, but does not lie in ${f(U_1)}$. Thus one may find a sequence ${x_n}$ in ${U_1}$ with ${f(x_n)}$ converging to ${z}$. By compactness of ${K_{r_*}^1}$ (which contains ${U_1}$), we may pass to a subsequence and assume that ${x_n}$ converges to a limit ${x}$ in ${K_{r_*}^1}$; then ${f(x)=z}$. By continuity, there is thus a ball ${B}$ centred at ${x}$ that is mapped to ${B(y,r)}$ for some ${r < r_*}$; this implies that ${B}$ lies in ${K_{r_*}}$ and hence in ${K_{r_*}^1}$ (since ${x \in K_{r_*}^1}$) and thence in ${U}$ (since ${r}$ is strictly less than ${r_*}$). As ${x}$ is adherent to ${U_1}$ and ${B}$ is connected, we conclude that ${B}$ lies in ${U_1}$. In particular ${x}$ lies in ${U_1}$ and so ${z=f(x)}$ lies in ${f(U_1)}$, a contradiction.

As ${f(U_1)}$ is equal to ${B(y,r_*)}$, we thus see that ${U_1}$ contains an element of ${f^{-1}(\{y\})}$. However, each element ${x}$ of ${f^{-1}(\{y\})}$ must be isolated since ${Df(x)}$ is non-singular. By compactness of ${K_{r_*}^1}$, the set ${K_{r_*}^1}$ (and hence ${U}$) thus contains at most finitely many elements of ${f^{-1}(\{y\})}$, and so there are finitely many components as claimed. $\Box$

Lemma 10 Every point in ${K_{r_*}^1}$ is adherent to ${U}$ (i.e. ${\overline{U} = K_{r_*}^1}$).

Proof: If ${x \in K_{r_*}^1}$, then ${\|f(x)-y\| \leq r_*}$. If ${\|f(x)-y\| then ${x \in U}$ and we are done, so we may assume ${\|f(x)-y\| = r_*}$. By differentiability, one has

$\displaystyle f(x') = f(x) + Df(x) (x'-x) + o(\|x'-x\|)$

for all ${x'}$ sufficiently close to ${x}$. If we choose ${x'}$ to lie on a ray emenating from ${x}$ such that ${Df(x)(x'-x)}$ lies on a ray pointing towards ${y}$ from ${f(x)}$ (this is possible as ${Df(x)}$ is non-singular), we conclude that for all ${x'}$ sufficiently close to ${x}$ on this ray, ${\|f(x')-y\| < r_*}$. Thus all such points ${x'}$ lie in ${K_{r_*}}$; since ${x}$ lies in ${K_{r_*}^1}$ and the ray is locally connected, we see that all such points ${x'}$ in fact lie in ${K_{r_*}^1}$ and thence in ${U}$. The claim follows. $\Box$

Corollary 11 There exists a point ${x_* \in K_{r_*}^1}$ with ${\|f(x_*)-y\| = r_*}$ (i.e. ${x_*}$ lies outside ${U}$) which is adherent to at least two connected components of ${U}$.

Proof: Suppose this were not the case, then the closures of all the connected components of ${U}$ would be disjoint. (Note that an element of one connected component of ${U}$ cannot lie in the closure of another component.) By Lemma 10, these closures would form a partition of ${K_{r_*}^1}$ by closed sets. By Lemma 8, there are at least two such closed sets, each of which is non-empty; by Lemma 9, the number of such closed sets is finite. But this contradicts the connectedness of ${K_{r_*}^1}$. $\Box$

Next, we prove

Proposition 12 Let ${x_* \in K_{r_*}^1}$ be such that ${\|f(x_*)-y\|=r_*}$, and suppose that ${x}$ is adherent to a connected component ${U_1}$ of ${U}$. Let ${\omega}$ be the vector such that

$\displaystyle Df(x_*) \omega = y - f(x_*) \ \ \ \ \ (5)$

(this vector exists and is non-zero since ${Df(x_*)}$ is non-singular). Then ${U_1}$ contains an open ray of the form ${\{ x_* + t \omega: 0 < t < \epsilon \}}$ for some ${\epsilon > 0}$.

This together with Corollary 11 gives the desired contradiction, since one cannot have two distinct components ${U_1, U_2}$ both contain a ray from ${x_*}$ in the direction ${\omega}$.

Proof: As ${f}$ is differentiable at ${x_*}$, we have

$\displaystyle f(x_* + t\omega) = f(x_*) + Df(x_*) t \omega + o(|t|)$

for all sufficiently small ${t}$; we rearrange this using (5) as

$\displaystyle f(x_* + t \omega) - y = (1-t) (f(x_*) - y) + o(|t|).$

In particular, ${f(x_*+t\omega) \in B(y,r_*)}$ for all sufficiently small positive ${t}$. This shows that all sufficiently small open rays ${\{ x_* + t \omega: 0 < t < \epsilon \}}$ lie in ${K_{r_*}}$, hence in ${K_{r_*}^1}$ (since ${x_* \in K_{r_*}^1}$), and hence in ${U}$. In fact, the same argument shows that there is a cone

$\displaystyle \{ x_* + t \omega': 0 < t < \epsilon; \|\omega' - \omega\| \leq \epsilon \} \ \ \ \ \ (6)$

that will lie in ${U}$ if ${\epsilon}$ is small enough. As this cone is connected, it thus suffices to show that ${U_1}$ intersects this cone.

Let ${\delta > 0}$ be a small radius to be chosen later. As ${Df(x_*)}$ is non-singular, we see if ${\delta}$ is small enough that ${f(x) \neq f(x_*)}$ whenever ${\|x-x_*\| = \delta}$. By continuity, we may thus find ${\kappa > 0}$ such that ${\|f(x)-f(x_*)\| > \kappa}$ whenever ${\|x-x_*\| = \delta}$.

Consider the set

$\displaystyle U' := \{ x \in U_1: \|x-x_*\| \leq \delta; \|f(x)-f(x_*)\| < \kappa \}.$

As ${x_*}$ is adherent to ${U_1}$, ${U'}$ is non-empty. By construction of ${\kappa}$, we see that we also have

$\displaystyle U' := \{ x \in U_1: \|x-x_*\| < \delta; \|f(x)-f(x_*)\| < \kappa \}$

and so ${U'}$ is open. By Corollary 4, ${f(U')}$ is then also non-empty and open. By construction, ${f(U')}$ also lies in the set

$\displaystyle D := \{ z \in B(y,r_*): \|z-f(x_*)\| < \kappa \}.$

We claim that ${f(U')}$ is in fact all of ${D}$. The proof will be a variant of the proof of Lemma 9. Suppose this were not the case. As ${D}$ is connected, this implies that there is an element ${z}$ of ${D}$ which is adherent to ${f(U')}$, but does not lie in ${f(U')}$. Thus one may find a sequence ${x_n}$ in ${U'}$ with ${f(x_n)}$ converging to ${z}$. By compactness of ${K_{r_*}^1}$ (which contains ${U'}$), we may pass to a subsequence and assume that ${x_n}$ converges to a limit ${x}$ in ${K_{r_*}^1}$; then ${f(x)=z}$. By continuity, there is thus a ball ${B}$ centred at ${x}$ contained in ${B(x_*,\delta)}$ that is mapped to ${B(y,r) \cap D}$ for some ${r < r_*}$; this implies that ${B}$ lies in ${K_{r_*}}$ and hence in ${K_{r_*}^1}$ (since ${x \in K_{r_*}^1}$) and thence in ${U}$ (since ${r}$ is strictly less than ${r_*}$). As ${x}$ is adherent to ${U_1}$ and ${B}$ is connected, we conclude that ${B}$ lies in ${U_1}$ and thence in ${U'}$. In particular ${x}$ lies in ${U'}$ and so ${z=f(x)}$ lies in ${f(U')}$, a contradiction.

As ${f(U') = D}$, we may thus find a sequence ${t_n > 0}$ converging to zero, and a sequence ${x_n \in U'}$, such that

$\displaystyle f(x_n) = f(x_*) + t_n (y - f(x_*)).$

However, if ${\delta}$ is small enough, we have ${\|f(x_n)-f(x_*)\|}$ comparable to ${\|x_n-x_*\|}$ (cf. (2)), and so ${x_n}$ converges to ${x_*}$. By Taylor expansion, we then have

$\displaystyle f(x_n) = f(x_*) + Df(x_*) (x_n-x_*) + o(\|x_n-x_*\|)$

and thus

$\displaystyle (Df(x_*)+o(1)) (x_n-x_*) = t_n Df(x_*) \omega$

for some matrix-valued error ${o(1)}$. Since ${Df(x_*)}$ is invertible, this implies that

$\displaystyle x_n-x_* = t_n (1+o(1)) \omega = t_n \omega + o(t_n).$

In particular, ${x_n}$ lies in the cone (6) for ${n}$ large enough, and the claim follows. $\Box$