# Generalization of Darboux's Theorem

Darboux's Theorem. If $f:[a,b]\to\mathbb R$ is differentiable and $f'(a)<\xi<f'(b)$, then there exists a $c\in (a,b)$, such that $\,f'(c)=\xi$.

Does any of the following generalizations

1. Let $U\subset\mathbb R^n$ connected and $f: U\to \mathbb R$ differentiable. Then $\nabla f[U]$ is connected,

2. Let $U\subset\mathbb R^n$ convex and $f: U\to \mathbb R$ differentiable. Then $\nabla f[U]$ is convex,

3. $H_k\big(\nabla f[U],\mathbb{Z}\big) \hookrightarrow H_k(U,\mathbb{Z})$, for all $k$,

hold?

-
@MarkMeckes $\nabla f[U]\subset \mathbb R^n$. – smyrlis Dec 22 '13 at 14:37
Yes, I realized that silly mistake and deleted my comment just before you posted your response. – Mark Meckes Dec 22 '13 at 14:38
What does $H_k$ mean, homology or Hausdorff measure? – Liviu Nicolaescu Dec 22 '13 at 14:50
Very nice question! Related MO question mathoverflow.net/questions/135946/… related blog post gilkalai.wordpress.com/2008/08/20/… – Gil Kalai Dec 22 '13 at 16:18
Related MSE post: Darboux's theorem of several variables. Dave L. Renfro's comment mentions some references. – Martin Sleziak Dec 26 '13 at 8:45

Consider $f:\mathbb{R}^{2} \rightarrow \mathbb{R}$ with $f(x,y)=\mathrm{e}^{x}\cos y$ then $\nabla (f)$ is nothing but $\mathrm{e}^{\bar{z}} :\mathbb{C} \rightarrow \mathbb{C}$, with image neither convex nor simply connected. This gives a negative answer to the second and the third part of your question.

Regarding the first part I do not know the complete answer. But I can say only the following: for every $V\in \mathbb{R}^{n}$, $\nabla f[U]\cdot V$ is a connected subset of $\mathbb{R}$, because the partial derivatives satisfies Darboux theorem; hence they send open connected sets to connected subset of $\mathbb{R}$. Moreover, as a consequence of chain rule $\nabla f[U]\cdot V$ is a partial derivative. In fact there is no a hyper plane which separates $\nabla f[U]$.

So it is interesting to consider the following question:

Let $A$ be a subset of $R^{n}$, such that $A\cdot V$ is connected for all $V$, does this implies that $A$ is connected?

-
Consider the region $A\subset \mathbb{R}^n$, $n\geq 2$, defined as the union of the closed ball of radius $1/4$ centered at the origin with the annulus $3/4\leq \Vert x\Vert \leq 1$. It is disconnected and for any vector $V$ of length $1$ the projection $A\cdot V$ is the interval $[-1,1]$. – Liviu Nicolaescu Dec 23 '13 at 13:37
@liviuNicolaescu thanks for the example. However it can be shown that this set can not be equal to $\nabla f[U]$, when $U$ is open connected set. In fact no sphere can separates $\nabla f[U]$. Without lose of generality assume the sphere which separates the image, is the unit sphere around 0. Let $a,b \in U$ and $\parallel \nabla f(a)\parallel <1$ and $\parallel \nabla f(b)\parallel >1$. Choose a unit speed curve $\gamma: [0,1] \rightarrow U$ which connect a to b and its velocity at end points is parallel to $\nabla f(a)$, $\nabla f(b)$ now apply Darboux to $f\circ \gamma$ – Ali Taghavi Dec 23 '13 at 18:43
So it is natural to ask:"Let $A$ be a subset of $\mathbb{R}^{n}$ which can be separated by no hyperplane or sphere, does it implies that $A$ is connected"? – Ali Taghavi Dec 23 '13 at 18:48

It turns out than none of the three potential generalisations holds.

Counterexamples for the last two questions are presented in the answer of Ali Taghavi, and in particular by function $f(x,y)=(\mathrm{e}^x\cos y,\mathrm{e}^x\sin y)$, as $f[\mathbb R^2]=\mathbb R^2\smallsetminus\{(0,0)\}$.

For the first question, a counterexample appears in:

Solution to the gradient problem of C.E. Weil, by Zoltán Buczolich

where the author gives a complete answer to the famous gradient problem of C. E. Weil. On an open set $G\subset \mathbb{R}^{2}$ he constructs a differentiable function $f:G\to\mathbb{R}$, for which there exists an open set $\Omega_{1}\subset\mathbb{R}^{2}$ such that $\nabla f({p})\in \Omega_{1}$ for a ${p}\in G$ but $\nabla f({q})\not\in\Omega_{1}$ for almost every ${q}\in G$. This also shows that the Denjoy-Clarkson property does not hold in higher dimensions.

-