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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?

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@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
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What does $H_k$ mean, homology or Hausdorff measure? –  Liviu Nicolaescu Dec 22 '13 at 14:50
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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
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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

2 Answers 2

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?

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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
up vote 11 down vote accepted

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.

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