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Let $f:U\longrightarrow \mathbb{R}$ a differentiable function where $U\subset\mathbb{R}^n$ open connected.

What can we say about the image of the derivative $f'(U)\subset \mathbb{R}^n$?

$f'(U)$ is connected?

If $n=1$ , $\;f'(U)$ is an interval by Darboux's Theorem, some reference?

Any hints would be appreciated.

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felipeuni
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2 Answers2

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EDIT 1 : It seems that the proof below is wrong, since $\epsilon$ should depend on $x$ and thus $S$ is not necessarily connected.
EDIT 2 : As answered in MathOverflow (in the link posted by @Martin Sleziak), it turns out that the generalization of Darboux's theorem doesn't hold.

For $\epsilon$ sufficiently small, consider : $$S = \{\ (\frac{f(x+te_1)-f(x)}{t},...,\frac{f(x+te_n)-f(x)}{t}) : x\in U \;,\, t\in (0,\epsilon_x)\ \}$$ Since $g:(x,t)$ $\rightarrow$ $(\frac{f(x+te_1)-f(x)}{t},...,\frac{f(x+te_n)-f(x)}{t})$ is continuous, S is connected. Moreover it's clear that $f'(U) \subset \overline S$ (*).

Now for $x \in U$ and $t\in (0,\epsilon),$ consider $h_x(t)=(f(x+te_1),...,f(x+te_n)).$ By the Mean value theorem, there exists $t_0 \in (0,t)$ such that : $$\frac{h_x(t)-h_x(0)}{t}=h_x'(t_0)$$ which is equivalent to saying : $$(\frac{f(x+te_1)-f(x)}{t},...,\frac{f(x+te_n)-f(x)}{t})=(\frac{\partial f}{\partial x_1}(x+t_0e_1),...,\frac{\partial f}{\partial x_n}(x+t_0e_n))$$ Hence $S \subset f'(U)$ (**).

By (*) and (**) we have : $S \subset f'(U) \subset \overline S$, therefore $f'(U)$ is connected.

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Yassine Marzougui
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    You say that By the Mean value theorem, there exists $t_0 \in (0,t)$ such that : $$\frac{h_x(t)-h_x(0)}{t}=h_x'(t_0)$$ which is equivalent to saying : $$(\frac{f(x+te_1)-f(x)}{t},...,\frac{f(x+te_n)-f(x)}{t})=(\frac{\partial f}{\partial x_1}(x+t_0e_1),...,\frac{\partial f}{\partial x_n}(x+t_0e_n))$$ - That's not correct! – Yiorgos S. Smyrlis Dec 21 '13 at 12:33
  • To clarify @Yiorgos' comment, the mean value theorem gives you a $t_i$ which makes the $i$th component equal, but there is no reason for the $t_i$'s to all have the same value. – ronno Dec 21 '13 at 12:35
  • Note that I applied the mean value theorem to $h_x$ and not to each of the components, in order to have the same $t_0$ – Yassine Marzougui Dec 21 '13 at 12:38
  • @Post No Bills, you are right, $\epsilon$ here depends on $x$, I introduced it to make sure $f(x+te_i)$ is well defined, I should then recheck if $S$ is still connected... – Yassine Marzougui Dec 21 '13 at 21:15
  • I'm not sure If I should delete the answer or not (I'm not sure of the protocol used in this site), anyway I just added a comment at the top and kept the answer in the case where the idea may inspire s.o for similar problems... – Yassine Marzougui Dec 26 '13 at 16:41
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The MathOverflow thread, linked in comments and answer, points to the paper Solution to the gradient problem of C.E. Weil by Buczolich, in which he constructed a differentiable function $f:\mathbb R^2\to\mathbb R$ such that $\nabla f(0)=0$ and $|\nabla f|\ge 1$ at all other points. I'd like to add that a shorter proof was found in A note on Buczolich's solution of the Weil gradient problem: a construction based on an infinite game by Malý and Zelený (unfortunately, their article is behind Springer's paywall). An interesting aspect of the latter paper is the game-theoretical formulation:

Let $B$ be an open ball in $\mathbb R^2$. The point-line game is a sequence of rounds. The first and the second players play points $a_k\in B$ and lines $p_k$, respectively, obeying the following rules. In the first round, the first player plays a point $a_1\in B$ and then the second player plays a line $p_1$ with $a_i\in p_1$. In the $k$th round, the first player plays a point $a_k\in B\cap p_{k-1}$ and then the second player plays a line $p_k$ passing through $a_k$. The first player wins if the sequence $(a_k)$ diverges, otherwise the second player wins.

Malý and Zelený prove

Theorem 1.2. The second player has a winning strategy.

and reprove Buczolich's result based on this theorem.

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    This is an incorrect quote of Buczolich's theorem: He only proved that $|\nabla f(p)|\ge 1$ for almost all points $p\in R^2$. – Moishe Kohan Oct 16 '17 at 18:08