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Let $f:R^N \to R^N$ be a differentiable mapping, and $J_f$ its Jacobian determinant. Suppose that $\exists a,b \in R^N : J_f(a)<0,J_f(b)>0$. Is it right that on every continous curve connecting $a$ and $b$ there exists a point $c$ such that $J_f(c)=0?$

Some discussions started here...

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    $\begingroup$ What am I missing? $J_f$ is a continuous function on $\mathbb{R}^N$, so it is continuous when restricted to any continuous curve connecting $a$ to $b$. That curve is just the image of some continuous $\gamma:[0,1]\to\mathbb{R}^N$. So $F=J_f\circ\gamma:[0,1]\to\mathbb{R}$ is continuous and satisfies $F(0)<0$ and $F(1)>0$, so by the intermediate value theorem it must vanish for some value in $(0,1)$. $\endgroup$ – Joe Silverman Mar 27 '14 at 20:43
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    $\begingroup$ @JoeSilverman I guess some people by "differentiable" mean just differentiable, not necessarily $C^1$. Personally, I also disapprove such maps :) $\endgroup$ – Alex Degtyarev Mar 27 '14 at 21:07
  • $\begingroup$ @AlexDegtyarev Ah, well, if $J_f$ is allowed to be (mildly) discontinuous, then can't one construct a counterexample by taking a standard example of a map $f$ that is differentiable, but not $C^1$. $\endgroup$ – Joe Silverman Mar 28 '14 at 1:27
  • $\begingroup$ @JoeSilverman This I don't know, as for ordinary derivatives $f'(x)$ there is an intermediate value theorem, even without the assumption that it's continuous. (Don't remember the exact name.) So, the question about the Jacobian may be not as simple as it seems. $\endgroup$ – Alex Degtyarev Mar 28 '14 at 8:52
  • $\begingroup$ @JoeSilverman The only hope that I see is that the restriction of the Jacobian to a curve is some kind of ordinary derivative. But standard proofs of such things usually assume class $C^1$, so one should go all the way over the proofs to see of they work in the more general setting. Boring :) $\endgroup$ – Alex Degtyarev Mar 28 '14 at 8:55
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The answer is negative. To construct a counterexample I used Jan Maly's paper THE DARBOUX PROPERTY FOR GRADIENTS, 1996. First let's consider $$ \varphi(x,y):= \begin{cases} \displaystyle \frac {2xy^4}{x^2+y^4} -x, (x,y) \ne(0,0); \\0,(x,y)=(0,0). \end{cases} $$ We observe that $\varphi(x,y)$ is everywhere differentiable, $\varphi^\prime_x(0,0)=-1$, $\varphi^\prime_x(0,y)=1 $ $\forall y\ne0$ and $\varphi^\prime_y(0,y)=0$. Now let's consider $f(x,y):=(\varphi(x,y),x+y)$. It's clear that $f:R^2\to R^2$ is everywhere differentiable and $$ J_f(0,y) = \begin{cases} 1, y\ne0; \\-1, y=0. \end{cases} $$ Thus intermediate value (Darboux) property for Jacobian determinant doesn't hold even on a straight line $x=0$.

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