I was wondering if the following was true:

Given $f: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$, a differentiable function where $\|f(x,y) \| > \|(x,y)\|^{2}$ for all $(x,y)$, then given any $M \in \mathbb{R}$ does there exists a point $(x_{0}, y_{0})$ where $\vert det(Df(x_{0},y_{0}) \vert >M$?

There is a $\mathbb{R}$ analogue of this which is true via the mean-value theorem, I however do not see how the mean value property helps in this case. The analogue is below:

If $f: \mathbb{R} \rightarrow \mathbb{R}$ is a differentiable function and $f(x)> x^{2}$ for all $x$, then given any $M \in \mathbb{R}$ there is an $x_{0}$ such that $\vert f'(x_{0}) \vert > M$.

We have as an application of the mean value theorem for any $x_{1}> x_{2}$, that there exists an $x_{0}$ where \begin{align*} \vert f'(x_{0}) &= \frac{\vert(f(x_{1})-f(x_{2}) \vert}{\vert x_{1}-x_{2} \vert} \\ &> \frac{ \vert f(x_{1}) \vert - \vert f(x_{2}) \vert }{x_{1}-x_{0}} \\ &> \frac{x_{1}^{2} -x_{1}^{2}}{x_{1}-x_{2}} \\ &= x_{2}+x_{1} \end{align*} The result then follows if we choose $x_{1}$ and $x_{2}$ to where $x_{1} + x_{2}> M$.

  • $\begingroup$ $Df(x_{0},y_{0})$ should be the Jacobian at the point $(x_{0},y_{0})$ $\endgroup$
    – user135520
    Dec 11, 2015 at 13:46
  • $\begingroup$ can you link to the analogue in $\mathbb{R}$? $\endgroup$ Dec 11, 2015 at 13:55
  • $\begingroup$ Sure, I made an edit above. $\endgroup$
    – user135520
    Dec 11, 2015 at 13:58

1 Answer 1


Let $f(x,y)= (x^2+y^2+1 \hskip 2pt ; \hskip 2pt 0)$, then $\det Df=0$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .