Local estimate for a $C^1$ map with negative definite Jacobian

Question

Let $U$ be an open set in ${\Bbb R^n}$ and $f=(f^1,\cdots,f^n):U\to{\Bbb R^n}$ be $C^1$. Suppose $f'(x_0)$ is negative definite for some $x_0\in U$. Show that there exists $\epsilon>0$ and a neighborhood $V$ of $x_0$ such that for any $y_1,\cdots, y_n\in V$ $$ \xi\cdot A\xi\leq -\epsilon|\xi|^2 $$ for all $\xi\in{\Bbb R^n}$ where $$ A=\left( \begin{matrix} \nabla f^1(y_1)\\ \nabla f^2(y_2)\\ \vdots\\ \nabla f^n(y_n) \end{matrix} \right). $$

---

If $y_1=\cdots=y_n=x_0$, then $A=f'(x_0)$ and the inequality is basically saying $A+\epsilon I_n$ is negative semi-definite. Here are my questions:

• If this case (which looks simpler) can be shown (I don't see how though), does it help the general cases?
• What does $A$ generally mean and how can the general cases be done?

Answer 1

Try proving by contradiction, and a diagonal argument. $A$ is essentially an approximation to $f'(x_0).$ Suppose the contrary of the problem, and observe that for each $m \in \mathbb{N},$ we see that for $\epsilon_m = 1/m, V_m = B(x_0,\epsilon_m),$ there exist points $y^{(m)}_1, \cdots y^{(m)}_n,$ and $\xi^{(m)} \in \mathbb{S}^{n-1},$ such that

$$ \xi^{(m)} \cdot A^{(m)} \xi^{(m)} \geq -\epsilon_n. $$ Now use the compactness of the unit sphere $\mathbb{S}^{n-1}$, and observe $A^{(m)} \to f'(x_0),$ through possibly a subsequence. This contradicts negative definiteness of $f'(x_0).$

$A^{(m)}$ really has the obvious meaning! Where did you find this problem?

Answer 2

I don't have a geometric interpretation for $A$. Maybe it has some use in a context not present in the question. Or it could be just something made up for an exercise. Right now I think of $A$ as "perturbed Jacobian".

The key point is simple: strict inequalities, applied to continuous functions, yield open sets. But one has to carefully set up the functions and inequalities to get the desired result.

In this case, I would define a map $F:U^n\times S^{n-1}\to\mathbb R$ as follows: $$F(y_1,\dots,y_n,\xi) = \xi \cdot A(y_1\dots,y_n)\xi$$ with $A$ as in your question. (Note that $U^n$ is the Cartesian product of $n$ copies of $U$, while $S^{n-1}$ is the unit sphere). Since $F$ is continuous, the set $\{F<-\epsilon\}$ is open in $U^n\times S^{n-1}$ for every $\epsilon$. For each $\xi\in S^{n-1}$ there is $\epsilon_\xi>0$ for which this set contains $ (x_0,\dots,x_0,\xi)$, hence it contains a neighborhood of this set. Such a neighborhood contains s product-type neighborhood $V_\xi^n\times W_\xi$ where $V_\xi$ is a neighborhood of $x_0$ in $\mathbb R^n$ and $W_\xi$ is a neighborhood of $\xi$ in $S^{n-1}$. From the open cover $$\{(x_0,\dots,x_0)\}\times S^{n-1}\subset \bigcup_{\xi\in S^{n-1}}V_\xi^n\times W_\xi$$ you can choose a finite subcover. Intersect $V_\xi$ over the members of this subcover, take the minimum of $\epsilon_\xi$, and you are done.

Although it's true that sequences + contradiction, as in Raghav's answer, is a quicker way.