Let $\;f \in C^3(\mathbb R^2;\mathbb R)\;$ and $\;f(a)=f(b)=0\;$ for some $\;a,b \in \mathbb R^2\;$.
We assume $\;a,b\;$ are non-degenerate, in the sense that the quadratic forms $\;D^2(f(a))y \cdot y\;$ and $\;D^2(f(b))y \cdot y\;$ are positive definite.
Our professor said that: "Since $\;a,b\;$ are non-degenerate zeros of $\;f\;$, it follows that there are positive constants $\;k,M\;$ such that: $\;D^2f(a+x)y\cdot y\ge M^2 {\vert y \vert}^2\;\;\forall \vert x \vert \le k\;$
Why is the above true and how does it follow from the non-degeneracy of $\;a,b\;$?
As seen here if the quadratic form $\;D^2(f(a))y \cdot y\;$ is positive definite then there exist some $\;\gamma \gt 0\;$ such that: $\;D^2(f(a))y \cdot y \ge {\gamma}^2 {\vert y \vert}^2\;$. Is this sufficient in order to claim that the lower bound holds also in a neighborhood of $\;a\;$?
I would appreciate if somebody could explain to me what I'm missing here.
Thanks in advance!
Let $A=(a_{jk})$ and $B=(b_{jk})$ be two symmetric $2 \times 2$ matrices and suppose that $A$ is positive definite. Hence ther is $c >0$ such that $Q_A(y) \ge c|y|^2$ for all $y \in \mathbb R^2$, where $Q_A$ denotes the quadratic form of $A$.
Let $d:= \frac{c}{8}$ and suppose that $ \max\{|a_{jk}-b_{jk}|: j,k \in \{1,2\}\} \le d$.
Then, for $y=(y_1,y_2) \in \mathbb R^2$:
$Q_A(y)-Q_B(y) \le \sum_{j,k=1}^2|a_{jk}-b_{jk}||y_j||y_k| \le 4d|y|^2 \le \frac{c}{2}|y|^2$.
This gives
$Q_B(y) \ge Q_A(y) -\frac{c}{2}|y|^2 \ge c|y|^2-\frac{c}{2}|y|^2=\frac{c}{2}|y|^2$,
hence $B$ is positive definite.