I post here because I encounter a problem to understand something. To begin, let consider $f : \mathbb{R}^n \rightarrow \mathbb{R}$ a n times differentiable function. Then, we have, by the Taylor's formula :
$$ \forall a, h \in \mathbb{R}^n \quad f(a+h)=f(a)+Df_a(h) + ... + \frac{1}{n!}D^nf_a(h^n) + o_o(||h||^n)$$
So, when $n=1$, it's easy to see that each term of the taylor expansion is "bigger" than the next one, i.e for $k \in \mathbb{N}$, we have : $D^kf_a(h^k)=f^{(k)}(a)h^k$ and then $o(h^k) = o(D^kf_a(h^k))$, but I've some trouble when $n>1$. Actually, for $n=2$ and the taylor formula at the order 2, we have :
$D^2f_a((x_1, x_2), (x_1,x_2)) = \frac{\partial ^2 f}{\partial x_1^2}(a) x_1^2 + 2\frac{\partial ^2 f}{\partial x_1 \partial x_2}(a) x_1x_2 + \frac{\partial ^2 f}{\partial x_2^2}(a) x_2^2$
And I want to show that : $o(||(x_1,x_2)||^2) = o(||D^2f_a((x_1, x_2), (x_1,x_2))||)$. By considering (and utilizating the equivalence of the norm in finite dimension), if I consider the norm $||(x_1,x_2)|| = \max (|x_1|, |x_2|)$, then I have : $D^2f_a((x_1, x_2), (x_1,x_2)) = O(||(x_1,x_2)||^2)$, but I whould like to have $||(x_1,x_2)||^2 = O(||D^2f_a((x_1, x_2), (x_1,x_2))||)$. But, still with the norm max, if I consider $x_1x_2$, I should have : $x_1x_2$ which verify : $x_1x_2 \geq C||(x_1,x_2)^2||$, but if I consider $(x_1,x_2) = (e, e^2)$, for $e<<0$, I have : $e^3 \geq Ce^2$, which is not true.
So, I'm kind of lost. And here, I've only considered $\mathbb{R}^n$, but what happen if I consider a vector space of infinite dimension ? (cause I could not use the equivalence of norm)
Without some assumptions on $f$, the equality $\|(x_1, x_2)\|^2 = O(D^2 f_a((x_1, x_2), (x_1, x_2))$ does not hold. For example, choose $f(x_1,x_2) = \frac 12(x_1^2 - x_2^2)$ and $a = 0\in \mathbb R^2$ so that $D^2 f_0((x_1, x_2), (x_1, x_2)) = x_1^2 - x_2^2$. For $(x_1, x_2) = (1, 1)$ we have $\|(x_1, x_2)\|^2 = 2$ but $D^2 f_0((x_1, x_2), (x_1, x_2)) = 0$.