Let $p\in S$ be an elliptic point of a surface $S$. I want to show that there exists a neighbourhhod $V$ of $p$ in $S$ such that all points in $V$ belong to the same side of the tangent plane $T_p(S)$. The book I'm reading (Do Carmo), proceeds as follows:
Let $\mathbf{x}$ be a parametrization at $p$, with $\mathbf{x}(0,0)=p$. We consider the distance function:
$$d(u,v)=\langle \mathbf{x}(u,v)-p,N(p) \rangle$$
Applying Taylor's Formula we obtain:
$$\mathbf{x}(u,v)=\mathbf{x}(0,0)+\mathbf{x}_u u +\mathbf{x}_v v+\frac{1}{2}(\mathbf{x}_{uu}u^2+2\mathbf{x}_{uv} uv + \mathbf{x}_{vv}v^2)+\overline{R}$$
where all derivatives are evaluated at $(0,0)$ and $\displaystyle\lim_{(u,v)\to (0,0)}\frac{\overline{R}(u,v)}{u^2+v^2}=0$. Rearranging the terms:
$$d(u,v)=\frac{1}{2}II_p(w)+R(u,v)$$
where $II_p(\cdot)$ is the ssecond fundamental form at $p$, $w=u\mathbf{x}_u+v\mathbf{x}_v$ and $R(u,v)=\langle \overline{R}(u,v),N(p)\rangle$.
The author claims that $\lim_{w\to \mathbf{0}}\frac{R}{|w|^2}=0$.
My question is Why is this? The norm $|w|$ is given by the first fundamental form $|w|^2=I_p(w)=Eu^2+2Fuv+Gv^2$, with $EG-F^2>0$. Hence:
$$\frac{\overline{R}}{|w|^2}=\frac{u^2+v^2}{Eu^2+2Fuv+Gv^2}\frac{\overline{R}}{u^2+v^2}$$
If I could show that the quotient $$\frac{u^2+v^2}{Eu^2+2Fuv+Gv^2}$$ is bounded in a neighbourhood of $(0,0)$, then I would be done, but I don't see why this should be the case.
Any ideas? (or alternative proofs of the initial claim?).
By the Spectral Theorem, you can orthogonally diagonalize the first fundamental form matrix $A=\begin{bmatrix} E&F\\F&G\end{bmatrix}$, and from this it follows that $A\mathbf x\cdot\mathbf x \ge c\|\mathbf x\|^2$, where $c$ is the smaller eigenvalue of $A$. Since the first fundamental form is everywhere positive definite and varies continuously on the surface, it is straightforward to check that your denominator is (on any compact set, hence locally in your surface) bounded below by $c(u^2+v^2)$ for some $c>0$.