I know that compact set has maximum.
My advisor said that solve this proposition by using the compactness property. I can't understand what he said at all.
Can anyone verify this proposition?
Please teach me as easy as every undergraduate can understand.
Let $O \in \mathbb{R^3}$ be any point of the space and call $S$ your closed surface of $\mathbb{R}^{3}$. Consider the function $f:S \to \mathbb{R}$ given by $f(x) = \|x - O \|^2$. Since $S$ is compact there is an absolute maximum $x_0 \in S$ for $f$. Let $\gamma(t)$ be any curve of $S$ through $x_0$ i.e. $\gamma(0)=x_0$. Then
$$\begin{cases} \frac{d f(\gamma(t))}{d t}|_{t=0} = 0\\ \frac{d^2f(\gamma(t))}{d t^2}|_{t=0} \leq 0 \end{cases}$$
Computing explicitly in $\mathbb{R}^3$ the above conditions implies $$\begin{cases} \gamma'(0)\cdot(\gamma(0)-O) = 0 \\ \gamma''(0)\cdot(\gamma(0) - O) + \gamma'(0)\cdot\gamma'(0) \leq 0\end{cases}$$
The first one tells to you that the non-zero vector $N := \gamma(0) - O$ is normal to $S$ at $x_0$ (due to the fact that $\gamma(t)$ is any curve through $x_0$). The second one is related to the shape operator $A_{\mathbf{n}}$ of $S$ at $x_0$, where $\mathbf{n}$ is a unit normal vector of $S$ at $x_0$ such that $N = \| N\| \mathbf{n}$. Indeed, $$\begin{aligned} \gamma''(0)\cdot(\gamma(0) - O) + \gamma'(0)\cdot\gamma'(0) &= \gamma''(0). N + \gamma'(0)\cdot\gamma'(0) \\ &= \gamma''(0). \|N \| \mathbf{n} + \gamma'(0)\cdot\gamma'(0) \\ &= \|N \| \gamma''(0). \mathbf{n} + \gamma'(0)\cdot\gamma'(0) \\ &= \|N \| \gamma'(0). (-\frac{d \mathbf{n}}{ds}) + \gamma'(0)\cdot\gamma'(0) \\ &= \|N \| \gamma'(0). dG (\gamma'(0)) + \gamma'(0)\cdot\gamma'(0) \\ &= \|N\| \gamma'(0) \cdot A_{\mathbf{n}}(\gamma'(0)) + \gamma'(0)\cdot\gamma'(0) \leq 0 \, . \end{aligned}$$
Then both eigenvalues of $A_{\mathbf{n}}$ are negative hence $x_0$ is an elliptic point of $S$. Indeed, let $k_1,k_2$ be the two eigenvalues of $A_{\mathbf{n}}$ and let $\gamma_1(t),\gamma_2(t)$ be two curves (through $x_0$) whose tangent vectors are the respective eigenvectors. Then for $i=1,2$: $$ \|N\| \gamma_{i}'(0) \cdot A_{\mathbf{n}}(\gamma_{i}'(0)) + \gamma_{i}'(0)\cdot\gamma_{i}'(0) \leq 0$$ so $$ \|N\| \gamma_{i}'(0) \cdot k_i \gamma_{i}'(0) + \gamma_{i}'(0)\cdot\gamma_{i}'(0) \leq 0$$ then $$\|N\|k_i + 1 \leq 0$$ which shows that $k_i \leq -\frac{1}{\|N\|} < 0$ for $i=1,2$.
Some comments about the shaper operator. The shape operator $A_{\mathbb{n}}$ is a linear map from the tangent space $T_pS \to T_pS$ defined as the derivative of the unit normal vector field $\mathbb{n}$. It was defined by Gauss in his famous work about the geometry of surfaces. Gauss defined it as follows: consider the map $G : S \to \mathbb{S}(1)$, where $\mathbb{S}(1) \subset \mathbb{R}^3$ is the unit sphere defined as: $$ G(p) := \mathbf{n}(p) $$ where $\mathbf{n}(p)$ is a unit normal vector field of $S$. Then the shape operator is defined as $$A_{\mathbf{n}}(\mathbf{v}_p) := -(d G)(\mathbf{v}_p) \, \,$$ Observe that the vector $-(d G)(\mathbf{v}_p)$ is perpendicular to $\mathbf{n}$ hence $$A_{\mathbf{n}}(\mathbf{v}_p) \in T_pS \, .$$ That is to say, $A_{\mathbf{n}}$ is a linear map of $T_pS$. Moreover, it is not difficult to see that it is symmetric w.r.t. the inner product of $T_pS$. So as we learn in linear algebra the 'symmetric matrix' $A_{\mathbf{n}}$ can be diagonalized. The two eigenvalues are since Gauss, classically denoted by $k_1,k_2$ and called principal curvatures. Their product $$ \kappa := k_1 k_2 = \det(A_{\mathbf{n}})$$ is the so called Gauss curvature of $S$. Finally, a point $p \in S$ is called elliptic point if $\kappa(p) > 0$.
