Proving Compactness of a Set of Trapped Energies I have a very simple question, but it requires a bit of background. Here it is:
Let $p(x,\xi)=|\xi|_g+V(x),$ where $(x,\xi)\in \mathbb{R}^{2n},$ $V\in C_c^\infty(\mathbb{R}^n),$ and $g$ is a Riemannian metric on $\mathbb{R}^{n}.$ Assume further that $\text{supp} V,\text{supp} (g_{ij}-\delta_{ij})\subset B(0,r_0)$ for some $r_0>0.$ Consider the set $K=K_1\cap p^{-1}(I),$ where $K_1\subseteq \mathbb{R}^{2n}$ is closed and $I\subseteq\mathbb{R}\setminus\{0\}$ is compact. More specifically, $K_1=\Gamma^+\cap\Gamma^-,$ where $$\Gamma^{\pm}=\{(x,\xi): X(t)\not\rightarrow\infty\text{ as }t\rightarrow\mp\infty\},$$ with $X$ being the $X$-component of the Hamiltonian flow generated by $p$ with initial condition $(x,\xi)$. 
Assume further that $$K\subset \{|x|<r_0\}.$$
All that I want to conclude is that $K$ is compact as a subset of $\mathbb{R}^{2n}$. Clearly, it's closed. But, I do not see how I can conclude that it's bounded. Evidently, it follows from  $K\subset \{|x|<r_0\},$ but I do not see it. That tells me that it's bounded in $x$, but why is it bounded in $\xi$?
 A: $p$ is proper because $V$ is compactly supported, so $p^{-1}(I)$ is compact. $K \subseteq p^{-1}(I)$, so $K$ is bounded.

EDIT The above is wrong. Here's how the argument should go:
If we have a sequence of points $(x_n, \xi_n)$ such that $\lvert \xi_n \rvert \to \infty$ as $n \to \infty$, then $\lvert \xi_n \rvert_g \to \infty$ (if I'm understanding the meaning of $\lvert \xi_n \rvert_g$ correctly: are we viewing $\xi_n$ as an element of $T_{x_n}(\mathbb{R}^n)$? Also: oops, the $n$ in $x_n$ is not the same as the $n$ in $\mathbb{R}^n$. Hopefully not too confusing.)
To see this, let $S$ be the unit sphere in $\mathbb{R}^n$. The map
$$Q := (x, \xi) \mapsto \lvert \xi \rvert_g - \lvert \xi \rvert : \mathbb{R}^n \times S \to \mathbb{R}$$
is continuous and has compact support, since
$$\operatorname{supp}(Q) \subseteq \left(\bigcup_{i,j} \operatorname{supp}(g_{i,j}-\delta_{i,j})\right) \times S.$$
Thus, let $M \in \mathbb{R}$ be the minimum value attained by $Q$. Since $\lvert \xi \rvert = 1$ for all $\xi \in S$ and $\lvert \xi \rvert_g > 0$ for all $\xi \in S$, we have $M > -1$.
Since our original sequence $(x_n, \xi_n)$ had $\lim_{n \to \infty} \lvert \xi_n \rvert = \infty$, we may assume without loss of generality that $\xi_n \neq 0$ for all $n$. Let $\xi'_n = \xi_n/\lvert \xi_n \rvert$ so that $\xi_n = \lvert \xi_n \rvert \xi'_n$ and $\xi'_n \in S$. Then
$$\lvert \xi_n \rvert_g - \lvert \xi_n \rvert = \Big\lvert \lvert \xi_n \rvert \xi'_n \Big\rvert_g - \Big\lvert \lvert \xi_n \rvert \xi'_n \Big\rvert = \lvert \xi_n \rvert \Big\lvert \xi'_n \Big\rvert_g - \lvert \xi_n \rvert \Big\lvert \xi'_n \Big\rvert\\ = \lvert \xi_n \rvert \Big( \lvert \xi'_n \rvert_g - \lvert \xi'_n \rvert \Big) = \lvert \xi_n \rvert Q(\xi'_n) \geq M \lvert \xi_n \rvert,$$
so $\lvert \xi_n \rvert_g \geq \lvert \xi_n \rvert + M \lvert \xi_n \rvert = (1+M) \lvert \xi_n \rvert$. Since $1+M$ is a positive constant and $\lvert \xi_n \rvert \to \infty$ as $n \to \infty$, we conclude that $\lvert \xi_n \rvert_g \to \infty$ as $n \to \infty$.
Now, suppose for contradiction that $K$ is unbounded. Pick some unbounded sequence of points $(x_n, \xi_n) \in K$. Since $K \subseteq \{x : \lvert x \rvert < r_0\}$, we must have $\lvert \xi_n \rvert \to \infty$ as $n \to \infty$. But then $\lvert \xi_n \rvert_g \to \infty$ as $n \to \infty$, so $p(x_n, \xi_n) \to \infty$ as $n \to \infty$ (this uses the fact that $V$ is bounded). But $I$ is bounded and $K \subseteq p^{-1}(I)$, so the sequence $p(x_n, \xi_n)$ must be bounded, contradicting the previous statement!
