# 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|

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 $$K\subset \{|x| but I do not see it. That tells me that it's bounded in $$x$$, but why is it bounded in $$\xi$$?

• Do you mean "$g$ is a Riemannian metric on $\mathbb{R}^n$"? – diracdeltafunk May 22 at 2:17
• Indeed, thanks for the catch! – user790311 May 22 at 20:37

$$p$$ 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!

• The problem is that I don't see why $p$ is proper. Of course, if $p$ is proper then I can conclude the compactness. – user790311 May 22 at 20:38
• You're absolutely right – indeed $p$ is not proper and I was being silly. Instead, we want to say something like "$p$ is proper in the $\xi$-direction", which I've explained in an edit to my answer. – diracdeltafunk May 23 at 0:09
• Thanks, I need some time to process all of this! FYI: I mean by $|\xi|_g^2$ the quantity $\sum\limits_{j=1}^n g^{ij}(x)\xi_i\xi_j.$ $\xi\in T^*_x(\mathbb{R}^n).$ – user790311 May 23 at 15:24
• Okay, so your general argument is that if it were unbounded in $\xi$ in the Euclidean sense, then it would be unbounded in the norm induced by $g$, in which case it would contradict that such a limit should be contained in $I$ (which is compact)? Is there intuition for passing to the Euclidean cosphere bundle? – user790311 May 23 at 16:17
• Thanks for the clarification. Yes, that's exactly the general argument. The intuition for passing to the "cosphere bundle" (in quotes cause I haven't heard that term before, but I'm sure you know what you're talking about) is that the function $\lvert \cdot \rvert_g$ is entirely determined by its values on this cosphere bundle, and the cosphere bundle has compact fibers, meaning it's easy to get bounds on the behavior of $\lvert \cdot \rvert_g$ over bounded regions of $x$-values. Since I wanted to show that $\lvert \cdot \rvert_g$ is "close enough to $\lvert \cdot \rvert$", this seemed useful. – diracdeltafunk May 23 at 19:43