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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 $\xi$?

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

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  • $\begingroup$ The problem is that I don't see why $p$ is proper. Of course, if $p$ is proper then I can conclude the compactness. $\endgroup$ – user790311 May 22 at 20:38
  • $\begingroup$ 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. $\endgroup$ – diracdeltafunk May 23 at 0:09
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    $\begingroup$ Thanks, I need some time to process all of this! FYI: I mean by $|\xi|_g^2$ the quantity $\sum\limits_{j=1}^n g^{ij}(x)\xi_i\xi_j.$ $\xi\in T^*_x(\mathbb{R}^n).$ $\endgroup$ – user790311 May 23 at 15:24
  • $\begingroup$ 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? $\endgroup$ – user790311 May 23 at 16:17
  • $\begingroup$ 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. $\endgroup$ – diracdeltafunk May 23 at 19:43

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