Uniform Boundedness and the Arzela-Ascoli Theorem in a Riemannian Manifold Let $(M, g)$ be a complete Riemannian Manifold and let $(x_n)$ be a sequence of curves in $\Omega = \{x \in C^1([0, 1], \ M): \ x(0) = p, \quad x(1) = q, \quad \dot{x}(0) = v, \quad \dot{x}(1) = w \}$ such that $g(\dot{x}_n, \dot{x}_n)$ is uniformly bounded. A paper I am currently reading makes the claims that:

*

*There exists a compact subset $K$ of $M$ containing the image of all the $x_n$

*Arzela-Ascoli theorem gives the existence of a subsequence of $(x_n)$ which converges uniformly to some continuous curve satisfying $x(0) = p$ and $x(1) = q$.

I've come up with some reasoning as to why this may be true (which I've put below), but I'm not entirely confident because I'm fairly new to functional analysis (especially in the context of manifolds). Any help is appreciated.

1) This feels fairly intuitive, but I'm not certain that my reasoning holds. If the uniform bound is given by $g(\dot{x}_n, \dot{x}_n) < R$, then the sequence of lengths of the curves—say $L(x_n)$—also satisfies $L(x_n) < R$. Since $M$ is complete, we can construct a ball of radius $R$ around $x_n(0) = p$ which must contain the image of each $x_n$ (otherwise, we contradict the bound). The compact set $K$ can then be taken as the closure of this ball.
2) Now assuming 1 holds, we have the uniform boundedness of $(x_n)$ because their images are contained in a compact subset of a metric space. I suppose that equicontinuity is implied by the fact that these curves are $C^1$ together with the uniform boundededness—but I'm not sure how to show that in the context of manifolds. Assuming Arzela-Ascoli works as usual in this context, we'd then get a subsequence of $(x_n)$ which converges uniformly to some continuous curve $x$ (with respect to the topology induced by the metric, I suppose). The fact that $(x_n) \subset \Omega$ together with uniform convergence should imply that $x$ satisfies the relevant boundary conditions.
 A: You are correct about (1), except that you should get $L(x_n) \le R$ since
$$ L(x_n) = \int_0^1 \| x_n'(t)\|\, \mathrm d t = \int_0^1 \sqrt{g(x_n'(t), x_n'(t))} \, \mathrm d t.$$
The closed ball of radius $\sqrt R$ is closed, hence compact since $(M, g)$ is complete.
For (2), the equicontinuity follows from the bound on $g(x_n', x_n')$. Now $\{x_n\}$ is a sequence of continuous maps $x_n : [0,1] \to K$. Using the bound and the definition of the distance on $(M, g)$, we have
$$ d( x_n(t), x_n(s)) \le L( x_n) = \int_t^s \| x_n'(\tau)\| \, \mathrm d\tau \le \sqrt R |t-s|. $$
In particular, for all $\epsilon >0$, choose $\delta = \epsilon /\sqrt R$. Then whenever $t, s\in [0,1]$ and $|t-s|<\delta$, we have $d( x_n(t), x_n(s)) <\epsilon$ for all $n$. Thus the family is equicontinuous.
Lastly, note that Arzela-Ascoli theorem can be generalized to compact domain, see here. I suppose the proof is similar to the case of functions. Another way (which is an overkill) is to apply Nash embedding theorem, so that $(M, g)$ is isometrically embedded in $\mathbb R^N$ for some $N$. Then one can think of $x_n : [0,1] \to \mathbb R^N$ and apply the simpler version on each coordinates $1, 2, \cdots, N$.
