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Let $(M, g)$ be a complete Riemannian Manifold of dimension $d$ and let $(x_n)$ be a sequence 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. I've recently read and proven that there exists a subsequence—say $(x_{n_k})$—which converges uniformly to a continuous curve $x: [0, 1] \to M$ with $x(0) = p$ and $x(1) = q$ (see my previous question here).

However, the paper I'm reading goes on to say that "the boundedness of $g(\dot{x}_n, \dot{x}_n)$ also implies that $x$ is of Sobolev class $H^1$ and that the convergence of $(x_{n_k})$ to $x$ is weak in the $H^1$ sense." For the sake of consistency, they've defined $H^1([0, 1], M)$ as the space of curves $\gamma: [0, 1] \to M$ such that, if $(U, \phi)$ is a local coordinate chart on $M$ and $I \subset [0, 1]$ closed is such that $\gamma(I) \subset U$, then $\phi \circ \gamma \in H^1(I, \mathbb{R}^d)$. I.e., all local representatives of $\gamma$ are continuous with weak derivative in $L^2([0,1], \mathbb{R}^d)$.

This goes a bit over my head, and all attempts at consulting the literature thus far have failed. I'm hoping that someone could explain why this is true. My thoughts are below.

Given the definition of $H^1$, it seems wise to work within charts and then (hopefully) glue everything together nicely in the end. To that end, choose a finite partition of $[0, 1]$, say $I_j$, and a corresponding family of charts $(U_j, \phi_j)$ covering the image of $x$ such that for sufficiently large $n$ we have $x_{n}(I_j) \subset U_j$. By construction, $\phi_j \circ x_{n} \in C^1(I_j, \mathbb{R^d})$. As far as I understand, $$||\phi_j \circ x_n||_{H^1} = ||\phi_j \circ x_n||_{L^2} + ||(\phi_j \circ x_n)'||_{L^2} = \left(\int_0^1 ||(\phi_j \circ x_n)(t)||^2_{\mathbb{R}^d}dt\right)^{1/2} + \left(\int_0^1 ||(\phi_j \circ x_n)'(t)||^2_{\mathbb{R}^d}dt\right)^{1/2}$$

If we can bound both integrals on the right independent of $n$, then since $H^1$ is a Hilbert space there would exist a weakly convergent subsequence. However, I'm not sure how to bring the uniform boundedness of the Riemannian metric in, how to conclude that the weak convergence is to the same $x$ as before (I imagine the two are related), or how to glue it all back together.

infinitylord
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    Choose your charts on compact neighbourhoods of $\Bbb R^n$ so that the metrics on $\Bbb R^n$ and their coordinate images in $M$ are equivalent (I mean bi-Lipschitz). Then since $|x_n'|$ is uniformly bounded on $M$ its also uniformly bounded in $\Bbb R^n$ since the two metrics are bilipschitz. You also already know that $|x_n|$ is uniformly bounded (since it converges uniformly!) so you can do the same thing. – s.harp Aug 10 '20 at 18:00
  • @s.harp Thanks for the response. Why is it that the metrics on $\mathbb{R}^n$ and their coordinate images in $M$ will be bi-Lipschitz equivalent in compact chart neighborhoods? – infinitylord Aug 11 '20 at 10:40

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Lets carry out the above comment with more detail and care.

Let $\Psi : U\to \Bbb R^d$ be a (arbitrarily fine) chart and restrict $x_n$ to be contained in the chart. We are interested in seeing why both $$\int \|\Psi\circ x_n\|^2\quad \text{ and }\quad \int\|(\Psi\circ x_n)'\|^2$$ are bounded above. The first integral is the simplest: We may assume our chart has domain a compact neighbourhood, meaning here that $\overline U$ is compact and that $\Psi$ can be extended to be a chart on a neighbourhood of $\overline{U}$. This implies that $\Psi(U)$ is a bounded set in $\Bbb R^d$ (say with upper bound $C$) and then: $$\int \|\Psi \circ x_n\|^2 ≤\int C^2 ≤ C^2$$ as the domain of $x_n$ is something in $[0,1]$ (remember we have restricted it). For the second integral we do the same step, this time with the differential $D\Psi$. By compactness of the domain we may assume the differential has bounded modulus (say its bound is $c$) and then: $$\int \|\Psi \circ x_n)' \|^2 = \int dt\, \|D_{x_n(t)}\Psi\ (x_n'(t))\|^2_{\Bbb R^d}≤\int dt \|D_{x_n}\Psi\|^2\,\|x_n'(t)\|^2_{M}≤c^2\int\|x_n'(t)\|^2_M$$ by assumption $\|x_n'(t)\|$ is uniformly bounded (in $t$ and $n$), hence there is some constant $D$ with $\|x_n'(t)\|≤D$, so in total $$\int\|(\Psi \circ x_n)'\|^2_{\Bbb R^d}≤ c^2 D^2.$$

This gives the statement. Apply it to your scenario by covering the path of $x$ with finitely many charts having the above property, then all but finitely many $x_n$ are contained in the union of these charts and you get that the sum of the above constants for every chart are a bound for all but finitely many of these integrals.

s.harp
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  • The only step that I'm a bit unsure of is $||D_{x_n} \Psi(x_n'(t))||{\mathbb{R}^d}^2 \le ||D{x_n} \Psi||M^2 ||x_n'(t)||_M^2$. Do you define the norm of the differential as $||D{x_n} \Psi||M^2 := \sup{v \in S^d} ||D_{x_n} \Psi(v)||$ where $S^d$ is the unit sphere in the tangent space of $M$ at $x_n(t)$? Then the boundedness of the norm is trivial at a point, and is given uniformly because of the compact domain? – infinitylord Aug 11 '20 at 13:46
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    $D_{x_n(t)}\Psi$ is a linear map $T_{x_n(t)}M\to T_{\Psi(x_n(t))}\Bbb R^d$. The norm is defined by: $$|D_{x_n(t)}\Psi|=\sup_{v\in T_{x_n(t)}M, |v|M≤1}|D{x_n(t)}\Psi(v)|_{\Bbb R^d}$$ The inequality you remark on then follows from sub-multiplicativity of this norm. The boundedness of this operator norm on $U$ then follows from the compact domain. – s.harp Aug 11 '20 at 14:14
  • Great, that makes sense to me. Thanks for your help! – infinitylord Aug 11 '20 at 14:51