I have been stuck on this problem for a week and I'm not sure whether or not it is hard or I'm just missing something obvious.

**General gist of the problem**

I have a variational problem on a Riemannian manifold where I am approximating a Lagrangian with a discrete Lagrangian. I am looking for some sort of guarantee that critical points of the discrete Lagrangian can be approximated by curves of a certain level of differentiability such that they end up approximating a solution of the original problem. A reference would be great as I don't mind following this up myself but I don't know where to search.

**My specific problem**

I have a complete Riemannian manifold $M$ and I have a functional $f : C^{2}_{v_0, v_1}([0,1], M) \rightarrow \mathbb{R}$ defined over $C^2$ curves $x : [0, 1] \rightarrow M$ satisfying $\dot{x}(0) = v_0 \in TM$ and $\dot{x}(1) = v_1 \in TM$ and we now add a few additional constraints such as $x(\frac{1}{3}) = x_1$ and $x(\frac{2}{3}) = x_2$. The functional is then defined as$$ f(x) = \int_0^1 \| \nabla_{\dot{x}(t)} \dot{x}(t) - V\|^2 dt$$

where $V$ is a bounded Lipschitz (in the sense there is a $k$ such that $\||V(y)\| - \|V(x)\| | \leq k d(x,y)$) smooth vector field. I now consider the discrete Lagrangian defined by the formula: $$ f_n(x_n) = \dfrac{1}{N} \left(\sum_{i=1}^{N-1} \|N^2(l_i^+ + l_i^-) - V\|^2 \right)$$

Here $x_i := x(\frac{i}{N})$, and the covariant derivative is approximated using the functions $l_i^+, l_i^-$ where I define $l_i^+$ by $\log_{x_i} x_{i+1}$ and $l_i^-$ by $\log_{x_i} x_{i-1}$. Again, the $x_i$ are constrained under the same constraints by requiring that $N l_0^+ = v_0$ and $N l_N^- = -v_1$. We can set $N$ to be a multiple of $3$ so that we can set $x_{\frac{N}{3}} = x_1$ and $x_{\frac{2N}{3}} = x_2$.

If I find a critical point of $f_n$, call it $y_n$, can I find a curve $y$ that's $C^2$ which agrees with $y_n$ at $t = \frac{i}{N}$ such that $\|f(y) - f_n(y_n)\| \rightarrow 0$?