Let $(M, \nabla)$ be a Riemannian manifold and $\gamma: [0,1] \to M$ a smooth curve and let $X$ be a vector field. I would like to prove
$$|X(\gamma(1))-X_{\parallel} | \le L(\gamma) \| \nabla X \|_{\infty}$$
where $X_{\parallel}$ is the parallel transport of $X(\gamma(0))$ onto $T_{\gamma(1)}M$ along $\gamma$.
I have no idea of how to do that.
For $t\in[0,1],$ let $X_\|(t)$ denote the parallel transport of $X(\gamma(t))$ into $T_{\gamma(1)}M$ along $\gamma$. By the relation between the connection and parallel transport, we have $$\frac{d}{dt}X_\|(t)=P_{t\to1}\nabla_{\dot{\gamma}(t)}X,$$where $P_{t\to1}$ denotes parallel transport from $\gamma(t)$ to $\gamma(1)$ along $\gamma$. Hence, $$\left|\frac{d}{dt}X_\|(t)\right|\le\left|\dot{\gamma}(t)\right|\cdot\|\nabla X\|_\infty,$$and the desired inequality follows by integration.