Checkpointed reverse sweep recomputes an intermediate

Given: The two $C^1$ transitions, initial state $x$, and uniform-block schedule with $T=2$ and $m=1$.

Uniform-block and binomial checkpoint schedules for reverse sweeps: uniform-block mode saves its scheduled checkpoints, restores a block start, replays the transitions needed to reach each step, records that step's local tape, and reverses the step.

Uniform-block and binomial checkpoint schedules preserve reverse adjoints with stated costs: when deterministic replays stay in the declared domains, uniform-block checkpointing returns the same terminal value and initial-state cotangent as the uncheckpointed reverse sweep.

The chain rule for total derivatives: $D(g\circ f)(a)=Dg(f(a))\circ Df(a)$: the derivative of a composition of totally differentiable maps is the ordered composition of their total derivatives.

Chain sum product and composition rules for Banach derivatives: on the real Banach space $\mathbb{R}$, multiplication $(a,b)\mapsto ab$ is bounded bilinear because $\mathrm{\mid }ab\mathrm{\mid }\le \mathrm{\mid }a\mathrm{\mid }\mathrm{\mid }b\mathrm{\mid }$, and the identity map has Fréchet derivative $1$. Applying the bounded-bilinear product rule to two identity maps gives $D(x\mapsto x^2)=2x$, a continuous derivative.

The derivatives of sine and cosine are cosine and minus sine: on $\mathbb{R}$, $D(\sin u)=\cos u$ and $D(\cos u)=-\sin u$ for every $u\in \mathbb{R}$; these one-dimensional derivatives are total derivatives.

Total differentiability gives a local $O(\mathrm{\parallel }h{\mathrm{\parallel }}_2)$ increment bound and therefore continuity: a totally differentiable map on $\mathbb{R}$ is continuous; in particular, $\cos$ is continuous because it is differentiable by [F5], so $u\mapsto \sin u$ is $C^1$.