Lemma: Linear independence of the actual ramp derivatives

Independent of Input (S-bv) and of Gate V (analytic argument). The actual ${\sigma }_{\epsilon }$ is explicit: XII-C, Section sec:freeze (imported by XVI-L, Scope), defines ${\sigma }_{\epsilon }(t)=\rho (t/\epsilon )$, where $\rho (s)=0$ for $s\le 0$, $\rho (s)=e^{-1/s}/(e^{-1/s}+e^{-1/(1-s)})$ for $0<s<1$, and $\rho (s)=1$ for $s\ge 1$; here $0<\epsilon <T$ is fixed but not numerically specified. For every open interval $J\subset (0,\epsilon )$ and every $m$, the functions ${\sigma }_{\epsilon },{\sigma }_{\epsilon }^{\prime },\dots ,{{\sigma }_{\epsilon }}^{(m)}$ are linearly independent over $\mathbb{R}$ on $J$. The actual ${\chi }_{\mathrm{I}}$ is not explicit: XVI-J/K/L give only ${\chi }_{\mathrm{I}}\in {C_c}^{\infty }(M)$, an independent compact switch.

Proof. Since ${{\sigma }_{\epsilon }}^{(k)}(t)={\epsilon }^{-k}{\rho }^{(k)}(t/\epsilon )$, it suffices to treat $\rho$ on $J^{\prime }\subset (0,1)$. Suppose ${\sum }_{k\le m}{c}_k{\rho }^{(k)}=0$ on $J^{\prime }$ with $c_m\ne 0$. If $m=0$, then $\rho =0$ on $J^{\prime }$, which is impossible because $\rho >0$ on $(0,1)$. Otherwise, $\rho$ is real-analytic on $(0,1)$ (a quotient of analytic functions with positive denominator), so the identity holds on all of $(0,1)$. Then $\rho {|}_{(0,1)}$ solves a constant-coefficient linear ODE of order $m$. Since $\rho$ is smooth on $\mathbb{R}$ and vanishes for $s\le 0$, ${\rho }^{(j)}(s)\to 0$ as $s\to 0^{+}$ for all $j$, so the solution has zero Cauchy data at $0$ and vanishes identically, a contradiction.