Let $T(\;\cdot\;,t)$ be a $C^1$-diffeomorphism on $\mathbb R^d$ for $t\in[0,\tau]$ with $T(\;\cdot\;,0)=\operatorname{id}_{\mathbb R^d}$, $\Omega\subseteq\mathbb R^d$ and $\Omega_t:=T(\Omega,t)$ and $\Gamma_t:=\partial\Omega_t$ for $t\in[0,\tau]$. How do we obtain the last displayed equation in the following excerpt taken from a paper (see p. 19)?
I don't get how they've obtained this from the assumption that $z_t:\Omega_t\to\mathbb R$ has a $C^1$-extension to a neighborhood of $\overline\Omega_t$ for all $t$. Since $z=z_0$, $\nabla z$ is the gradient of $z_0$. Wouldn't we need something like $z_t=\left.g\right|_{\Omega_t}$ for all small enough $t$ and some $g\in C^1(O)$, where $O$ is an open subset containing all $\Omega_t$?
When the paper writes $$ \lim_{t\to 0} \frac{z_t(x) - z(x)}t + \lim_{t\to 0} \frac{z_t(T(x,t)) - z_t(x)} t$$ we need to know that $z_t(x)$ exists. But we only know that $x \in \Omega_0$, not that it exists in $\Omega_t$. But if we know $z$ has a $C^1$ extension to a neighborhood of $\bar\Omega_0$, then we can say that for $0 \leq t \leq t_0$ we have $z(x)$ is defined for $x \in \Omega_t$. Note that the choice of $t_0$ can depend upon $z$, so the negative answer to this question If $T_t$ is a diffeomorphism with $T_0=\text{id}$ and $O$ is a neighborhood of $\Omega$, then $T_t(\Omega)\subseteq O$ for small $t$ doesn't apply.