Let $p \in \mathrm{PAC}([0,1],\mathbb{R}^n)$ be a piecewise absolutely continuous function defined over $[0,1]$. That is, there exists a partition $(t_k)_{k=0,....,N}$ of $[0,1]$ such that the restriction of $p$ is absolutely continuous over $[t_k,t_{k+1}]$ for $k=0,...,N-1$.
My question is that can we deduce that $p\in \mathrm{W}^{1,\infty}$ ? that is, $p\in L^{\infty}([0,1],\mathbb{R}^n)$ and $\dot p\in L^{\infty}([0,1],\mathbb{R}^n)$ ?
No. It's not even true that the space of absolutely continuous functions is a subset of $W^{1,\infty}([0,1];\mathbb R^n)$.
Here's a counter-example. Let $n=1$ and $u(x) = \sqrt{x}$. This function is absolutely continuous on $[0,1]$. However, $u'(x)= \frac 1 {2\sqrt x}$ which not not bounded in $[0,1]$.