Here the link to the same question I posted on MSE with no answer.

Let $(X,d)$ be a complete and separable metric space and let $I:=(0, + \infty)$. I recall the definition of absolutely continuous curve in this setting: we say that $u \in AC(I;X)$ if there exists $g \in L^1(I)$, $g \ge0$ a.e. s.t. $$ \tag{1} d(u_s,u_t) \le \int_s^t g(r)dr \quad \forall \, s,t \in I, \quad s \le t.$$ It is well known that for $u \in AC(I;X)$ the limit $$ \tag{2}\lim_{h \to 0} \frac{d(u_{t+h}, u_t)}{h}$$ exists for a.e. $t \in I$ and therefore defines a function $|u'|: I \to \mathbb{R}$ which can be proven to be an element of $L^1(I)$ besides being obviously non negative. In particular $|u'|$ is the minimal $g$ we can put into the definition of absolute continuity, meaning that if $g \in L^1(I)$,$g \ge 0$ satisfies (1), then $|u|'\le g$ a.e. in $I$.

Hence, given $u \in AC(I;X)$, we obtain the existence of a set $A_u \subset I$ of full measure where $u$ is metrically differentiable (i.e. the limit in (2) exists).

Now I came to my question: if $v_0 \in X$ and $u \in AC(I;X)$, one can consider the absolutely continuous real valued function $f^{u}_{v_0} : I \to [0, + \infty)$ defined as $$f^{u}_{v_0}(t):=d^2(u_t,v_0) \quad t \in I.$$ Being absolutely continuous, it is differentiable in time for a.e. $t \in I$, say in a full measure set $A^u_{v_0}$ that, in principle, depends both on $u$ and $v_0$.

Is it possible to show that actually this set depends only on $u$?

For example, if $X$ is a real separable Hilbert space, this is true. This is true even in Wassestein spaces. In both cases $A^u_{v_0}=A_u$. I am wondering if this is true in general.


Consider the separable metric space $X=[0,1]\times\{0,1\}$ endowed with the $\ell_1$-metric $d:X\times X\to\mathbb R$ defined by $$d\big((x,i),(y,j)\big)=|x-y|+|i-j|. $$ It seems that this metric space yields a counterexample to your question.


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