# Points of differentiability of squared distance from a point in metric spaces

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.