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

Question

I posted this same question on MSE with no answer.

Let $I:=(0, + \infty)$ and let $(X,d)$ be a complete and separable metric space. In this setting we say that $u : I \to X$ is absolutely continuous if there exists $g \in L^1(I)$, $g \ge0$ s.t. $$ \tag{1}\label{1} d(u_s,u_t) \le \int_s^t g(r)dr \quad \forall \, s,t \in I, \quad s \le t.$$ It is well known (see Duda - Absolutely Continuous Functions with Values in a Metric Space (DOI)) that such a $u$ is metrically differentiable.

Now for an absolutely continuous $u : I \to X$ and $v \in X$, consider the real-valued function $f_{u,v} : I \to [0, + \infty)$ defined by $$f_{u,v}(t):=d^2(u_t,v) \quad t \in I.$$ Then $f$ is also absolutely continuous, so it is differentiable for almost all $t \in I$. Let $A_{u,v}$ be the full-measure domain of differentiability for $f_{u,v}$.

Can we show that $A_{u,v}$ depends only on $u$?

For example, this is true if $X$ is a real separable Hilbert space, and even in Wasserstein spaces. Is this true in general?

Answer 1

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.