4
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.