Absolutely Continuous Curves and Metric Derivative

Throughout this article $(X,d)$ will be a given complete metric space.

We recall that a map $f:(a,b)\to \mathbb{R}$ is said to absolutely continuous on $(a,b)$ if for every $\varepsilon>0$ there exists $\delta>0$ such that for every finite family of pairwise disjoint intervals $(x_i,y_i)\subset (a,b)$, if
$$
\sum_{i=1}^N (y_i-x_i)<\delta,
$$
then
$$
\sum_{i=1}^N |f(y_i)-f(x_i)|<\varepsilon.
$$

We have the following equivalent characterization of absolute continuity.

Proposition 1. A map $f:(a,b)\to \mathbb{R}$ is absolutely continuous if and only if there exists a function $g\in L^1(a,b)$ such that for every $a<s\le t<b$,
$$
f(t)-f(s)=\int_s^t g(r)\ dr.
$$

Now we give the definition of absolutely continuous curves on $(X,d)$.

Definition 1 (Absolutely continuous curves). Let $(X,d)$ be a complete metric space and let $v:(a,b)\to X$ be a curve. We say that $v$ belongs to $AC^p(a,b;X)$, for $p\in[1,+\infty]$, if there exists $m\in L^p(a,b)$ such that
$$
d(v(s),v(t))\leq \int_s^t m(r)\ dr
\qquad \forall, a<s\leq t<b.\tag{1}
$$

In the case $p=1$ we are dealing with absolutely continuous curves and we will denote the corresponding space simply with $AC(a,b;X)$.

Any curve in $AC^p(a,b;X)$ is uniformly continuous; if $a>-\infty$ (resp. $b<+\infty$) we will denote by $v(a+)$ (resp. $v(b-)$) the right (resp. left) limit of $v$, which exists since $X$ is complete. The above limit exist even in the case $a=-\infty$ (resp. $b=+\infty$) if $v\in AC(a,b;X)$. Among all the possible choices of $m$ in (1) there exists a minimal one, which is provided by the following theorem.

Theorem 1 (Metric derivative). Let $p\in[1,+\infty]$. Then for any curve $v$ in $AC^p(a,b;X)$ the limit

$$
|v’|(t):=\lim_{s\to t}\frac{d(v(s),v(t))}{|s-t|}
$$

exists for $\mathcal{L}^1$-a.e. $t\in(a,b)$. Moreover the function $t\mapsto |v’|(t)$ belongs to $L^p(a,b)$, it is an admissible integrand for the right hand side of (1), and it is minimal in the following sense:
$$
|v’|(t)\leq m(t)\qquad \text{for }\mathcal{L}^1\text{-a.e. }t\in(a,b),
$$

for each function $m$ satisfying (1).

Proof : Since the space $(a,b)$ is separable and curve $v$ is continuous, the space $v((a,b))$ is also separable and then let $(y_n)\subset X$ be dense in $v((a,b))$. Let

$$
d_n(t):= d(y_n,v(t)).
$$

Now we first claim that all functions $d_n$ are absolutely continuous. For all $s,t\in (a,b),\ s\le t$, since $v\in AC^p(a,b;X)$, there exists $m\in L^p(a,b)$ such that
$$
|d_n(s)-d_n(t)|=|d(y_n, v(s))-d(y_n,v(t))|\le d(v(s),v(t))\le \int_s^t m(r)\ dr
$$
For all $\varepsilon>0$, by the absolute continuity of the integral for $m\in L^p(a,b)$, there exists $\delta>0$ such that
$$
\int_E m(r)\ dr< \varepsilon
$$
as long as $\mathcal{L}^1(E)<\delta$. Hence for every finite family of pairwise disjoint intervals $(x_i,y_i)\subset (a,b)$, if
$$
\sum_{i=1}^N (y_i-x_i)<\delta,
$$
then

$$
\sum_{i=1}^n |d_n(y_i)-d_n(x_i)|\le\int_{\bigcup_{i=1}^n (x_i,y_i)} m(r)\ dr <\varepsilon.
$$

Therefore all functions $d_n$ are absolutely continuous in $(a,b)$ and then the function

$$
h(t):= \sup_{n\in \mathbb{N}} |d_n^\prime (t)|
$$

is well-defined $\mathcal{L}^1-$a.e. in $(a,b)$. Let $t\in (a,b)$ be a point where all functions $d_n$ are differentiable and notice again that
$$
|d_n(s)-d_n(t)|=|d(y_n, v(s))-d(y_n,v(t))|\le d(v(s),v(t))\quad \text{ for all } n\in \mathbb{N}
$$
hence
$$
\sup_{n\in \mathbb{N}} |d_n(s)-d_n(t)|\le d(v(s),v(t))
$$
and then
$$
h(t)=
\sup_{n \in \mathbb{N}} \liminf_{s \to t} \frac{|d_n(s)-d_n(t)|}{|s-t|}\le \liminf_{s \to t} \frac{d(v(s), v(t))}{|s-t|}.
$$
For each function $m$ satisfying (1), by Lebesgue differentiation theorem, we further get
$$
h(t)\le \liminf_{s\to t}\frac{1}{t-s} \int_s^t m(r)\ dr=m(t) \quad \mathcal{L}^1-a.e.
$$
And therefore $h\in L^p(a,b)$. On the other hand, for all $\varepsilon>0$, since $(y_n)$ is dense in $v((a,b))$, there exist $(y_{n_k})\to v(s)$ and then
$$
d(y_{n_k},v(t))\to d(v(s),v(t))\quad \text{as } k\to \infty.
$$

Hence
$$
|d_{n_k}(s)-d_{n_k}(t)|=|d(y_{n_k},v(s))-d_{n_k}(y_{n_k},v(t))|\to d(v(s),v(t)),\quad \text{as } k\to \infty,
$$
which means for all $\varepsilon>0$, the exists a sufficiently large $k$ such that
$$
d(v(s),v(t))< |d_{n_k}(s)-d_{n_k}(t)|+\varepsilon.
$$
Therefore,
$$
d(v(s),v(t))=\sup_{n\in \mathbb{N}} |d_{n}(s)-d_{n}(t)|.
$$
Moreover, since $d_n$ are absolutely continuous, for all $s\le t$,
$$
d_n(t)-d_n(s)=\int_s^t d_n^\prime (r)\ dr\le \int_s^t h(r)\ dr,
$$
hence
$$
d(v(s),v(t))=\sup_{n\in \mathbb{N}} |d_{n}(s)-d_{n}(t)|\le \int_s^t h(r)\ dr \quad \forall s,t\in (a,b),\ s\le t.
$$

Therefore, by Lebesgue differentiation theorem,
$$
\limsup_{s\to t}\frac{d(v(s),v(t))}{|s-t|}\le h(t)
$$
at any Lebesgue point $t$ of $h$. Combining the results above, we get

$$
h(t)\le \liminf_{s \to t} \frac{d(v(s), v(t))}{|s-t|}\le \limsup_{s\to t}\frac{d(v(s),v(t))}{|s-t|}\le h(t),
$$
therefore, the limit
$$
|v’|(t):=\lim_{s\to t}\frac{d(v(s),v(t))}{|s-t|}
$$

exists for $\mathcal{L}^1$-a.e. $t\in(a,b)$,
$$
|v’|(t)=h(t)\leq m(t)\qquad \text{for }\mathcal{L}^1\text{-a.e. }t\in(a,b).
$$
and $|v’|(t)=h(t) \in L^p(a,b)$ which is an admissible integrand for the right hand side of (1). $\square$

Reference

Ambrosio, Luigi, Nicola Gigli, and Giuseppe Savaré. Gradient flows: in metric spaces and in the space of probability measures. Basel: Birkhäuser Basel, 2005.

The cover image of this article was taken in Innsbruck, Austria.