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Given a metric space $(X, d)$ and a function $f:I:=(a,b)\subseteq \mathbb{R} \rightarrow X$, we can define the metric derivative of $f$ at $t\in I$ as $$ f'(t) := |\frac{d}{dt}|f(t) := \lim_{\delta\rightarrow 0} \frac{d(f(t+\delta),f(t))}{|\delta|} \in \mathbb{R} $$

Of course, the limit in $t$ exists only for those $f$ such that the function $g_t(\delta) := d(f(t+\delta),f(t)) $ is differentiable at $\delta=0$ (since $f'(t) = g_t'(0)$).

My question is: if the functions $g_t$ are not differentiable, but just weakly differentiable, can we provide a notion of weak metric derivative? I tried to have a look around and I didn't find this. Is it sufficient to require that "the $g_t$ are weakly differentiable at $0$"? Could someone hint whether there are some subtleties one should pay attention to?

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  • $\begingroup$ Should it not be $\lvert\delta\rvert$ instead of $\delta$ in the denominator in the definition of the metric derivative? $\endgroup$
    – Lorago
    Commented Nov 5, 2023 at 13:45
  • $\begingroup$ "Of course, the limit in $t$ exists only for those $f$ such that..." Not quite. If $f(t)=|t|$ then $f$ is metrically differentiable at $0$ but $g_0=f$ is not differentiable at $0$. (Also, using the notation $f'$ for metric derivative is a bit confusing since that notation is typically used for the actual derivative.) $\endgroup$
    – M W
    Commented Nov 5, 2023 at 20:23
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    $\begingroup$ Also, what do you mean by "weakly differentiable"? Normally weak derivatives are only uniquely defined up to a set of measure $0$, so what does it even mean to be "weakly differentiable at $0$"? $\endgroup$
    – M W
    Commented Nov 5, 2023 at 20:29
  • $\begingroup$ Thanks for the useful observations. As for the first: yes, I was definitely sloppy there. For the second, I gave a (not very thoughtful) hint, but indeed that doens't make much sense. Adjustments such as "weakly differentiable in a neighbourhood of $0$", despite making sense, seem a bit too unrelated from the concept I would like to define. $\endgroup$
    – rod
    Commented Nov 5, 2023 at 23:15
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    $\begingroup$ Yes there is in a sense. For example, the Sobolev space $W^{1,p}(I,X)$ can be defined as the space of all functions $f\in L^p(I,X)$ such that for all Lipschitz function $\phi\colon X\to\mathbb{R}$ we have $\phi\circ f\in W^{1,p}(I)$ and the derivative is a.e. bounded by some $g\in L^p(I)$. Then instead of $X\to\mathbb{R}$ you can use an isometric embedding $X\to\ell^\infty(X)=\ell^1(X)^*$ and formulate the weak metric derivative from here (recall McShane extension lemma). However, you could get into some technicality as $\ell^\infty$ is usually "too big" as a dual Banach space to work with. $\endgroup$
    – user10354138
    Commented Nov 9, 2023 at 17:44

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