Let $ A $ be an $ \mathcal{H}^1$-measurable subset of $ \mathbb{R} $ and $ \gamma \colon A \subseteq \mathbb{R} \to \ell^\infty $ be a Lipschitz mapping with the Lipschitz constant $ L $. Also, assume that for all $ n \in \mathbb{N} $ and $ \mathcal{H}^1$-a.e. $ t \in A $,
$$
\gamma\_n'(t) = 0,
$$
where $ \gamma\_n $ is the $ n^{th} $ component of $ \gamma $.


I want to prove that
$$
\lim\_{h \to 0}{\frac{\|\gamma(t+h) - \gamma(t)\|\_\infty}{\vert h \vert}} = 0,
$$
at $ \mathcal{H}^1$-a.e. $ t \in A $.


Any suggestions or ideas are greatly welcomed. Thanks.


Edit: If $ A $ happens to be an interval, the proof is extremely easy because each component is going to be constant. The problem is $ A $ is not necessarily an interval.



Is the speed of a curve in $ \ell^\infty $ zero a.e. if the derivative of each component is zero a.e.?
Is ${\cal H}^1$ one-dimensional Hausdorff measure? So this is just Lebesgue measure?


Then I think the answer is yes, specifically the desired limit is zero at every Lebesgue point of $A$. If $t$ is a Lebesgue point then for any $\epsilon > 0$ we can find $r > 0$ such that $\frac{\mu(A \cap I)}{\mu(I)} \geq 1-\epsilon$ for any interval $I$ centered at $t$ of length at most $2r$. Then any function which has Lipschitz number at most $L$ and whose derivative is zero on $A$ will satisfy $\frac{|f(t+h)-f(t)|}{|h|} \leq 2L\epsilon$ for all $|h| \leq r$. So all components will satisfy this inequality, hence you also get it when you take the sup norm.


