Let $\gamma\_{\varepsilon} \rightharpoonup \gamma$ in $W^{1,\infty}(0,1)$. Then for any fixed $s \in \mathbb (0,1)$ does the limit $\lim\_{\varepsilon \rightarrow 0} \frac{\gamma\_{\varepsilon}(s\varepsilon)}{\varepsilon}$ exist?


I am on the fence as to whether or not it does. I rewrite it as $\lim\_{\varepsilon \rightarrow 0} \frac{s\gamma\_{\varepsilon}(s\varepsilon)}{s\varepsilon}$. Then possibly, the limit is $s D\gamma(0)$.



Convergence of Difference Quotients
You must have assumed $\gamma(0)=0$. Let $f\_\epsilon\in L^\infty$ be the derivative of $\gamma\_\epsilon$. Then
$$\frac1\epsilon\gamma\_\epsilon(s\epsilon)=\int\_0^sf\_\epsilon(t\epsilon)dt.$$
Your assumption is that $f\_\epsilon$ converges weak-star to some $f\in L^\infty$. Take for instance
$$f\_\epsilon(x)=\sin\frac{x}\epsilon,$$
which converges to $f\equiv0$ weak-star. Then $\frac1\epsilon\gamma\_\epsilon(s\epsilon)$ converges towards
$$\int\_0^s\sin t dt,$$
which does depend upon $s$. So your guess is wrong.


