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 $s D\gamma(0)$.
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.
