1
$\begingroup$

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 $s D\gamma(0)$.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thank you for your answer, this example is sufficient enough to address my query. $\endgroup$
    – dcs24
    Jul 6 '12 at 15:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.