Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $U$ be bounded, with a $C^1$ boundary. Show that a ''typical'' function $u\in L^p(U) (1\leq p < \infty)$ does not have a trace on $\partial U$. More precisely there does not exist a bounded linear operator $$T:L^p(U)\rightarrow L^p(\partial U) $$ such that $Tu=u|_{\partial U}$ whenever $u\in C(\bar{U})\bigcap L^p(U)$

share|improve this question
    
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. –  Julian Kuelshammer Dec 28 '12 at 10:34
    
There is another answer at this website, the link is: math.stackexchange.com/questions/332599/pde-question-in-evans –  user76346 May 6 '13 at 8:46

1 Answer 1

Hint: Construct an $L^p$ bounded sequence of continuous functions that look "nice" in the interior of $U$, but grow fast near the boundary. In order to construct this sequence explicitly, you can use the $C^1$ boundary condition in order to pass to local coordinates.

share|improve this answer
    
Another option is to build such functions out of $\operatorname{dist } (x,\partial U)$. That would work for rough domains too. –  user53153 Dec 28 '12 at 16:09

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.