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Let $\Omega$ an open and bounded domain in $R^n$ . Let $u \in W^{1,p} (\Omega) \cap L^{\infty}(\Omega)$ $(2 \leq p < \infty)$ . Let $B \subset \subset \Omega$ a open ball and consider $u_1$ the restriction of $u$ on $B$. Then we have $u_1 \in W^{1,p} (B) \cap L^{\infty}(B)$. Let $T : W^{1,p} (B) \rightarrow L^{p}(\partial B) $ the trace operator. Can I say that $T(u_1) \in L^{\infty}(\partial B)$ ? Intuitively this is true .. But I dont know how to proof or counter example... Someone can give me a help? I am not good with the trace operator...

thanks in advance

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  • $\begingroup$ Please have a look here. It gonna helpful. math.stackexchange.com/questions/1079409/… $\endgroup$
    – spatially
    Dec 25, 2014 at 0:01
  • $\begingroup$ Why do you ask all of your questions using the word intuitive? $\endgroup$
    – Tomás
    Dec 26, 2014 at 11:53
  • $\begingroup$ Hi Tomás . I write this when the affirmation seems to be true in simple cases (for example : smooth functions) $\endgroup$
    – math student
    Dec 26, 2014 at 14:51

1 Answer 1

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Let $\phi_n\in C^\infty(B)$ be a sequence satisfying: $\phi_n\to u_1$ in $W^{1,p}(B)$ and $\|\phi_n\|_\infty\le M$ for all $n$, where $M$ is a positive constant (see here).

Note that $\phi_n\to u_1$ in $W^{1,p}(B)$, therefore, $(\phi_n-M)^+\to (u_1-M)^+$ in $W^{1,p}(B)$.

By one hand, $T((\phi_n-M)^+)=0$ for all $n$. On the other hand, once $(\phi_n-M)^+\to (u_1-M)^+$ in $W^{1,p}(B)$, we have that $T((\phi_n-M)^+)\to T((u_1-M)^+)$, thus $T((u_1-M)^+)=0$.

The last paragraph does implies that $$\inf\{\alpha\in\mathbb{R}:\ T((u_1-\alpha)^+)=0\}<\infty.$$

To conclude, use this post.

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