# simple exercise - Sobolev Spaces

Let $\Omega \subset R^n$ a bounded and open set . Let $\psi : \Omega \rightarrow [ - \infty , + \infty]$ a function . Let $\eta \in H^{1,p} (\Omega)$ $( 1\leq p< \infty )$ . Let $u \in H^{1,p} (\Omega)$ with $u \geq \psi$ a.e in $\Omega$ and $u - \eta \in H^{1,p}_{0} (\Omega)$. Supose that $\psi$ and $\eta$ are essentially bounded above by a constant $M$ in $\Omega$. Show that the function $min (u , M )$ satisfies $min (u,M) \geq \psi$ a.e in $\Omega$ and $min(u,M) - \eta \in H^{1,p}_{0} (\Omega)$ . Someone can give me a hint ?

• Hi Tomas I am trying to prove the lemma 3.24 in the book of heinonen . I dont know how to show that $min ( u , M ) - \eta \in H^{1,p}_{0}( \Omega )$. My advisor said to me to use "trace". But in the book of heinonen dont appear about the "trace". I never studied "trace" =\ ... . I am trying to prove without "trace" .... – math student Mar 16 '13 at 0:22
• Did you alread showed that $\min (u,M)$ is differentiable in the weak sense? There is no Lemma 3.24 in Heinonen. – Tomás Mar 16 '13 at 1:04
• Hi Tomas. I make a mistake. It's theorem 3.24 and not lemma.. sorry . Using theorem 1.20 of heinonen I have the weak diferentiability of the min (u , M) – math student Mar 16 '13 at 17:51
• To use trace theory, is necessary some regularity in the boundary of the domain. Do we have such regularity? Also, take a look here in trace part: en.wikipedia.org/wiki/Sobolev_space . Consult Evan's book too. – Tomás Mar 16 '13 at 19:34

Let's prove that $\min(u,M)-\eta\in H_0^{1,p}$. Because $u-\eta\in H_0^{1,p}$, we can find a sequence $v_n\in C_0^\infty(\Omega)$ such that $v_n\rightarrow u-\eta$ in $H^{1,p}$. Take any sequence $u_n$ such that $u_n\rightarrow u$ in $H^{1,p}$. Note that $v_n-u_n\rightarrow -\eta$ in $H^{1,p}$, so we can suppose that $v_n=u_n-\eta_n$, where $u_n\rightarrow u$ in $H^{1,p}$ and $\eta_n\rightarrow\eta$ in $H^{1,p}$.
Let $f_n=\min (u_n,M)-\eta_n$. Note that $f_n\in H_0^{1,p}$.
I will leave you to prove that $f_n\rightarrow \min(u,M)-\eta$ in $H^{1,p}$ (remember that $\min(u_n,M)=\frac{u_n+m-|u_n-M|}{2}\Big)$. Now there is only one problem to solve, to wit, $f_n$ does not need to be in $C_0^1$ (why?). But you can remedy this easily by taking for each $n$, $g_n$ close to $f_n$ in $H^{1,p}$. From here, you can conclude.
• Hi Tomás. Why $f_n \in H^{1,p}_{0}$ ? Nao consegui provar isso ... . Nao sei se é tao trivial q nao estou conseguindo enxergar.. – math student Mar 16 '13 at 23:38