Intuitive question about the boundary values of a Sobolev function

Question

Let $B_R$ a ball of radius $R$ in $R^n.$ Let $u \in H^1(B_R)$ and let $u^{1} \in H^1(B_R)$ with $u^1 - u^{+} \in H^1_{0}(B_R)$. ($u^{+}$ denotes the positive part of the function $u$). Let $v(y) : = max \{ u(y) , u^{1} (y)\}$. Intuitively $ v - u^{+} \in H^1_{0}(B_R)$ because $u \leq u^{+}$. I tried a lot to prove this, but I am getting anywhere. Someone could help me to prove or give a counterexample ?

Answer 1

You can write $$ v -u^+= \max(u,u^1)-u^+ = \max(u^+-u^1+u^-,0) + u^1-u^+. $$ Now by assumption $u^1-u^+\in H^1_0(B_R)$. Moreover, $u^+-u^1=0$ on the boundary, and $\max(u^-,0)=0$ on the boundary. Hence the trace of $v-u^+$ is zero. And $v-u^+\in H^1_0(B_R)$.