Hello i have a functional $I:X\rightarrow \mathbb{R}$ defined by $$I(u)=\int_{\Omega} \varphi(|u(x)|) u(x) v(x) dx $$
and we define $u^-=\min\{0,u\}$
does we have $$I(u)=-\int_{\Omega} \varphi(|u^-(x)|)u^-(x) v(x) dx$$ or not ?
Thank you
Negative part of a function
1
$\begingroup$
functional-analysis
analysis
1 Answers
1
It's a bit tricky to give a precise answer as you haven't specified what $X$ or $\varphi$ are, but I still doubt that this identity is true. Consider a function $u \ge 0$ on $\Omega$. Then $u^- =0$, and so $$ \int_{\Omega} \varphi(|u^-|) u^- v =0 $$ whereas $$ \int_\Omega \varphi(|u|) u v = \int_\Omega \varphi(u) u v. $$ If there exists a single nonnegative $u$ for which the latter integral doesn't vanish, then you have a contradiction.