# simple exercise sobolev spaces

let B a open Ball in R^n . Let $\psi \in C_{o}^{\infty} (B)$ . Define $\varphi = max ( 0 , \psi ) = {\psi}^{+}$. let $1 < p <q < \infty$ . Consider $s = q/p$ . Then ${\varphi}^{s} \in W^{1,p} (B)$

The affirmation is true or false?

Depends on the interpretation of "$\psi\in C_0^\infty(B)$". If it simply means that $\psi$ is an infinitely differentiable function that is zero on the boundary, then the statement is false. Indeed, $\psi$ can oscillate rapidly at small scale near the boundary, which will make its derivative non-integrable, and consequently the derivatives of $\varphi$ and $\varphi^s$ as well.
But if $\psi\in C_0^\infty(B)$ includes the continuity of derivatives up to the boundary, then the statement is true. Indeed, since the derivatives of $\psi$ are bounded, it is a Lipschitz function. The maximum of Lipschitz functions is also Lipschitz. Finally, raising a bounded Lipschitz function to a power greater than $1$, we get a Lipschitz function again. Thus $\varphi^s\in W^{1,\infty}(B)\subset W^{1,p}(B)$.
• My interpretation is your second interpretation . I didnt understand your afirmation : raising a bounded Lipschitz function to a power greater than 1, we get a Lipschitz function again. Thus $\varphi^s$∈W1,∞(B)⊂W1,p(B) – math student Feb 19 '13 at 2:34
• @LeandroTavares This follows from the chain rule: $\nabla (\varphi^s) = s\varphi^{s-1} \nabla \varphi$. In this product every term is bounded. – user53153 Feb 19 '13 at 2:37
• why the coordinates of $∇ \varphi$ is bounded? – math student Feb 19 '13 at 2:44
• @LeandroTavares Because $\varphi$ is a Lipschitz function. – user53153 Feb 19 '13 at 2:45