norm in sobolev space $W_{1,p}$

Question

$(||u||_{Lp}^p+||\triangledown u||_{Lp}^p)^{(1/p)}$ is norm in $W_{1,p}$ , Where $u \in W_{1,p}$ . I need to prove triangle inequality for that.

Any help would be appreciated!

Answer 1

More in general, let $u,v \in W^{k,p}(U)$. Then if $1\le p < \infty$, Minkowski's inequality implies \begin{align} \|u+v\|_{W^{k,p}(U)} &= \left(\sum_{|\alpha|\le k} \|D^\alpha u + D^\alpha v \|_{L^p(U)}^p \right)^{1/p} \\ &\le \left(\sum_{|\alpha|\le k} (\|D^\alpha u\|_{L^p(U)}+\| D^\alpha v \|_{L^p(U)})^p \right)^{1/p} \\ &\le \left(\sum_{|\alpha|\le k} \|D^\alpha u\|_{L^p(U)}^p \right)^{1/p} + \left(\sum_{|\alpha|\le k} \|D^\alpha v \|_{L^p(U)}^p \right)^{1/p} \\ &= \|u\|_{W^{k,p}(U)} + \|v\|_{W^{k,p}(U)}. \end{align}

We used: definition, triangle inequality for the $L^p$ norm and then the Minkowski inequality.