Improve the estimate constant

Question

In Brezis proposition 9.3

I am trying to show $(i)$ implies $(ii)$.

if $u \in W^{1,p}(\Omega)$, then there are $g_{1},g_{2}, ... , g_{N} \in L^{p}(\Omega)$ , $g_{i} = \frac{\partial{u}}{\partial{x_{i}}}$ such that

$\int_{\Omega} u\frac{\partial{\phi}}{\partial{x_{i}}} = - \int_{\Omega}g_{i}\phi$, for tall $\phi \in C_{c}^{\infty}(\Omega)$, for tall $i = 1, ... , N$. There, applying Holder inequality

$\bigg|\int_{\Omega} u\frac{\partial{\phi}}{\partial{x_{i}}}\bigg| = \bigg|\int_{\Omega}g_{i}\phi \bigg| \leq ||\phi||_{L^{p'}(\Omega)}||\frac{\partial{u}}{\partial{x_{i}}}||_{L^{p}(\Omega)}$.

How to arrive $||\nabla u||_{L^{p}(\Omega)}$ instead $||\frac{\partial{u}}{\partial{x_{i}}}||_{L^{p}(\Omega)}$ ?

Answer 1

Observe the following fact: for every $p \geq 1$ it holds that $||\frac{\partial{u}}{\partial{x_{i}}}||_{L^{p}(\Omega)} \leq ||\nabla u||_{L^{p}(\Omega)}$.

This is because $|a_i| \leq \sqrt{a_1^2 + \dots a_n^2}$ for all $1 \leq i \leq n$ and thus $|a_i|^p \leq \bigg(\sqrt{a_1^2 + \dots a_n^2} \bigg)^p$.