Let $\Omega\subset \Bbb R^d$ be an open set, $u \in W^{1,p}(\Omega)$ with $1\leq p<\infty $ such that for every $\varphi\in C_c^\infty(\Omega)$ every unit vector $e\in \mathbb{S}^{d-1}$, \begin{align}\label{eq:cont-weak-derivatve} \left|\int_{\Omega} u(x) \nabla \varphi (x)\cdot e ~\mathrm{d}x \right| \leq C\|\varphi\|_{L^{p'}(\Omega)}. \end{align} where $1/p+1/p'=1$ and $C>0$.
Can I conclude that \begin{align}%\label{eq:cont-weak-derivatve} \int_{\Omega} |\nabla u(x)|^p \mathrm{d}x \leq C^p. \end{align} with $|z|= (z_1^2+\cdots+z_d^2)^{1/2}$. when I take $e= e_i$ canonical coordinates I get \begin{align} \int_{\Omega} |\partial_{x_i}u(x)|^p \mathrm{d}x \leq C^p. \end{align} Note that $(|z_1|^p+\cdots+|z_d|^p)^{1/p}=1$ then $|z_i|\leq 1$ so that $|z|\leq d^{1/2}$. Thu, in general $|z|^p\leq d^{p/2}(|z_1|^p+\cdots+|z_d|^p)$. Take $z=\nabla u$ then this implies
\begin{align} \int_{\Omega} |\nabla u(x)|^p \mathrm{d}x \leq d^{1+p/2}C^p . \end{align} But I don't want the fact $d^{1+p/2}$