Let $ \Bbb S^{d-1}=\{(x\_1,\cdots ,x\_d): x\_1^2+ \cdots +x\_d^2=1\}\subset \Bbb R^d$ be the unit
sphere. Let $\nabla u= (\partial\_{x\_1}u,\cdots, \partial\_{x\_d}u)$ be the gradient of a function $u\in C\_c^\infty(\Omega)$ with $\Omega \subset \Bbb R^d$ open. For $e\in \Bbb S^{d-1}$, we write $$\nabla u(x)\cdot e = \partial\_{x\_1}u(x) e\_1+\cdots+\partial\_{x\_d}u(x) e\_d.$$


Then, for $p\geq 1$, what is the following quantity?


\begin{split}
I(u) =\sup\_{e\in \mathbb{S}^{d-1}}\int\_\Omega |\nabla u(x)\cdot e|^pdx 
\end{split}


I have tried to use the fact that for all $z\in \Bbb R^d$ we have, $$|z|=\sup\_{e\in \mathbb{S}^{d-1}}|z\cdot e|$$ And only got the upper estimate $$I(u)\leq \int\_\Omega |\nabla u(x)|^pdx$$
In fact I am expecting an equality here.



Optimizing the gradient norm on the unit sphere
I do not believe there is a simple formula to express $I(u)$, but for sure for most of the functions the inequality
$$
I(u)<\int\_\Omega |\nabla u|^p\, dx
$$
is sharp. For example if $\Omega=B$ is a ball and $u(x)=f(|x|)$ is a radial function, then $\nabla u$ is a vector field orthogonal to the sphere $\mathbb{S}^{d-1}$. Then $|\nabla u(x)|=|\nabla u(x)\cdot e|$ only for points $x$ on the line passing through $0$ and parallel to $e$ for all other points the inequality is sharp.


To have equality you would have to have supremum under the sign of the integral, but that would be a different expression.


