Optimizing the gradient norm on the unit sphere 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.
 A: 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.
A: More of a comment than an answer, really, but too long for the comment box: For a fixed smooth function $u$ the map
$$
e\mapsto U(e):=\int_\Omega |\nabla u(x)\cdot e|^p dx
$$
is differentiable as a function of $e\in\mathbb R^{d}$, and its differential in the direction $h$ is simply
$$
DU(e).h
=\int_\Omega p|\nabla u(x)\cdot e|^{p-2} \nabla u(x)\cdot h\, dx
=\left(\int_\Omega p|\nabla u(x)\cdot e|^{p-2} \nabla u(x)\,dx\right)\cdot h
$$
A maximizer $e$ on the unit sphere must then satisfy the first-order optimality condition $ DU(e)\cdot h=0$ for all tangent directions $h\in T_e\mathbb S^{d-1}\Leftrightarrow h\cdot e=0$, which means here that $DU(e)$ must be colinear to $e$. In other words,
$$
e=C \int_\Omega p|\nabla u(x)\cdot e|^{p-2} \nabla u(x)\,dx
$$
for some normalization constant $C=\frac{\pm1}{\left|\int_\Omega p|\nabla u(x)\cdot e|^{p-2} \nabla u(x)\,dx\right|}$ (provided the denominator does not vanish).
Of course we have a $\pm$ degree of freedom due to the invariance $U(e)=U(-e)$.
I don't know how much one can exatract from this integral condition, but at least it is clear that the reasonable guess $e=C\int |\nabla u|^{p-2}\nabla u$ is too naive and does not work (since it does not satisfy a priori this integral condition).
Note that for $p=2$ the solution is obviously given by $e= C\int \nabla u$, the average gradient (provided it is not zero, of course), so the functional is somehow the "directional $TV$ norm" $I(u)=\int |\partial_e u|$ in the average (most varying) direction $e=C\int \nabla u$.
Interesting functional!
A: A bit of a longer comment and expansion on Piotr's answer: if you restrict to the case $p$ is even, you can write your integral as
$$ \int |\nabla u \cdot e|^p = \int (\nabla u \otimes \cdots \otimes \nabla u) \cdot (e \otimes \cdots \otimes e) $$
where the tensor product is taken over $p$ times. Since the arguments are manifestly symmetric, you have
$$ I(u) \leq \sup_{T \in \mathcal{S}^P\mathbb{R}^d, |T| = 1} \int (\otimes^p \nabla u) \cdot T =: J(u) $$
Here $\mathcal{S}^p\mathbb{R}^d$ is the set of symmetric rank $p$ tensors over $\mathbb{R}^d$, with induced inner product. The optimization of $J(u)$ is a linear problem and can be easily solved with
$$ T = \frac{\int \otimes^p \nabla u}{| \int \otimes^p \nabla u|}$$
and $J(u) = | \int \otimes^p \nabla u |$.
If you are lucky enough that this $T$ is a pure tensor (of the form $e \otimes e \otimes \cdots \otimes e$) then $I(u) = J(u) = | \int \otimes^p \nabla u |$. However, for generic $u$ this cannot be expected, and you get $I(u) < J(u)$ in this case.
As a final note, you have that
$$ \left| \int \otimes^p \nabla u \right| \leq \int |\otimes^p \nabla u| = \int |\nabla u|^p $$
The inequality step is strict as long as $\nabla u$ is not almost everywhere parallel to a fixed vector. So this method, while it doesn't give you the exact value of $I(u)$, can at times give you a slightly improved upper bound.
A: LONG COMMENT: My collaborators and I have in fact studied the function
$$ x\in \mathbb{R}^d \mapsto \left(\int_{\mathbb{R}^d} |x\cdot\nabla f|^p\,dx\right)^{1/p} $$
quite extensively, for example in this paper. Here are a few highlights:

*

*It defines a norm on $\mathbb{R}^d$. The unit ball of this norm is called the $L^p$ polar projection body of the function $f$. The unit ball of the dual norm is called the $L^p$ projection body. This generalizes the definition of the $L^p$ projection and polar projection bodies of a convex body, which were studied here.

*If we denote the norm by $\|\cdot\|_{f,p}$, then the sup of $\|e\|_{f,p}$ over all unit vectors $e$ is half the diameter of the polar projection body, where the diameter is defined to be the maximum distance between two parallel supporting hyperplanes.

*Also, note that the quantity
$$ \int_{S^{d-1}} \|e\|_{f,p}^{-n}\,de = \int_{S^d}\left(\int_{\mathbb{R}^d} |e\cdot\nabla f|^p\,dx\right)^{-n/p}\,de $$
is the volume of the polar projection body. Although this does not define a norm on any function space, we proved that it satisfies a sharp affine Sobolev inequality. We call the inequality affine, because this quantity is invariant under the action of $SL(d)$ acting on $\mathbb{R}^d$. It is also not hard to show that this inequality implies the standard sharp Sobolev inequality on $\mathbb{R}^d$ that uses the standard Euclidean norm.

A paper that discuss this function, when $f$ is a probability density function can be found here:
E. Lutwak, D. Yang, G. Zhang. Moment-entropy inequalities, Annals of Probability 32 (2004) 757-774.
