Obtain $L^p$ norm by using "Riesz Representation"

Question

Let $\Omega\subset \mathbb R^N$ be open bounded with smooth boundary. Let $u\in W^{1,2}(\Omega)$ be given. Hence, we have $$ \int_\Omega|{\nabla u}|^2<\infty. $$ Let $\nu\in \mathcal S^{N-1}$ be given. Here $\mathcal S^{N-1}$ denotes the $N$ dimensional unit sphere. We have $$ \int_\Omega |\nabla u \cdot \nu|^pdx\leq \int_\Omega |\nabla u|^pdx<\infty $$ by $|\nu|=1$. (Note $\nu$ is only a unit vector here, not a function)

My question 1: do we have $$ \sup_{\nu\in \mathcal S^{N-1}}\int_\Omega |\nabla u \cdot \nu|^pdx =? \int_\Omega |\nabla u|^pdx $$ My question 2: can I replace $$ \int_\Omega |\nabla u|^pdx $$ by $$ \int_\Omega |\nabla u|^pd\mu $$ where $\mu$ is a positive finite Radon measure, and still have the same result as in question 1?

Thank you!

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PS: I feel the Riesz Representation has sth to do about my questions... but I am not sure...

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PPS: If my question 1 does not work. Would it be helpful to partition $\Omega$ into a countably many small pieces and do $\sup$ at each of them and sum up? i.e., will this work? $$ \sum_{n=1}^\infty \sup_{\nu\in\mathcal S^{N-1}}\int_{\Omega_n}|\nabla u\cdot \nu|^p dx = \int_\Omega |\nabla u|^pdx $$ where $\bigcup_{n=1}^\infty \Omega_n=\Omega$ and $\Omega_n$ is disjoint with each other.

Answer 1

As comments already revealed, the answer is negative. Consider the function $$f(x, y)=\frac{x^2+y^2}{2}$$ on the square domain $\Omega=[0,1]\times [0,1]$. You have that $$ \int_\Omega |\nabla f(x, y)|^2\, dxdy= \frac23, $$ but if $\nu=(\cos \theta, \sin \theta)$, then $$ \int_\Omega |\nabla f(x, y)\cdot \nu|^2\, dxdy = \frac13 + \frac{\cos \theta\sin\theta}{2}, $$ whose maximum is $\frac13+\frac14$, strictly smaller than $\frac23$.

Answer 2

As noted by @GiuseppeNegro, we do not get exact equality. But one can get "asymptotic equality" as follows:

As you noted yourself, we have $$ \sup_{|v|=1} \int |\nabla f \cdot v|^2 dx \leq \int |\nabla f|^2 dx, $$ simply by Cauchy Schwarz.

Conversely, by taking $v =e_i $ to be an element of the standard basis, we get $$ \sup_{|v|=1} \dots \geq \int |\partial_i f|^2 dx $$ for every $i \in \{1,\dots n\} $. Now, sum over $i $ to get $$ \sup \dots \geq \frac {1}{n} \int |\nabla f|^2 dx. $$