Norm on $W^{1,\infty}(\mathbb{R}^N;\mathbb{R}^N)$ I found in two different sources two different definitions of the norm on the space $W^{1,\infty}(\mathbb{R}^N;\mathbb{R}^N)$ the space of bounded and Lipschitz-continuous applications from $\mathbb{R}^N$ into itself.
First definition of norm: 
$||\theta||=\sup_{x\neq y}\left \{|\theta(x)|+\frac{|\theta(x)-\theta(y)|}{|x-y|} \right\}$
Second definition:
$||\theta||=||\theta||_{\infty}+\mathrm{ess sup}_{y\in\mathbb{R}^N} ||D_y \theta (y)||$
These are from Antoine Henrot - Shape Variation and Optimization, EMS, 2005.
It is said that they are the same, but I don't understand why...
The third one is:
$||\theta||=\mathrm{ess sup}_{0\leq |\alpha|\leq 1}||D^\alpha \theta||_{\infty}$
This is from Pascal Frey - Introduction a l'optimisation de forme et application a mecanique des fluides, 2014.
I don't really understand what is exactly the space $W^{1,\infty}(\mathbb{R}^N;\mathbb{R}^N)$. And I don't see why do we have the same $x$ in the first definition of the norm. And I also can't see why do we need the essential supremum on the third one. Isn't it enough to use the maximum???
I didn't ever saw the notation $D_y \theta(y)$?? In the mentioned book says that it is a Frechet derivative and the norm involved is the operatorial one.
P.S. The norm $|.|$ is the euclidean norm of $\mathbb{R}^N$ or any other equivalent norm?
I would appreciate if someone could make a clean and clear definition of the space $W^{1,\infty}(\mathbb{R}^N;\mathbb{R}^N)$ and of its usual norm:
$||\theta||_{W^{1,\infty}(\mathbb{R}^N;\mathbb{R}^N)}$
 A: Let's think in $\mathbb{R}^1$ for convenience. Your function $\theta$ is Lipschitz; this implies that $\theta'$ is bounded wherever it exists, since
$$
|\theta'(y)| = \lim_{h\rightarrow 0}\left|\frac{\theta(y+h)-\theta(y)}{h}\right|\leq L.
$$
Furthermore, a Lipschitz function is differentiable almost everywhere (Rademacher). If we define
$$
\| f \|_{\infty} = \text{esssup}_{x\in\mathbb{R}} |f(x)|,
$$
where $|\cdot|$ is the absolute value function,
we have $\|\theta\|_{\infty}<\infty$ and $\|\theta'\|_{\infty}<\infty$. Therefore it makes sense to define
$$
\|\theta \| = \|\theta\|_{\infty} + \|\theta'\|_{\infty}.
$$
You can check this is indeed a norm (satisfies the triangle inequality), and it is precisely your second definition. Note that the essential supremum is the same as the maximum when it comes to $\theta$ because $\theta$ is defined everywhere; this is not the case for $\theta'$.
To see intuitively that this is equivalent to the first definition, suppose $\theta'$ exists everywhere and the values $\|\theta\|_{\infty}$ and $\|\theta'\|_{\infty}$ are attained by $|\theta(\alpha)|$ and $|\theta'(\beta)|$ respectively (without loss of generality $\beta\neq \alpha$). The second norm is therefore $\|\theta \| = |\theta(\alpha)| + |\theta'(\beta)|$. The first norm
$$
\|\theta\|
=\sup_{x\neq y}\left \{|\theta(x)|+\frac{|\theta(x)-\theta(y)|}{|x-y|} \right\}
=\sup_{y\neq \alpha}\left \{|\theta(\alpha)|+\left|\frac{\theta(\alpha)-\theta(y)}{\alpha-y} \right|\right\}
=|\theta(\alpha)|+\sup_{y\neq \alpha}\left \{\left|\frac{\theta(\alpha)-\theta(y)}{\alpha-y} \right|\right\}
,
$$
and
$$
\frac{\theta(\alpha)-\theta(y)}{\alpha-y}
= \theta'(\eta)
$$
for some $\eta \in (\alpha,y)$ (Mean Value Theorem). Therefore
$$
\sup_{y\neq \alpha}\left \{\left|\frac{\theta(\alpha)-\theta(y)}{\alpha-y} \right|\right\}
=
\sup_{\eta}\left \{\left|\theta'(\eta) \right|\right\},
$$
and $\|\theta \| = |\theta(\alpha)| + |\theta'(\beta)|$ indeed.
One has to be more careful making this argument because the suprema need not be attained, and $\theta$ may only be differentiable almost everywhere, but the main idea is there. The generalisation to $\mathbb{R}^N$ can also be done similarly, by considering directional (Fréchet) derivatives. I hope this helps!
