Decompositing Sobolev Norms on Subspaces

Question

Notation: Let $\Omega = \Omega_1\cup\dots\cup \Omega_n$ be disjoint open bounded subsets of $\mathbb{R}^d$. Let $\|\cdot\|_{k,\infty:A}$ be the Sobolev norm given by $$ \|u\|_{k,\infty:A} \triangleq \max_{|\alpha|\leq k} \|D^{\alpha}u\|_{L^{\infty}}, $$ where $u$ is $k$-times weakly differentiable on $A$, and $A$ is a stand-in for any of the sets $\Omega,\Omega_1,\dots,\Omega_n$.

What is the relationship between $\|u\|_{k,\infty:\Omega}$ and $\|u\|_{k,\infty:\Omega_1},\dots,\|u\|_{k,\infty:\Omega_n}$?

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Hypothesis:

What I (think) I have is that $$ \|u\|_{k,\infty:\Omega} \leq \sum_{i=1}^n \|u\|_{k,\infty:\Omega_i} $$

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Reasoning:

Is my reasoning correct? Something is bothering me about it... $$ \begin{aligned} \|u\|_{k,\infty:\Omega} & = \max_{|a|\leq k}\operatorname{ess-sup}_{x \in \Omega} \|D^{\alpha} u\|\\ & = \max_{|a|\leq k}\max_{i=1,\dots, n}\operatorname{ess-sup}_{x \in \Omega_i} \|D^{\alpha} u\|\\ & = \max_{i=1,\dots, n}\max_{|a|\leq k}\operatorname{ess-sup}_{x \in \Omega_i} \|D^{\alpha} u\|\\ & = \max_{i=1,\dots, n}\|u\|_{k,\infty:\Omega_i}\\ & = \sum_{i=1}^n \|u\|_{k,\infty:\Omega_i}. \end{aligned} $$

Have I missed something?

Answer 1

Your system $$ \begin{aligned} \|u\|_{k,\infty:\Omega} & = \max_{|a|\leq k}\operatorname{ess-sup}_{x \in \Omega} \|D^{\alpha} u\|\\ & = \max_{|a|\leq k}\max_{i=1,\dots, n}\operatorname{ess-sup}_{x \in \Omega_i} \|D^{\alpha} u\|\\ & = \max_{i=1,\dots, n}\max_{|a|\leq k}\operatorname{ess-sup}_{x \in \Omega_i} \|D^{\alpha} u\|\\ & = \max_{i=1,\dots, n}\|u\|_{k,\infty:\Omega_i}\\ & = \sum_{i=1}^n \|u\|_{k,\infty:\Omega_i}. \end{aligned} $$ is correct with the exception of the last equality, which is wrong.

But your hypothesis is correct, because the inequality $$ \max_{i=1,\dots, n}\|u\|_{k,\infty:\Omega_i} \leq \sum_{i=1}^n \|u\|_{k,\infty:\Omega_i}. $$ holds.

In my opinion, the best description of $\|u\|_{k,\infty:\Omega}$ is $$ \|u\|_{k,\infty:\Omega} = \max_{i=1,\dots, n}\|u\|_{k,\infty:\Omega_i} = \|(\|u\|_{k,\infty:\Omega_i})_{i=1}^n\|_{\ell^\infty}, $$ where we used the $\ell^\infty$ norm on $\Bbb R^n$.