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In Gilbarg-Trudinger's section on the maximum principle for weak solutions, the boundary sup of Sobolev functions defined as follows:

Let $\Omega$ be a bounded domain in $\mathbb{R}^n$. For $u, v \in H^1(\Omega)$, $u \leq v$ on $\partial \Omega $ if $(u-v)^+ \in H^1_0 (\Omega)$. Define $$\sup_{\partial \Omega} u = \inf \; \{ k \in \mathbb{R} \; | \; u \leq k \text{ on } \partial \Omega \}, \quad u \in H^1(\Omega).$$

Here, of course, $H^1(\Omega) = W^{1,2}(\Omega)$ and $H^1_0(\Omega) = W^{1,2}_0(\Omega)$.

My question is whether the following inequalities are valid.

Let $u, v \in H^1(\Omega)$.

  1. $\sup_{\partial \Omega} (u+v) \leq \sup_{\partial \Omega} u + \sup_{\partial \Omega} v.$
  2. For $k \in \mathbb R$ with $\sup_{\partial \Omega} u \leq k$, $u \leq k$ on $\partial \Omega$.
  3. If $u \leq v$ on $\partial \Omega$, then $\sup_{\partial \Omega} u \leq \sup_{\partial \Omega} v.$

I tried to prove these inequalities by the definition, but it does not seem straightforward. I also looked for some questions on MathStackExchange and MathOverflow using some relevant keywords, but there are no results on these inequalities. (Addition: This might be related.)

I initially guessed that the inequalities are valid, and in fact those are true when the domain has Lipschitz boundary. The proof is done using the trace operator. See the comments and the linked Math.SE post above. However, I am not sure there is any proof without using the trace operator or without the assumption on regularity of the boundary. Could you provide any good idea?

Thanks!

P.S I posted the same question on MathOverflow (link)

Yongmin Park
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  • Do you know how to prove the corresponding properties for functions on a common domain with the usual definition of $\le$ and $\sup$? – GReyes Aug 22 '20 at 00:46
  • @GReyes Of course. But such a quite straightforward argument does not seem to be applied to my cases. In fact, even I do not know whether there is transitivity for the relation "$\leq$ on $\partial \Omega$". – Yongmin Park Aug 22 '20 at 03:25
  • I may be wrong here but suppose $u\le v$ and $v\le w$ on $\partial\Omega$. Then $u\le v$ a.e. in $\partial \Omega$ and $v\le w$ a.e. in $\partial \Omega$ (because $(u-v)^+$ and $(v-w)^+$ have zero trace). Therefore $u\le w$ a.e. there (just take the union of the null-sets where $u>v$ which is a null-set as well). Then $(u-w)^+$ has zero trace on $\partial\Omega$ and $u\le w$ according to the definition. – GReyes Aug 22 '20 at 22:38
  • @GReyes I think we need to assume $\Omega$ is a Lipschitz domain. Then, we can prove that $u \leq v$ on $\partial \Omega$ if and only if $Tu \leq Tv$ a.e. on $\partial \Omega$, where $T$ is the trace operator. You actually used this property, right? Moreover, all the inequalities are trivial using this property. However, I wonder there is another proof without using the trace operator... – Yongmin Park Aug 22 '20 at 23:21

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