I'm reading proof which begins with a statement that confuses me:

Prop: Let $u \in H^1(\Omega)$ and suppose that $$\sup_{\partial \Omega} u = M = \inf \{m\in \mathbb{R} : u \leq m \text{ on } \partial \Omega \text{ in } H^1(\Omega)\} < \infty$$ Then for any $k \geq M$, $\max(u-k,0) \in H^1_0(\Omega)$ and $\max(u-k,0)\geq 0$ on $\Omega$ in the $H^1$ sense (this means there's a non-negative sequence of Lipsichtz continuous functions converging to $\max(u-k,0)$.)

The proof starts:

To show $\max(u-k,0) \in H^1_0(\Omega)$ it suffices to prove the existance of a sequence $v_n \in H^{1,\infty}_0(\Omega)$ (i.e. lipschitz continuous functions) such that $$v_n \rightharpoonup \max(u-k,0) \in H^1(\Omega)$$

Why does it suffice to show weak convergence in $H^1$?? Also why weak convergence in $H^1$ not $H^1_0$?


This follows from the Lemma of Mazur. It implies that there is a sequence of convex combinations of the sequence $v_n$ which converges strongly to $\max(u-k,0)$. These convex combinations are still Lipschitz continuous.


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