# Does weak convergence in $H^1$ imply strong convergence in $H^1_0$?

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.