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Say $v \in H^1_0(\Omega)$ for some bounded Lipschitz domain $\Omega \subset \mathbb{R}^n$ satisfying $|v(x)| \leq 1$ for almost any $x \in \Omega$. I wonder whether there exists a sequence $( v_j )_{j \in \mathbb{N}} \subset \mathcal{C}^\infty_0(\Omega)$ with $|v_j(x)| \leq 1$ for almost any $x \in \Omega$ and all $j \in \mathbb{N}$ such that $v_j \to v$ in $H^1(\Omega)$.

I considered setting $v = 0$ near the boundary and mollifying the resulting function. This should yield convergence in $L^2$ but only local convergence of the gradients in $L^2$. I then tried to use a partition of unity, but it seems that I then loose the compact support.

I am happy about any kind of suggestion. Thanks in advance.

neca
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1 Answers1

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Yes, this is possible.

Sketch of a proof:

This can be done with the help of the accepted answer to this question, but it requires some modifications.

Let $\varepsilon>0$. In the method in the linked answer one can also choose a function $\phi:\Bbb R\to\Bbb R$ such that $\phi(x)=x$ for $x\in [-1,1]$ and $|\phi(x)|<1+\varepsilon$ for all $x\in \Bbb R$.

Then there is a smooth approximation $u_\varepsilon \in C_0^{\infty}(\Omega)$ that satisfies $\|u_\varepsilon - v\|_{H_0^1(\Omega)}<\varepsilon$. ( Note that the $C_0^{\infty}(\Omega)$-property is not explicitly stated there, but will automatically happen if we use the same construction.)

Finally, we can define $$v_j := \frac1{1+1/j} u_{1/j},$$ which satisfies your assumptions.

harfe
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