# Cutoff function in the proof of inner regularity of Poisson equation

In chapter 11 of Jost partial differential equations he wants to prove the following theorem of interior regularity for the Poisson equation:

Theorem 11.2.1: Let $$u\in W^{1,2}(\Omega)$$ be a weak solution of $$\Delta u=f$$, with $$f\in L^2(\Omega)$$. For any $$\Omega'\subset\subset\Omega$$, then $$u\in W^{2,2}(\Omega')$$ and $$\lVert u \rVert_{W_{2,2}(\Omega')}\leq const(\lVert u \rVert_{L_2(\Omega)}+\lVert f \rVert_{L_2(\Omega)})$$

where $$\Omega \subseteq \mathbb{R}^n$$ is a bounded open domain. To prove this theorem he uses this cutoff function: $$\eta(x)= \begin{cases} 1 & x \in \Omega'\\ 0 & d(x,\Omega')\geq \delta\\ 1-\frac{1}{\delta}d(x,\Omega') & 0\leq d(x,\Omega')\leq \delta \end{cases}$$ where $$\delta=d(\Omega',\partial \Omega)$$. It is easy to prove that $$\eta$$ satisfies the following properties:

• $$0\leq \eta \leq 1$$
• $$\eta=1$$ on $$\Omega'$$
• $$\eta\in W_0^{1,2}(\Omega)$$
• $$\lvert \nabla\eta\rvert\leq \frac{2}{\delta}$$

Then he takes $$v=\eta^2 u$$ and he plugs it in the definition of weak solution: $$\int_\Omega \nabla u \cdot \nabla v=-\int_\Omega fv \qquad \forall v\in H_0^1(\Omega)=W_0^{1,2}(\Omega)$$ I cannot see why $$v=\eta^2 u\in H_0^1(\Omega)$$. I know that $$\eta, u\in H_0^1(\Omega)$$, but $$H_0^1(\Omega)$$ is not an algebra.

I know that I can take a $$C^\infty_c(\Omega)$$ cutoff function using mollifications, but I would like to understand why we can use this simpler cutoff function.

Thank you

Edit: Since $$\eta$$ is lipschitz $$\eta\in W^{1,\infty}_0(\Omega)$$. Now from this lemma (Differentiation of a product of Sobolev functions) I deduce that also $$\eta^2\in W^{1,\infty}_0(\Omega)$$. I don't know if it can help

The point is that you have more information about $$\eta$$ than just the fact that it is in $$H^1_0(\Omega)$$. Applying the product rule, we have \begin{align} |\nabla(\eta^2 u)| =& |2 \eta u \nabla \eta + \eta^2 \nabla u |\\ \leq & 2| \eta| |u|| \nabla \eta| + |\eta|^2| \nabla u | \\ \leq & \frac{4}{\delta}|u| + |\nabla u| \, . \end{align} Thus $$$$\int_\Omega|\nabla(\eta^2 u)|^2 \leq C \lVert u \rVert_{H^1_0(\Omega)}^2 \, ,$$$$ for some constant $$C$$ dependent on $$\delta$$.
• Thank you, but I don't know why I can apply the product rule in this case. All the versions I know have stronger hypothesis. In other words I think I miss a theorem like: if $f,g\in H_0^1(\Omega)$ and $(\nabla f) g+f(\nabla g)\in L^2(\Omega)$ then $fg\in H_0^1(\Omega)$ e $\nabla (fg)= (\nabla f) g+f(\nabla g)$ Aug 23, 2022 at 15:39
• You can see that this holds true by mollifying $f$ and $g$ applying the product rule and passing to the limit. Aug 24, 2022 at 22:56