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


1 Answer 1


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 \begin{equation} \int_\Omega|\nabla(\eta^2 u)|^2 \leq C \lVert u \rVert_{H^1_0(\Omega)}^2 \, , \end{equation} for some constant $C$ dependent on $\delta$.

  • $\begingroup$ 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 $\nabla (fg)= (\nabla f) g+f(\nabla g)$ $\endgroup$
    – Luca.b
    Aug 23, 2022 at 15:39
  • 1
    $\begingroup$ You can see that this holds true by mollifying $f$ and $g$ applying the product rule and passing to the limit. $\endgroup$ Aug 24, 2022 at 22:56

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