The following questions concerns a problem I am treating in my Masters dissertation.

Let $\Omega $ be an open, bounded domain in $\mathbb{R}^3$. Then the norm

$$ \Vert u\Vert^2 = \Vert u\Vert_2^2 + \Vert\nabla u\Vert_2^2 + \Vert\Delta u\Vert_2^2 $$

equivalent to the usual norm in $H^2 (\Omega)$. Indeed, I know from this question that $(\Vert u\Vert_2^2 + \Vert\Delta u\Vert_2^2)^{1/2}$ is equivalent to the norm in $H^2(\Omega)$. From this follows that for some $c > 0$

\begin{align*} c \Vert u\Vert_{H^2}^2 & \leq \Vert u \Vert_2^2 + \Vert\Delta u\Vert_2^2 \\ & \leq \Vert u\Vert_2^2 + \Vert\nabla u\Vert_2^2 + \Vert\Delta u\Vert_2^2 \\ & \leq \Vert u\Vert_{H^2}^2 \end{align*}

Now, let $$ V = \left\{u \in H^2(\Omega) \ : \ \frac{\partial u}{\partial n} = 0 \text{ on } \partial \Omega\right\} $$ and $$ \tilde V = \left\{ u \in V \ : \ \int_\Omega u \ dx = 0 \right\}. $$ How can I show that
$$ \Vert u\Vert = \Vert\nabla u\Vert_2 + \Vert\Delta u\Vert_2 $$ is an equivalent norm on $\tilde V$? Does it follow from the Poincaré-Wirtinger inequality?

  • 1
    $\begingroup$ The hint for your actual question is Poincaré's inequality. But you have to be careful - the linked question treats the case of Sobolev spaces on $\mathbb{R}^d$, for the version on bounded domains you need some boundary regularity. Without it, this equivalence of norms can fail (Mazya's book on Sobolev spaces contains some cautionary examples). $\endgroup$
    – MaoWao
    Commented Dec 9, 2019 at 14:10
  • $\begingroup$ @MaoWao By boundary regularity you mean that the boundary must be regular or that the function must have some regularity on the boundary? In my case I am dealing with an smooth, bounded open set. $\endgroup$ Commented Dec 9, 2019 at 14:30
  • 1
    $\begingroup$ I meant regularity of the boundary. If it is smooth, you are fine. $\endgroup$
    – MaoWao
    Commented Dec 9, 2019 at 14:42
  • $\begingroup$ I still can't write the argument, could you help me @MaoWao? $\endgroup$ Commented Dec 9, 2019 at 16:42
  • 2
    $\begingroup$ $c\|u\|_{H^2}^2 \leq \|u-0\|_2^2+\|\Delta u\|_2^2 \leq C(\|\nabla u\|_2^2 + \|\Delta u\|^2_2) \leq C \|u\|_{H^2}^2$ for all $u \in \tilde V$ $\endgroup$
    – Cahn
    Commented Dec 9, 2019 at 20:13

1 Answer 1


As you already noted, $(\|u\|_2+\|\Delta u\|_2)^{1/2}$ is an equivalent norm on $H^2$, hence there is a constant $c>0$ such that $$c\|u\|_{H^2}^2 \leq \|u\|_2^2+\|\Delta u\|_2^2, \quad \forall u \in H^2.$$ This is also true for $u \in \tilde V \subset H^2$. Since the elements in $\tilde V$ have zero mean, we can make use of the following Poincare inequality, $$\|u\|_2^2 \leq C_P\|\nabla u\|^2, \quad \forall u \in \tilde V. $$

Putting these two inequalities together, we have for all $u \in \tilde V$,

$$c\|u\|_{H^2}^2 \leq \|u\|_2^2+\|\Delta u\|_2^2 \leq C_P\|\nabla u\|_2^2 + \|\Delta u\|^2_2 \leq C (\|\nabla u\|_2^2 + \|\Delta u\|^2_2) \leq C \|u\|_{H^2}^2,$$

where $C:=\max\{1,C_P\}$. Hence, $(\|\nabla u\|_2^2 + \|\Delta u\|_2^2)^{1/2}$ is an equivalent norm on $\tilde V$.


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