# An equivalent norm in a subspace of $H^2 (\Omega)$

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?

• 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). Commented Dec 9, 2019 at 14:10
• @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. Commented Dec 9, 2019 at 14:30
• I meant regularity of the boundary. If it is smooth, you are fine. Commented Dec 9, 2019 at 14:42
• I still can't write the argument, could you help me @MaoWao? Commented Dec 9, 2019 at 16:42
• $c\|u\|_{H^2}^2 \leq \|u-0\|_2^2+\|\Delta u\|_2^2 \leq C(\|\nabla u\|_2^2 + \|\Delta u\|^2_2) \leq C \|u\|_{H^2}^2$ for all $u \in \tilde V$
– Cahn
Commented Dec 9, 2019 at 20:13

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 $$\tilde V$$.