I am reading Brenner and Scott's book on finite elements, chapter 12 at the moment. I have come up to something that seems simple, but that I am having trouble figuring out by myself to do with the relationship of different norms.
I have got the spaces
$H^{1}(\Omega)^{n} = \{ w \in L^{2}(\Omega)^{n}\ |\ \partial_{i}w \in L^{2}(\Omega)^{n}\}\\ H^{1}_{0}(\Omega)^{n} = \{ w \in H^{1}(\Omega)^{n}\ |\ w = 0 \text{ for } x \in \partial\Omega\}$
where $\Omega \subset \mathbb{R}^{n}$ is a polygonal domain, and $H^{1}_{0}$ is the subset of $H^{1}$ functions with vanishing trace. I also define
$H(\text{div}) = \{ w \in L^{2}(\Omega)^{n}\ |\ \text{div}(w) \in L^{2}(\Omega). \} $
The corresponding norms are
$\displaystyle || w ||_{L^{2}(\Omega)^{n}} = \left(\sum_{i=1}^{n}||w_{i}||_{L^{2}(\Omega)}^{2}\right)^{1/2}\\ \displaystyle || w ||_{H^{1}(\Omega)^{n}} = \left(|| w ||_{L^{2}(\Omega)^{n}}^{2} + \sum_{i=1}^{n}||\partial_{i}w||_{L^{2}(\Omega)^{n}}^{2}\right)^{1/2} \\ ||w||_{H(\text{div})} = \left( ||w||_{L^{2}(\Omega)^{n}}^{2} + ||\text{div}(w)||_{L^{2}(\Omega)}^{2}\right)^{1/2}.$
There is an inequality (Brenner and Scott, pg. 336) which seems very simple, but I can't justify, which is
$\displaystyle \frac{b(w,p)}{||w||_{H^{1}(\Omega)^{n}}} \leq \sqrt{n}\frac{b(w,p)}{||w||_{H(\text{div})}}$
for $w \in H_{0}^{1}(\Omega)$. Since the only thing that changes in here is the norm, I am now trying to show that
$\displaystyle ||w||_{H^{1}(\Omega)^{n}} \geq \frac{1}{\sqrt{n}}||w||_{H(\text{div})}\tag{1}$
which I can't. I give some more details below, but the inequality (1) is what I am trying to prove
I am not sure how to use the fact that $w\in H_{0}^{1}$, and have only gotten so far as to write out the different norms in full. I have that
$$||w||_{H^{1}}^{2} = ||w||_{L^{2}}^{2} + \sum_{i=1}^{n}||\partial_{i}w||_{L^{2}}^{2} \\ \qquad \qquad \qquad \quad = \sum_{i=1}^{n}\left( ||w_{i}||_{L^{2}}^{2} + \sum_{j=1}^{n}||\partial_{i}w_{j}||_{L^{2}}^{2} \right)$$
and
$$||w||_{H(\text{div})}^{2} = ||w||_{L^{2}}^{2} + ||\sum_{i=1}^{n}\partial_{i}w_{i}||_{L^{2}}^{2} \\ \qquad\qquad\qquad\quad \leq \sum_{i=1}^{n}||w_{i}||_{L^{2}}^{2} + \left(\sum_{i=1}^{n}||\partial_{i}w_{i}||_{L^{2}}\right)^{2}.$$
I am not sure if this is a good start, as with this direct method of comparison I don't know how to use the fact that $w =0$ on $\partial\Omega$. Any hints as to how I can show inequality (1) would be appreciated
Keeran
There is less than meets the eye here. By Cauchy-Schwarz, for any vector $ (v_1,\dots,v_n)$ we have $$\left|\sum_{i=1}^n v_i\right|^2 \le n \sum_{i=1}^n v_i^2$$ In particular,
$$\left|\sum_{i=1}^n \partial_i w_i\right|^2 \le n \sum_{i=1}^n (\partial_i w_i)^2 \le n \sum_{i=1}^n \|\partial_i w\|^2$$ Integration yields $$ \|\mathrm{div}\, w\|_{L^2}^2 \le n \sum_{i=1}^n \|\partial_i w\|_{L^2}^2 $$ Therefore, $$ \|w\|_{H(\mathrm{div})}^2 \le n\|w\|_{H^1}^2 $$ Boundary values of $w$ are irrelevant.