Showing a bilinear form is coercive and continuous

Question

Let $\Omega \subset \mathbb{R}^3$ be open and bounded. Define $H$ to be the completion of the space divergence-free smooth functions with compact support in $\Omega$ in the $(L^2(\Omega))^3$ norm. Then define another space $V$ as $V = H \cap (H^1(\Omega))^3$ equipped with the $(H^1(\Omega))^3$ norm.

I am following a book that proves that the Stokes problem has a unique weak solution. In their proof they define a bilinear map $h: V \times V \rightarrow \mathbb{R}$ by $$h(u,v) = \langle \nabla u, \nabla v\rangle$$ where $\langle f, g \rangle = \int f g$ is the usual pairing. They claim that $h$ is coercive and continuous since $$h(u,u) = \|\nabla u\|^2 \geq c_1 \|u\|_{H^1}^2 \tag{1}$$ $$|h(u,v)| = |\langle \nabla u, \nabla v\rangle| \leq c_2 \|u\|_{H^1}\|v\|_{H^1} \tag{2}$$

I have a few questions:

1. What is the definition for the norms in $(L^2(\Omega))^3$ and $(H^1(\Omega))^3$? Are these the usual $L^2$ and $H^1$ norms but the integrand is replaced by the $\mathbb{R}^3$ norm of the function?
2. How is $h(u,u) = \|\nabla u\|^2$? I see how this follows if $\| \cdot \| = \|\cdot \|_{L^2}$, but they did not specify so I assumed $\|\cdot \| = \| \cdot \|_V$.
3. Where do the inequalities in (1) and (2) come from?

Answer 1

Ad 1: You can identify $(L^2(\Omega))^3$ with the Bochner space $L^2(\Omega;\mathbb{R}^3)$.
Ad 2: When they write $h(u,u) = \Vert \nabla u\Vert^2$, the norm on the right hand side is the $L^2$-norm simply by definition $h(u,u) = \int \nabla u\cdot\nabla u$, and the norm on $V$ is the $(H^1(\Omega))^3$ norm.
Ad 3: For inequality (1) I would use some kind of Poincare inequality and for (2) Hölder's inequality.
EDIT: For $u\in V$ it holds (Why can we use the Poincare inequality here?) $$ \Vert u \Vert_{L^2}^2 \leq C^2 \Vert \nabla u \Vert_{L^2}^2 $$ for some constant $C>0$. Adding $\Vert \nabla u \Vert_{L^2}^2$ to both side yields $$ \Vert u \Vert_{H^1}^2 \leq (C^2+1) \Vert \nabla u \Vert_{L^2}^2.$$ For (2), we can use Hölder's inequality and obtain for all $u,v\in V$ $$\vert \langle \nabla u, \nabla v \rangle \vert \leq \Vert \nabla u \Vert_{L^2} \Vert \nabla v \Vert_{L^2}.$$ From there (2) follows immediately.