Consider $\Omega=[0,1]^3$ and the bilinear form
$B(u,v)=\int_{\Omega} (\nabla u)^T A \nabla v +uv \,dx, \; \; u,v \in H^1(\Omega).$
The matrix $A \in \mathbb{R}^{n \times n}$ is symmetric and positive definite. I want to show that $B$ is coerzive and bounded. So I have to show that there are constants $\delta$ und $C$ such that:
$B(v,v) \geq \delta ||v||_{H^1(\Omega)}^2 $ and $B(u,v) \leq C ||u||_{H^1(\Omega)} ||v||_{H^1(\Omega)}$.
The norm is defined as
$||v||_{H^1(\Omega)}:= \left(\int_{\Omega}v^2 \,dx + \int_\Omega v_x^2 \,dx +\int_\Omega v_y^2 \, dx +\int_\Omega v_z^2 \, dx \right)^\frac{1}{2},$ where $v_x$ denotes the partial derivative $\frac{\partial v}{\partial x}.$
My progress so far:
$B(v,v)= \int_\Omega a_{11}v_x^2 +a_{22}v_y^2+a_{33}v_z^2 +2a_{21}v_x v_y+2a_{31}v_x v_z+2a_{32}v_y v_z \, dx + \int_\Omega v^2 \, dx,$ where $a_{ij}$ denotes the matrix elements in $A$. The scalars $a_{11}, a_{22}, a_{33}$ are non-negative.
I dont see, how this could help me. Are there any tips to prove, that B is coerzive and furthermore bounded?
Notice that by the Cauchy-Schwarz inequality we have $$ |x^TAy|\leq |x||Ay|\leq \| A\|_{op}|x||y|,\qquad \forall x,y\in \mathbb{R}^3, $$ where $\| A\|_{op}= \max_{|z|=1} |Az|$, is the operator norm of $A$. On the other hand, by definition of $A$ being positive definite, $$ x^T A x > 0, \qquad \forall x\in \mathbb{R}^3\setminus\{0\}. $$ Restricting to the sphere, and using the homogeneity of the left-hand-side, we see that $$ x^TA x\geq m|x|^2, \qquad \forall x\in \mathbb{R}^3, $$ where $0<m := \min_{|z|=1} x^TAx$.
Use these with $x=\nabla u$ and $y=\nabla v$ to get the desired estimates on $B$.