Hi everyone I'm stuck with a proof and would be happy if anyone could help me out.
Let V be a Hilbert Space and $A:V\times V \rightarrow \mathbb{R}$ symmetric, elliptic (coercive with constant $\alpha1$) and continuous (with constant $\alpha2$) bilinear form. We define of the space $X := V \times V$ the bilinear form $B : X \times X \rightarrow \mathbb{R}$ with \begin{align*} B((u_1,u_2),(v_1,v_2)) := A(u_1,v_1) + A(u_1,v_2)+ A(u_2,v_2) \ \forall (u_1,u_2),(v_1,v_2) \in X \end{align*} Show that $B$ is coercive with constant ? . Let $X_N\subset X $ be a linear subspace. Is $B$ also coercive on $X_N$?
Here is my try:
By definition $B$ is coercive on $X$ if
\begin{align*} B(z,z)\geq \beta ||z||^2_X \ \forall z \in X, \ \beta \in \mathbb{R^+} \end{align*} If $z \in X$ arbitrary we have $z=(u,v)$ for some arbitrary $u,v \in V$. Then we have
\begin{align*} B(z,z)=B((u,v),(u,v)) = A(u,u) + A(u,v)+ A(v,v) \end{align*}
Now we can use that $A$ is coercive with constant $\alpha_1$ to get
\begin{align*} B(z,z)=B((u,v),(u,v)) \geq \alpha_1||u||^2_V + \alpha_1||v||^2_V + A(u,v) \\ \geq (\alpha_1 + \frac{A(u,v)}{||u||^2_V + ||v||^2_V})(||u||^2_V + ||v||^2_V) \end{align*}
Note that the left term defines a norm on $X$ so we have $||z||^2_X := (||u||^2_V + ||v||^2_V)$. This follows by the properties of the given $||\cdot||_V$ Norm on $V$. Therefore the coercivity constant $\beta$ of $B$ must be
\begin{align*} \beta = \inf_{u,v \neq 0}(\alpha_1 + \frac{A(u,v)}{||u||^2_V + ||v||^2_V}) \end{align*}
If $B$ is coercive we need $\beta>0 \forall u,v \in V$.
Is this proof right so far? Now I dont know how to show that $\beta>0$ since I dont know about $A(u,v)$.
You can estimate $$ |A(u,v)| \le \alpha_2 \|u\|_V\|v\|_V \le \frac{\alpha_2}2(\|u\|_V^2 + \|v\|_V^2). $$ Then you get coercivity of $A$ under the condition $\alpha_1 \ge\frac{\alpha_2}2$.
However, one can do better. Since $A$ is symmetric, it holds $$ B(z,z) = A(u,u) + A(v,v) + \frac12 ( A(u,v) + A(v,u))\\ = \frac34 A(u+v,u+v) + \frac14 A(u-v,u-v). $$ This decomposition is inspired by the eigenvalue decomposition of $\pmatrix{1&\frac12\\\frac12&1}$. This proves $$ B(z,z) \ge \frac{\alpha_1}4 ( 3\|u+v\|_V^2 + \|u-v\|_V^2). $$ Since $\|u\| \le \frac12\| (u+v) + (u-v)\|$ and $\|v\|_V\le \frac12\|(u+v)-(u-v)\|_V$, this proves that $B$ is coercive.