Let $\tau=[-1,1]^2$, consider the finite element $\left(\tau, Q_{1}, \Sigma\right)$, $Q=span\{1,x,y,x^2−y^2 \},$ $\Sigma=\{w(−1,0),w(1,0),w(0,−1),w(0,1)\}$.
Show that the unisolvent element leads to a finite element space, which is not $H^1-$conforming.
I have proved that this is a unisolvent finite element. What should I need to do to prove it leads to a finite element space which is not $H^1-$ conforming.
I know $H^1$-conforming means something like $V_h\subset H^1$.
Denote $V_h$ the finite element space. Recall that if $V_h\subset H^1(\Omega)$ and $Q$ consists of continuous functions, then $V_h\subset C(\Omega)$.
Thus, now we only need to show that $V_h$ is not contained in $C(\Omega)$. It's not difficult to find a function in $V_h$ but not continuous.
For example, let us consider $\Omega=\Omega_1\cup \Omega_2$ where $\Omega_1=[-1,0]\times [-1,1]$ and $\Omega_2=[0,1]\times [-1,1]$. Let $v_h\in V_h$ be defined as $v_h|_{\Omega_1}=x$ and $v_h|_{\Omega_2}=x^2-y^2$. Then $v_h|_{\Omega_1}(0,0)=0=v_h|_{\Omega_2}(0,0)$ whcih imples that $v_h\in V_h$. Note that $v_h|_{\Omega_1}(0,-1)=0\neq -1=v_h|_{\Omega_2}(0,-1)$. Thus $v_h\notin C(\Omega)$.