Let $\Omega=(a,b)$ and $a:H^1(\Omega)\times H^1(\Omega)\to \mathbb{R}$ to be the bilinear form defined by, $$a(\sigma,\tau)=\int_a^b\sigma \tau.$$ Consider also the bilinear form $b:H^1(\Omega)\times H_0^1(\Omega)\to \mathbb{R}$, given by, $$b(\tau,v)=\int_a^b \tau' v'.$$ I would like to show that there exists a constant $C>0$ such that, $$a(\tau,\tau)\geq C\|\tau\|^2_{H^1(\Omega)},$$ for all $\tau\in \ker(b)=\{\tau \in H^1(\Omega):b(\tau,v)=0,\forall v\in H_0^1(\Omega)\}.$
My attempt:
First, I tried to rewrite $\ker(b)$. Notice that, if $$\int_a^b\tau'v'=0, \quad\forall v\in H_0^1(\Omega),$$ then by using integration by parts we obtain that, $$\int_a^b \tau''v=0, \quad\forall v\in H_0^1(\Omega).$$ So, $\tau''\in [H_0^1(\Omega)]^{\perp}=\{0\}$. Thus, $\tau'=\text{constant}$. Any idea to follow this argument?