Proof that bilinear form in $H_0^1$ is coercive

Question

Let $$B(u,v)=\int_I uv + \int_I u'v'$$ where $u,v\in H_0^1(I)$ for a given interval $I=[a,b]\subset\mathbb{R}$.

How can I prove that the bilinear form $B$ is coercive, i.e., that $$B(u,u)\ge C\Vert u\Vert_{H_0^1(I)}^2\ \ ? $$

Here, $\Vert u\Vert_{H_0^1(I)}:=\int_I u^2 +\int_I (u')^2$. I have tried using that this norm is equivalent to $\Vert u\Vert_{L^2}+\Vert u'\Vert_{L^2}$.

Answer 1

Your bilinear form is $B:H_0^1(I)\times H_0^1(I)\to\mathbb{R}$ given by $$B(u,v)=\int_I uv + \int_I u'v'.$$ You want to prove that $B$ is coercive. In other words, you want to prove that there is $C>0$ such that $$|B(u,u)|\geq C\|u\|_{H_0^1(I)}^2.\tag{1}$$ To do this, you have to specify what is the norm in $H_0^1(I)$ (I think that there is a typo in your post). Well, there are three usual possibilities:

$\displaystyle\|u\|_{H_0^1(I)}=\left(\int_Iu^2+\int_I(u')^2\right)^{1/2}$ which can be written as $\displaystyle\|u\|_{H_0^1(I)}=\sqrt{\|u\|_{L^2(I)}^2+\|u'\|_{L^2(I)}^2}$;

$\displaystyle\|u\|_{H_0^1(I)}=\left(\int_Iu^2\right)^{1/2}+\left(\int_I(u')^2\right)^{1/2}$ which can be written as $\displaystyle\|u\|_{H_0^1(I)}=\|u\|_{L^2(I)}+\|u'\|_{L^2(I)}$;

$\displaystyle\|u\|_{H_0^1(I)}=\int_I(u')^2$ which can be written as $\displaystyle\|u\|_{H_0^1(I)}=\|u'\|_{L^2(I)}$.

These three norms are all equivalents and thus you can use any of them to prove the desired inequality $(1)$.

Well, using the first possibility above, we have $$B(u,u)=\int_I u^2 + \int_I (u')^2=\|u\|_{H_0^1(I)}^2$$ which imply $(1)$ with $C=1$ (and thus the the equivalence that you said is not needed).