Show that a map is a continuous bilinear form on $H^1(0,1)$ space

Question

Let $u,v \in H^1(0,1) = \{f : (0,1) \longrightarrow \mathbb{R}, f,f' \in L^2(0,1) \}$, show that $$a(u,v) = \int_0^1 (u'v' + uv)\; dx$$ is a continuous bilinear form.

Answer 1

The form $a$ is bilinear because the integral is linear. The continuity is a little more complicated: $$a(u,v)=\int_{0}^1 u'v'+\int_{0}^1 uv\leq \left(\int_{0}^1(u')^2 \right)^{1/2}\left(\int_{0}^1(v')^2 \right)^{1/2}+\left(\int_{0}^1u^2 \right)^{1/2}\left(\int_{0}^1v^2 \right)^{1/2}$$ $$\leq \left(\int_{0}^1u^2 +\int_{0}^1(u')^2\right)^{1/2}\left(\int_{0}^1v^2 +\int_{0}^1(v')^2\right)^{1/2}=||u||_{H'}||v||_{H'}$$ This prove the continuity of $a$. The complicated inequality that appears can it be justified by the following simple form $$(AB+CD)^2\leq (A^2+C^2)(B^2+D^2)$$

Answer 2

Bilinearity: $r \in \mathbb{R}$ $a(u+rw,v)=\int_{0}^{1} (u+rw)'v'+(u+rw)vdx= \int_{0}^{1} u'v'+uvdx+r\int_{0}^{1} w'v'+wvdx= a(u,v)+ra(w,v)$ analogously $a(u,v+rw)=a(u,v)+ra(u,w)$ Continuity: Consider $H^1(0,1)$ with the following norm: $$||u||_{H^1}=||u||_{L^2}+||u'||_{L^2}$$ then $$|a(u,v)|=|\int_{0}^{1} u'v'+uvdx| \leq \int_{0}^{1} |u'v'+uv|dx \leq \int_{0}^{1} |uv|dx +\int_{0}^{1} |u'v'|dx$$ By Hölder inequality $$|a(u,v)|\leq (\int_{0}^{1} |u|^2dx)^{\frac{1}{2}} (\int_{0}^{1} |v|^2dx)^{\frac{1}{2}}+(\int_{0}^{1} |u'|^2dx)^{\frac{1}{2}}(\int_{0}^{1} |v'|^2dx)^{\frac{1}{2}}$$ Since $(\int_{0}^{1} |u|^2dx)^{\frac{1}{2}}\leq ||u||_{H^1}$ and $(\int_{0}^{1} |u'|^2dx)^{\frac{1}{2}}\leq ||u||_{H^1}$, analogously for $v$.Then we have $$|a(u,v)|\leq 2||u||_{H^1}||v||_{H^1}$$ Therefore, a(.,.) is continuous.