Show that $|a(u,v)|\leq M\|u\|\|v\|$ and $|a(v,v)|\geq \nu\|v\|^2$ on $H_0^1(\Omega )$.

Question

Let $H_0^1(\Omega )$ where $\Omega \subset \mathbb R^d$ with the scalar product $$\left=\left_{L^2}+\sum_{i=1}^n\left<\partial _i f,\partial _i g\right>_{L^2}.$$ Let $$a(u,v)=\int_\Omega \nabla u\cdot \nabla v+\int uv$$ and $$L(v)=\int_\Omega fv$$ for $f\in L^2(\Omega )$. for $u,v\in H_0^1$. I want to show that $$|a(u,v)|\leq M\|u\|_{H_0^1}\|v\|_{H_0^1},\quad |a(v,v)|\geq \nu\|v\|^2\quad \text{and}\quad |L(v)|\leq N\|v\|_{H_0^1}$$ for certain $M,N$ and $\nu$ positive, with $u,v\in H_0^1(\Omega )$ ?

My attempts

For $L(v)$, we have that $$|L(v)|\leq \int_{\Omega }|f||v|\leq \|f\|_{L^2}\|v\|_{L^2}\leq \|f\|_{L^2}\|v\|_{H_0^1}=C\|v\|_{H_0^1}.$$

1) Is it correct ?

For $a(u,v)$, $$|a(u,v)|\leq \int |\partial u\cdot \nabla v|+\int |uv|\leq \int \|\nabla u\|\|\nabla v\|+\|u\|_{L^2}\|v\|_{L^2}\leq \|\nabla u\|_{L^2}\|\nabla v\|_{L^2}+\|u\|_{L^2}\|v\|_{L^2}$$

2) How can I conclude ?

For $a(v,v)$ I get $$|a(v,v)|=\|\nabla v\|_{L^2}^2+\|v\|_{L^2}^2=\|v\|_{H_0^1}^2$$

3) Is it correct ?

Answer 1

1) Yes.

2) Note that on the right-hand side of your estimate you have $\lVert \nabla u \rVert_{L^2}^2\leq \lVert u \rVert_{H_0^1}^2$ and $\lVert u \rVert_{L^2}^2\leq \lVert u \rVert_{H_0^1}^2$, see the definition of the $H_0^1$ norm that you wrote yourself below.

3) Yes.