Considering the following bilinear form : $$a(\phi,\psi)=\intop_{\varOmega}\left(\sum_{g=1}^{G}D_{g}\nabla\phi_{g}\cdot\nabla\psi_{g}+\sum_{g,h=1}^{G}\varSigma_{gh}\phi_{g}\psi_{h}\right)\,\mathord{d}\mathbf{r}$$
and assuming the hp:
$\mathord{D_{g}}$ for $\mathord{g=1,\ldots,G}$ are real, scalar, bounded measurable functions on $\varOmega$, with bound $\mathord{M}$. Moreover, each $\mathord{D_{g}}$ is positive with $\mathord{D_{g}\geq\delta}$ for some constant $\mathord{\delta>0.}$
$\mathord{\varSigma_{gh}}$ for $\mathord{g,h=1,\ldots,G}$ are real, scalar, bounded measurable functions on $\varOmega$, with bound $\mathord{M}$.
I want to prove that for $\mathord{\lambda=MG^{2}+\delta}$ we have:
$$a(\phi,\phi)\geq\delta\left\Vert \phi\right\Vert _{H^{1}(\varOmega)}^{2}-\lambda\left\Vert \phi\right\Vert _{L^{2}(\varOmega)}^{2}\quad\forall\phi\in H_{0}^{1}(\varOmega)$$
with the usual norm for $H_{0}^{1}$ and $L^{2}(\varOmega)$. So basically i want to prove that this bilinear form is weak coercive.
Since $D_{g}\geq\delta$ we have that :
$$ \intop_{\varOmega}\sum_{g=1}^{G}D_{g}\nabla\phi_{g}\cdot\nabla\phi_{g}\,\mathord{d}\mathbf{r}=\intop_{\varOmega}\sum_{g=1}^{G}D_{g}\bigl|\nabla\phi_{g}\bigr|^{2}\,\mathord{d}\mathbf{r}\geq\intop_{\varOmega}\delta\sum_{g=1}^{G}\bigl|\nabla\phi_{g}\bigr|^{2}\,\mathord{d}\mathbf{r}$$
but what about the second part? I think that i have to use the bound $M$ but i don't understand how i can pass from 2 index $g,h$ to the norm in $L^{2}(\varOmega)$. I should arrive at an inequality of the type :
$$\begin{array}{ccl} a(\phi,\phi)+\lambda\left\Vert \phi\right\Vert _{L^{2}(\varOmega)}^{2} & \geq & {\displaystyle \intop_{\varOmega}\delta\sum_{g=1}^{G}\bigl|\nabla\phi_{g}\bigr|^{2}\,\mathord{d}\mathbf{r}+\intop_{\varOmega}\bigl(\lambda-M\,G^{2}\bigr)\sum_{g=1}^{G}\bigl|\phi_{g}\bigr|^{2}\,\mathord{d}\mathbf{r}}\\ & \geq & {\displaystyle \min(\delta,\lambda-M\,G^{2})\left\Vert \phi\right\Vert _{H^{1}(\varOmega)}^{2}} \end{array} $$
Any help would be very appreciated.
It's a common trick to bound off-diagonal terms of this kind using the inequality $|ab|\leq \frac 1 2(a^2+b^2)$, which is a special case of the AM-GM inequality or Young's inequality and follows directly from $(|a|-|b|)^2\geq 0$.
First, by the triangle inequality and the bound on $\Sigma_{g,h}$, we have $$ \left|\sum_{g,h=1}^G \Sigma_{gh}\phi_g\phi_h\right|\leq \sum_{g,h=1}^G |\Sigma_{gh}|\cdot|\phi_g\phi_h|\leq M\sum_{g,h=1}^G|\phi_g\phi_h|. $$ Now we can use the elementary inequality $|ab|\leq \frac 1 2 a^2+\frac 1 2 b^2$to get $$ \sum_{g,h=1}^G |\phi_g\phi_h|\leq \frac 1 2\sum_{g,h=1}^G (\phi_g^2+\phi_h^2)=G\sum_{g=1}^G \phi_g^2=G\|\phi\|^2. $$ Therefore $$ \sum_{g,h=1}^G \Sigma_{gh}\phi_g\phi_h\geq -MG\|\phi\|^2=-(MG+\delta)\|\phi\|^2+\delta\|\phi\|^2. $$ I don't know how your source ended up with $G^2$ instead of $G$, but it doesn't hurt to prove something stronger.