Coercive bilinear form from Poisson equation with Robin boundary conditions

I need to find if the bilinear form arising from the following problem is coercive or not : $$-v\Delta u = f \quad \text{in } \Omega$$ $$v\frac{\partial u}{\partial n} +hu = 0 \quad \text{on } \partial\Omega$$

where $v >0$ and $h>0$ are constants, which is Poisson's equation with a Robin boundary condition. This is for a mathematics FEM class.

What I've done so far is find the variational form of the problem : $$v\int_{\Omega} \nabla u\cdot \nabla w + h\int_{\partial \Omega}u w= \int_{\Omega} fw$$ for all $w \in H^1(\Omega)$. So my bilinear form is $$a(w,w)=v\int_{\Omega} \nabla w\cdot \nabla w + h\int_{\partial \Omega}w^2$$ and I want to show that $\exists \alpha >0$ such that $$a(w,w)\ge\alpha \|w\|^2_{H^1(\Omega)}$$ I've searched a lot on this site and others and the answer seems to be it is indeed coercive : here (Last Question), here and here.

However, these answers use concepts that we have not covered in class (sequences of functions in $H^1$, compactness argument, Friedrich inequality namely). So I was wondering if the coercivity could be proven in a simpler way.

What's more, I asked my teacher and he told me that with what we've covered in class we are not able to show it is coercive, so that I should instead look for a counter-example, which seems contradictory to me because I can't find a counter-example if the bilinear form really is coercive...

• Looks coercive to me, assuming $n$ is the exterior normal. – user147263 Nov 2 '15 at 6:30
• Yes, $n$ is the exterior normal vector. – philb Nov 3 '15 at 13:46