Coercivity of bilinear form I want to show that there is a unique solution for 
$$-u''=f$$
with boundary condition
$$-u'(0)+u(0)=u'(1)=0$$
so I define bilinear form
$$a(v,w) = \int\limits_0^1 {v'} w'dx + v(0)w(0)$$
so I should show that this bilinear form is coercive.
then I should show that
$$a(v,v)\ge\alpha \|v\|^2_1$$ 
 A: An estimate you need is established using the standard trick underlying the Sobolev spaces techniques. The trick reduces to using some suitable integral representation
valid for all elements of the Sobolev space you are employing. In your trivial case,
it is a representation 
$$
v(x)=v(0)+\int\limits_0^x v'(y)\,dy \quad\forall\,x\in (0,1),
$$
valid for all elements $v\in H^1(0,1)$. Remember that $H^1$ is embedded into $C[0,1]$, 
more precisely, every element of $H^1(0,1)$ is an equivalence class that does contain a function which is absolutely continuous on $[0,1]$. The integral representation immediately implies an estimate
$$
\int\limits_0^1 v^2(x)\,dx \leqslant 2v^2(0)+2\int\limits_0^1\Biggl(\int\limits_0^1 |v'(y)|\,dy\Biggr)^2dx =2v^2(0)+2\Biggl(\int\limits_0^1 |v'(y)|\,dy\Biggr)^2\leqslant\\
\leqslant 2v^2(0)+2\!\int\limits_0^1 |v'(y)|^2\,dy =2a(v,v)\quad \forall\,v\in H^1(0,1),
$$
where the Cauchy-Bunyakovsky inequality has been employed. Thus you get an estimate you need
$$
\|v\|_1^2=\int\limits_0^1 |v'(x)|^2\,dx+ \int\limits_0^1 v^2(x)\,dx \leqslant 3a(v,v)
$$
with constant $\alpha=1/3$. Of course, this value of $\alpha$ is not sharp. But finding the sharp value of $\alpha$ is another interesting problem.
