Lax Milgram Lemma. Prove coercivity Consider the following problem:
\begin{cases}
(\mu u'-au)'=f \\
u(0)=u(1)=0
\end{cases}
The domain being the interval $\Gamma$=(0,1)
Where $\mu$ is a positive function in $C^1$($\Gamma$) and a is a $C^1$($\Gamma$)function. f belongs to $L^2$($\Gamma$).
If we consider a test function v in V= $H^1_0$ we can write the weak form of the problem:
$\int_\Gamma \mu u'v'd\Gamma -\int_\Gamma auv'd\Gamma$=$\int_\Gamma fv d\Gamma$
For the proof of continuity:
1
I got stuck trying to prove the coercivity of the bilinear term:
2
We need to set a condition on a to prove the coercivity. Can we say a should be positive so we have coercivity? and we consider the norm of $\mu$ in $C^1$
to be the constant of coercivity? thank you.
 A: Let $$B(u,v):= \int_\Gamma( -\mu u'v' + auv') d\Gamma \, .$$
Here is your proof of continuity again, slightly sharpened: 
$$ |B(u,v)| \le \|\mu\|_{L^\infty} \|u'\|_{L^2} \|v'\|_{L^2} + \|a\|_{L^\infty} \|u\|_{L^2} \|v'\|_{L^2}\\
\le (\|\mu\|_{L^\infty} + \|a\|_{L^\infty}) \|u\|_V \|v\|_V
$$
Coercivity means $B(u,u) \ge \epsilon \|u\|_V^2$ for all $u \in V$, for some $\epsilon > 0$. 
You have an error in the second line of your coercivity proof. 
Using $2uu' = (u^2)'$, one obtains by integration by parts
$$B(u,u) = \int_\Gamma \mu (u')^2 + \int_\Gamma a \frac{1}{2} (u^2)'
=  \int_\Gamma \mu (u')^2 - \int_\Gamma a' \frac{1}{2} u^2 
$$
Now one has to assume that $\mu(x) \ge \delta > 0$ for all $x$ for some $\delta$.
First consider the case where $\boxed{a' \le 0}$ on $\Gamma$. Then this may be estimated from below by $\int_\Gamma\mu (u')^2 \ge \delta \|u'\|_{L^2}^2$.  Then this expression is coercive on $V$ since by Poincare's inequality
$$
\pi^2\|u\|^2_{L^2} \le \|u'\|^2_{L^2}.
$$ 
So no additional condition on $\mu$ is required. 
Next consider the case where $\max a' = a_1 > 0$. Then we may estimate
$$B(u,u) =  \int_\Gamma \mu (u')^2 - \int_\Gamma a' \frac{1}{2} u^2
\ge \delta \|u'\|_{L^2}^2 - \frac{a_1}{2} \|u\|_{L^2}^2 
$$
and this is coercive if e.g. $\delta \pi^2 - \frac{a_1}{2} > 0$. 
