I have a question about the use of Poincaré inequality. If we consider an open interval $I=(l_0,l_1) \subset \mathbb{R}$ and define the Sobolev space
$$ H_{l_0}^2:=\{u\in H^2(I):u(l_0)=u'(l_0)=0\}.$$
How do I prove using Poincaré inequality that $\|u\|_{L^2(I)}\leq C\|u'\|_{L^2(I)}$ for all $u \in H^2_{l_0}$.
My attemps:
Since $u\in H^2(I)$ I have that $u\in H^1(I)$ and I know that $\|u\|_{L_2(I)}\leq \|u\|_{H^1(I)}$.
Now, using the Generalized Poincaré Inequality I obtain:
$$ \|u\|_{L^2(I)}^2\leq \|u\|_{H^1(I)}^2\leq C\left(\|u'\|_{L^2(I)}^2+\int_{\{l_0,l_1\}}(\gamma u)^2\,d\sigma\right).$$
I think that the term $\int_{\{l_0,l_1\}}(\gamma u)^2\,d\sigma$ dissapears but I don't know how to get to that part.
Can anyone help me with this problem?
It's simpler than that: Recall that in one dimension Sobolev functions of order 1 (you even have order 2) are absolutely continuous and we have that the fundamental theorem of calculus holds. Now for $u\in H^2_{l_0}$ by definition we have $$ u(t)= u(l_0)+ \int_{l_0}^t u'(s)\, ds=\int_{l_0}^{t} u'(s)\, ds, \qquad l_0\leq t\leq l_1, $$ so that taking absolute values, and using Cauchy-Schwarz, $$ |u(t)|\leq (t-l_0)^{1/2}\| u'\|_2. $$ Now square and integrate over $[l_0,l_1]$ to get $$ \| u\|_2^2 \leq \| u'\|_2^2 \int_{l_0}^{l_1} (t-t_0)\, dt = \dfrac{(l_1-l_0)^2}{2}\| u'\|_2^2. $$ This is what you want with $C=(l_1-l_0)/\sqrt{2}$.