How to prove this subspace of $W^{2,2}$ is Hilbert? Let $I=(l_0,l_1)\subset \mathbb{R}$
$$H^2_{l_0}=\{u\in H^2(I): u(l_0)=u'(l_0)=0\},$$
endowed with the inner product $$\langle u,v\rangle_{H^2_{l_0}}:=\left\langle u'',v''\right\rangle_{L^2(I)}=\int_I u''v''\,dx,$$ for real valued functions.
Here $H^2(I):=W^{2,2}(I)$
I have proven that $\left<\cdot,\cdot\right>_{H^2_{l_0}}$ defines an inner product, it also defines a norm. So far I have seen $H^2_{l_0}$ is a pre-Hilbert space, but I'm having troubles proving the completeness with the metric induced by the inner product.
I really appreciate your help.
 A: Take a Cauchy Sequence $(u_m)_m $ in $H^2_{l_0}(I)$. By poincaré inequality twice, there is $C_1,C_0>0$ such that
$$C_0||u_m-u_m\|_{L^2} \leq C_1||u'_m-u'_m\|_{L^2} \leq  ||u''_m-u''_m\|_{L^2} \quad \forall n,m \in \mathbb{N} $$
Because $L^2(I)$ is complete there is a $w_0,w_1,w_2 \in L^2(I)$ such that
\begin{align}
\|u_m''-w_2\|_{L^2} & \to 0 \\
\|u_m'-w_1\|_{L^2} & \to 0 \\ 
\|u_m-w_0\|_{L^2} & \to 0 \\
\end{align}
We are going to prove that $w''_0=w_2$.
\begin{align}
\int_0^{l_0} w_0 v'' &= \int_0^{l_0} (w_0-u_m) v''+\int_0^{l_0} u_mv'' \\
 & = \underbrace{\int_0^{l_0} (w_0-u_m)v''}_{\to 0 ~ \text{when} ~ m \to \infty  } + \underbrace{ u_m(x)v'(x)-u'_m(x)v(x) |^{l_0}_{0}}_{0} +\int_0^{l_0} u''_mv \quad (\text{Integration by parts).} \\
 & = \lim_m\int^{l_0}_0 u''_m v = \int^{l_0}_0 w_2 v \quad \forall  
v \in C^{\infty}_0(I)\end{align}
Hence, $w_0''=w_w.$
Observe that the Poincare inequality gives
$w_0'=w_1$. Additionally, $\|u_m-w_0\|_{H^2}\to 0$ when $m \to \infty$. Therefore  $w_0 \in H^2(I)$. Now we must check the boundary conditions.
\begin{align} 
  (w(x)v'(x)-w'(x)v(x)) |^{l_0}_{0} &= \int_{0}^{l_0} w''(x)v(x)-w(x)v''(x) \\
&= \lim_m \int_{0}^{l_0} u_m''(x)v(x)-u_m(x)v''(x) \\
&= \lim _m ~ (u_m(x)v'(x)-u_m'(x)v(x) )|^{l_0}_{0} \\
&= \lim_m (u_m(0) v'(0)-u_m'(0)v(0)) \quad \forall v \in H^2(I) \tag{*}
\end{align}
Because $H^2(I)$ is embedded into $C^{1}(\bar I)$. You can select $v$ with $v(0)=v(0)=v'(0)=0$ and $v(l_0) \neq 0 $. By (*) results $w(0)v(l_0)=0$, consequently $w(l_0)=0$, analogously $w'(l_0)=0$. Then, $w_0 \in H^2_{l_0}(I)$. In other words, $H^2_{l_0}(I)$ is closed.
