FEM solution interpolates the exact solution Consider the problem 
$$ 
  -u'' = f \ \ \text{in} \  (0,1), \\
  u(0) = (1) = 0.
$$ 
Assume that the Green's functions of the nodal values $G(x_j, \cdot)$ lie in $V_h = \{ v \in C([0,1]) : v \ \text{is  linear on each interval } [x_{j-1}, x_j] \}$.
I need to show, that the 

FEM solution $u_h$ is identical to the interpolant $I_h u$ of the exact solution.

I can show that the finite difference method give the same result as the finite element method and then show, that FDM solution is identical to the interpolant $I_h$. 
How can I show it using Green's function?
 A: Let $0=x_0<x_1<...<x_n=1$ be a partition of $[0,1]$.
Let $V_h$ be the space of functions $v$ such that $v \in C^0([0,1])$, $v$ is a linear polynomial on $[x_i,x_{i+1}]$ and $v(0)=v(1)=0$.
Let $u_h \in V_h$ be the finite element solution, that is,
$$\int_0^1 u_h'(x)v'(x)\,\mathrm{d}x=\int_0^1 f(x)v(x)\,\mathrm{d} \quad\forall v \in V_h$$
Also, given $v\in V_h$, we get the following from the weak form of the PDE
$$\int_0^1 u'(x)v'(x)\,\mathrm{d}x=\int_0^1 f(x)v(x)\,\mathrm{d}x$$
Subtracting, we get (1)
$$\int_0^1 (u-u_h)'(x)v'(x)\,\mathrm{d}x=0 \quad\forall v \in V_h$$
Take the following green's function
$$
g_x(t)=
\begin{cases}
    \frac{x^2-x}{x},& \text{if } x\geq 1\\
    0,              & \text{otherwise}
\end{cases}
$$
Consider (2)
$$
\int_0^1 (u-u_h)'(t)g_x'(t)\,\mathrm{d}t=\int_0^x(u-u_h)'(t) \times 1\,\mathrm{d}t+\int_x^1 (u-u_h)'(t) \times 0\,\mathrm{d}t = (u-u_h)(x)
$$
But, $g_{x_i} \in V_h$, so by (2) and (1) rspectively,
$$
(u-u_h)(x_i)=\int_0^1 (u-u_h)'(t)g_{x_i}'(t)\,\mathrm{d}t=0
$$
Thus, as $u(x_i)=u_h(x_i) \forall i$ and $u_h$ is linear on the intervals $[x_i,x_{i+1}]$ we get $u_h=I_h u$.
