In chapter 9 (Green's Functions for Time-Independent Problems) of Haberman's text 'Applied Partial Differential Equations', the author presents the following fundamental result
$$ L[G(x,x_s)] = \delta(x - x_s). \tag{1} $$
So, in words, if the boundary-value problem \begin{align} Lu &= f(x) \\ u(0) = 0 \quad &\mathrm{and} \quad u(L) = 0 \end{align} admits a solution in terms of a Green's function, the linear differential operator $L$ applied to $G$ yields the Dirac Delta. Now, consider the following specific example -
\begin{align} \frac{d^2u}{dx^2} &= f(x) \\ u(0) = 0 \quad &\mathrm{and} \quad u(L) = 0. \end{align}
The associated Green's function is then given by
$$ G(x,x_0) = -\frac{2}{L} \sum_{n = 1}^{\infty} \frac{ \sin (\,n\pi x/L)\sin(n\pi x_0/L) }{(n\pi/L)^2}. \tag{2}$$
However, upon applying the differential operator $d^2/dx^2$ to the Green's function, it is easy to see that if $L = 1$ and $x_0 = 1/2$, the resultant will have peaks at regularly spaced values, not just $x_0 = 1/2$. Therefore, strictly speaking,
$$ LG(x,x_0) \neq \delta(x - x_0).$$
Am I missing something?