Show that $\mathcal{A}$ defined by (3) is symmetric, (self-adjoint), on $C_0^2([0,L])$ i.e. show
$\langle\mathcal{A}f,g\rangle=\langle f,\mathcal{A}g\rangle,$ for all $f,g \in C_0^2([0,L])$
Hint: Write out the definition of $\mathcal{A}$ and $\langle , \rangle$ and integrate by parts twice.
The only definition of $\mathcal{A}$ that I know is $\mathcal{A}f(x)=-f''(x)$
The inner product is defined as $\int_0^Lf(x)g(x)dx$
So as far as I understand is to show that
So $\langle \mathcal{A}f(x),g(x) \rangle = \int_0^L \mathcal{A}f(x)g(x)dx = -\int_0^Lf''(x)g(x)dx$.
Use IBP let $u=g(x), du=g'(x)dx, v=f'(x), $ and $dv=f''(x)dx$.
So $ -\int_0^Lf''(x)g(x)dx = -\left( g(x)f'(x)- \int_0^Lf'(x)g'(x)dx \right)$.
Use IBP again let $u=g'(x), du=g''(x)dx, v=f(x), $ and $dv=f'(x)dx$.
So $-\left( g(x)f'(x)- \int_0^Lf'(x)g'(x)dx \right)=-\left( g(x)f'(x)- \left( g'(x)f(x) - \int_0^Lf(x)g''(x)dx \right)\right)$.
It can be seen that $-\left( g(x)f'(x)- \left( g'(x)f(x) - \int_0^Lf(x)g''(x)dx \right)\right) = $
$-g(x)f'(x)+ g'(x)f(x) + \int_0^Lf(x)(-g''(x))dx =$
$-g(x)f'(x)+ g'(x)f(x) + \int_0^Lf(x)(\mathcal{A}g(x))dx \overset{?}{=}\int_0^Lf(x)(\mathcal{A}g(x))dx$