Integral transform as continuous eigenfunction expansion So I'm taking a course in partial differential equations where we have access to video lectures online. I recently watched a video about using integral transforms to solve PDEs on infinite domains (which I have seen before, just not in this manner). In the video he solves the equation
$$ \frac{\partial u(x,t)}{\partial t} - a \frac{\partial^2 u(x,t)}{\partial^2 x} = \kappa (x,t), \text{    where } x \in R, \text{ and } t > 0$$
using separation of variables $u(x,t) = X(x)T(t)$, which leads to the problem $ X''(x) + k^2X(x)=0 $. This has solutions $f_k(x)=Ae^{ikx}, \text{with }k \in R$. He says that we now want to expand $u$ in terms of its eigenfunctions, but unlike in the case of a finite domain, where we have a discrete number of eigenfunctions, we now have a continuous number, so we need to change the sum to an integral, so he writes
$$ u(x,y) = \int_R{\hat{u}(k,t)}f_k(x)dk =  \int_R{\hat{u}(k,t)}Ae^{ikx}dk, $$
where we get $\hat{u}(k,t)$ by projecting $u(x,t)$ onto $f_k(x)$: $\hat{u}(k,t) = <f_k|u> \;=\int_R{u(x,t)f_k(x)}dx $. 
How is this mathematically rigorous? The ODE problem we had wasn't even as SL problem, so how do we know that $f_k(x)$ forms an orthogonal basis in $L^2(R)$? 
He also continues to say that we want to normalize these eigenfunctions, and the way he does this also seems questionable to me: 
$$ <f_k|f_{k'}> \; = \int_R{|A|^2e^{(k-k')ix}}dx = 2\pi |A|^2 \delta(k-k'). $$
He wants to normalize them in the meaning that he wants $<f_k|f_{k'}>$ to be $\delta(k-k')$, which gives $A = \frac{1}{\sqrt{2\pi}}$ (among other solutions).
All of this seems to lead up to what I've seen and been taught before: the way of solving this kind of equation with Fourier transforms, but I want to know how this can be made mathematically rigorous. 
Thanks!
 A: The problem $Lf=-f''$ is a well-posed Sturm-Liouville equation on $\mathbb{R}$ because no endpoint conditions at $\pm\infty$ are possible. The operator has singular points at $\pm\infty$. So $L$ is selfadjoint on a natural domain.
Singular equations with two singular endpoints can be tricky, and it usually requires introducing a matrix to connect the proper solutions on the right with those on the left. However, that is not the case here. The "eigenfunctions" $e^{i\lambda x}$ for $-\infty < \lambda < \infty$ are not technically eigenfunctions because they do not belong to $L^2(\mathbb{R})$. However, the general theory provides a way to expand functions in terms of these in an orthogonal manner as described. The building block operator is obtained by considering the operator on a semi-infinite interval such as $[0,\infty)$ and imposing a basis of endpoint conditions such as
$$
     (i)\;\; f(0)=0,\; f'(0)=1, \\
     (ii)\;\; f(0)=1,\; f'(0)=0.
$$
There are resulting expansions on $[0,\infty)$ of the form
$$
         f = \frac{2}{\pi}\int_{0}^{\infty}\left(\int_{0}^{\infty}f(y)\sin(\lambda y)dy\right)\sin(\lambda x)d\lambda \\
         f = \frac{2}{\pi}\int_{0}^{\infty}\left(\int_{0}^{\infty}f(y)\cos(\lambda y)dy\right)\cos(\lambda y)d\lambda
$$
These are then glued together in a vector type form to give you the full Fourier expansions in terms of $e^{i\lambda x}$.
Because of the translation and scale invariances of $L$, there are no weights in the expansions. However, for a Sturm-Liouville problem on $[0,\infty)$ such as
$$
          Lf = -\frac{1}{w}\left[\frac{d}{dx}\left(p\frac{df}{dx}\right)+qf\right],
$$
there are spectral weights that pop up to give expansions for problems with conditions $(i)$, $(ii)$ of the form
$$
     f = \int_{-\infty}^{\infty}\left(\int_{0}^{\infty}f(y)\varphi_{\lambda}(y)w(y)dy\right)\varphi_{\lambda}(x)d\mu(\lambda)
$$
The weight $w$ comes from the Sturm-Liouville problem. The measure $\mu$ is a spectral density measure, and may have discrete and continuous parts. And, it's trickier tying these together on the half-lines, for example. Otherwise, they're essentially the same, including the $\varphi_{\lambda}$ which are classical eigenfunctions satisfying the one set of endpoint conditions. I'm leaving out the possibility of conditions at $\infty$ because these are not relevant to Physics; they can exist in general, but not in Quantum problems, basically by axiom. Then, as expected
$$
         Lf = \int_{-\infty}^{\infty}\left(\int_{0}^{\infty}f(y)\varphi_{\lambda}(y)w(y)dy\right)\lambda\varphi_{\lambda}(x)d\mu(\lambda).
$$
There is a general Fourier transform arising out of this:
$$
          \hat{f}(\lambda)=\int_{0}^{\infty}f(y)\varphi_{\lambda}(y)w(y)dy
$$
And you have a Plancherel Theorem,
$$
       \int_{0}^{\infty}|f|^2w dx = \int_{-\infty}^{\infty}\left|\hat{f}(\lambda)\right|^2d\mu(\lambda).
$$
So the formalism of Quantum is justifiable in the realm of Sturm-Liouville equations. In fact, Dirac undoubtedly used this formalism in his formulation of Quantum. His notation suggests it.
