$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\supp}{\operatorname{supp}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\mes}{\operatorname{mes}}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bC}{\mathbb{C}}$
In Chapter 4 we considered Fourier Series which appeared as decomposition by eigenfunctions of operator $-\frac{d^\ }{dx^2}$ in $L^2(-l,l)$. Then we considered more general decompositions. The only restriction was that operator must be self-adjoint and the whole spectrum pure point (better discrete, that means of finite multiplicity, without points of accummulation).
However what to do when spectrum is continuous (or when there are both pure point and continuous spectrum). Fourier transform and Fourier integral (see Chapter 5 give an answer: Fourier integral is a decomposition by $e^{ikx}$ with $k\in \bR$ ($e^{i\boldsymbol{k}\cdot\boldsymbol{x}}$ in multidimensional case). Thus $u_k(x):=e^{ikx}$ with $k\in \bR$ can be considered as generalized eigenfunctions of $-\frac{d^\ }{dx^2}$ in $L^2(\bR)$: they satisfy equation \begin{gather} -\frac{d^\ }{dx^2} u= k^2 u \label{eq-13.A.1} \end{gather} (so $k^2$ are generalized eigenvalues but they do not belong $L^2(\bR)$. Natural question: why we cannot take $k\in \bC\setminus \bR$ here? They also satisfy (\ref{eq-13.A.1}) and also do not belong $L^2(\bR)$.
Intuitive answer: $u_k(x)$ with $k\in \bR$ almost belong to $L^2(\bR)$ which is not the case for $k\in \bC\setminus \bR$. More precisely, as $k\in \bR$ we have $u_k \in \mathcal{S}'(\bR)$, that is $u_k$ is a temperate distribution that is they have no more than polynomial growth (see Chapter 11), which is not the case for $k\notin \bR$.
General and abstract definition rests on the notion of rigged Hilbert space (aka nested Hilbert space, equipped Hilbert space, Gelfand triple) see Wikipedia, which belongs to advanced Real Analysis/