$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$
All systems we considered in the previous Section were orthogonal i.e. \begin{equation} (X_n, X_m)=0\qquad \forall m\ne n \label{eq-4.3.1} \end{equation} with \begin{equation} (X,Y):=\int_0^l X(x)\bar{Y} (x)\,dx,\qquad \|X\|^2:=(X,X). \label{eq-4.3.2} \end{equation} where $\bar{Y}$ means complex-conjugate to $Y$.
Exercise 1. Prove it by direct calculation.
However, instead we show that this nice property (and the fact that eigenvalues are real) is due to the fact that those problems are self-adjoint.
Consider $X,Y$ satisfying Robin boundary conditions \begin{align} &X'(0)-\alpha X(0)=0,\label{eq-4.3.3}\\ &X'(l)+\beta X(l)=0\label{eq-4.3.4} \end{align} with $\alpha,\beta\in \mathbb{R}$ (and $Y$ satisfies the same conditions). Note that \begin{align} (X'',Y)=&\int X''(x)\bar{Y}(x)\,dx \notag\\ =&-\int X'(x)\bar{Y}'(x)\,dx + X' (l)\bar{Y}(l)- X' (0)\bar{Y}(0)\notag \\ =&-(X',Y') -\beta X (l)\bar{Y}(l)-\alpha X(0)\bar{Y}(0).\qquad \label{eq-4.3.5} \end{align} Therefore if we plug $Y=X\ne 0$ an eigenfunction, $X''+\lambda X=0$, satisfying conditions (\ref{eq-4.3.3}) and (\ref{eq-4.3.4}), we get $-\lambda \|X\|^2$ in the left-hand expression (with obviously real $\|X\|^2\ne 0$) and also we get the real right expression (since $\alpha,\beta\in \mathbb{R}$); so $\lambda$ must be real: all eigenvalues are real.
Further, for $(X,Y'')$ we obtain the same equality albeit with $\alpha,\beta$ replaced by their complex conjugate $\bar{\alpha},\bar{\beta}$ \begin{gather*} (X,Y'')=\int X(x)\bar{Y}''(x)\,dx = -(X',Y') -\bar{\beta} X (l)\bar{Y}(l)-\bar{\alpha} X(0)\bar{Y}(0) \end{gather*} and therefore due to assumption $\alpha,\beta\in \mathbb{R}$ \begin{equation} (X'',Y)= (X,Y''). \label{eq-4.3.6} \end{equation} But then if $X,Y$ are eigenfunctions corresponding to different eigenvalues $\lambda$ and $\mu$ we get from (\ref{eq-4.3.6}) that $-\lambda(X,Y)=-\mu (X,Y)$ and $(X,Y)=0$ due to $\lambda\ne \mu$.
Remark 1. For periodic boundary conditions we cannot apply these arguments to prove that $\cos(2\pi nx/l)$ and $\sin(2\pi nx/l)$ are orthogonal since they correspond to the same eigenvalue; we need to prove it directly.
Remark 2.
So,
We need to discuss domain and the difference between symmetric and self-adjoint because operator $X\mapsto -X'$ is \alert{unbounded}.
Consider linear space $\mathsf{H}$, real or complex. From the linear algebra course's standard definition:
For complex linear space replace $\mathbb{R}$ by $\mathbb{C}$.
Assume that on $\mathsf{H}$ inner product is defined:
Definition 1.
For Hilbert space we will need another property (completeness) which we add later.
Definition 2.
Consider a finite orthogonal system $\{u_n\}$. Let $\mathsf{K}$ be its linear hull: the set of linear combinations $\sum_n \alpha_nu_n$. Obviously $\mathsf{K}$ is a linear subspace of $\mathsf{H}$. Let $v\in \mathsf{H}$ and we try to find the best approximation of $v$ by elements of $\mathsf{K}$, i.e. we are looking for $w\in \mathsf{K}$ s.t. $\|v-w\|$ minimal.
Theorem 1.
Proof. [3]: Obviously $(v-w)$ is orthogonal to $u_n$ iff (\ref{eq-4.3.7}) holds. If (\ref{eq-4.3.7}) holds for all $n$ then $(v-w)$ is orthogonal to all $u_n$ and therefore to all their linear combinations.
[4]-[5]: In particular $(v-w)$ is orthogonal to $w$ and then \begin{equation*} \|v\|^2= \|(v-w)+w\|^2=\|v-w\|^2+ 2\Re \underbracket{(v-w,w)}_{=0}+\|w\|^2. \end{equation*}
[1]-[2]: Consider $w'\in \mathsf{K}$. Then $\|v-w'\|^2=\|v-w\|^2+\|w-w'\|^2$ because $(w-w')\in \mathsf{K}$ and therefore it is orthogonal to $(v-w)$.
Now let $\{u_n\}_{n=1,2,\ldots,}$ be an infinite orthogonal system. Consider its finite subsystem with $n=1,2,\ldots, N$, introduce $\mathsf{K}_N$ for it and consider orthogonal projection $w_N$ of $v$ on $\mathsf{K}_N$. Then \begin{equation*} w_N= \sum_{n=1}^N \alpha_n u_n \end{equation*} where $\alpha_n$ are defined by (\ref{eq-4.3.7}). Then according to [4] of Theorem 1 \begin{equation*} \|v\|^2 =\|v-w_N\|^2+\|w_N\|^2\ge \|w_N\|^2=\sum_{n=1}^N |\alpha_n |^2\|u_n\|^2. \end{equation*} Therefore series in the right-hand expression below converges \begin{equation} \|v\|^2 \ge \sum_{n=1}^\infty |\alpha_n |^2\|u_n\|^2. \label{eq-4.3.8} \end{equation} Really, recall that non-negative series can either converge or diverge to $\infty$.
Then $w_N$ is a Cauchy sequence. Indeed, for $M>N$ \begin{equation*} \|w_N-w_M\|^2= \sum_{n=N+1}^M |\alpha_n |^2\|u_n\|^2\le \varepsilon_N \end{equation*} with $\varepsilon_N\to 0$ as $N\to \infty$ because series in (\ref{eq-4.3.8}) converges.
Now we want to conclude that $w_N$ converges and to do this we must assume that every Cauchy sequence converges.
Definition 3.
Remark 3. Every pre-Hilbert space could be completed i.e. extended to a complete space. From now on $\mathsf{H}$ is a Hilbert space.
Then we can introduce $\mathsf{K}$--a closed linear hull of $\{u_n\}_{n=1,2,\ldots}$ i.e. the space of \begin{equation} \sum_{n=1}^\infty \alpha_n u_n \label{eq-4.3.9} \end{equation} with $\alpha_n$ satisfying \begin{equation} \sum_{n=1}^\infty |\alpha_n |^2\|u_n\|^2<\infty. \label{eq-4.3.10} \end{equation} (Linear hull would be a space of finite linear combinations).
Let $v\in \mathsf{H}$. We want to find the best approximation of $v$ by elements of $\mathsf{K}$. But then we get immediately
Theorem 2. If $\mathsf{H}$ is a Hilbert space then Theorem 1 holds for infinite systems as well.
Definition 4. Orthogonal system is complete if equivalent conditions below are satisfied:
Remark 4.
Our next goal is to establish completeness of some orthogonal systems and therefore to give a positive answer (in the corresponding frameworks) to the question in the end of the previous Section 4.1: can we decompose any function into eigenfunctions? Alternatively: Is the general solution a combination of simple solutions?