Check if the orthogonal system is complete

Question

According to my lecture notes:

The orthogonal system $\sin{kx}, k=1,2, \dots $ in $[0, \pi]$ is a complete system in $L^2[0, \pi]$.

$1, \cos{kx}, \sin{kx}$ in $[0, 2 \pi] $is a complete system in $L^2[0,2 \pi]$.

$\sin{2x}, \sin{3x}, \dots, \sin{nx}, \dots n \geq 2$ in $[0,\pi]$ is an orthogonal system, BUT it isn't complete, because:

If we pick $X(x)= \sin x$, the Fourier coefficients of it in respect to the orthogonal system $\sin{2x}, \sin{3x}, \dots, \sin{nx}, \dots n \geq 3$ are $A_k=0, k=2,3, \dots$

$$A_k=\frac{\int_0^{\pi} \sin{nx} \sin{kx} dx}{\int_0^{\pi} \sin^2{kx}dx}=0$$

$\sin x$ should converge to $0$ in $L^2$ norm.

$$$$

So does this mean that we have to pick a specific $X$, then calculate $A_k$ and check if $X$ converges to $A_k$ in $L^2$ norm?

I tried to show that the orthogonal system $\sin{kx}, k=1,2, \dots $ in $[0, \pi]$ is a complete system in $L^2[0, \pi]$, as follows:

Let $X(x)= \sin x$.

$$A_k= \frac{\int_0^{\pi} \sin{kx} \sin x dx}{\int_0^{\pi} \sin^2{kx}dx}=\frac{2}{\pi} \int_0^{\pi} \sin{kx} \sin{x} dx=\frac{1}{\pi} \int_0^{\pi} \left[ \cos((k-1)x)-cos((k+1)x)\right]dx$$

$$A_k=\left\{\begin{matrix} 0 &, k \neq 1 \\ 1 & , k=1 \end{matrix}\right.$$

Is it right so far? Do we have to check now if $\sin x \sim A_k$ in $L^2[0, \pi]$ ? Or do we have to do something else?

EDIT: I found also the following in my notes.

$$f \sim \sum_{n=1}^{\infty} a_n x_n$$

$$S_N(x)= \sum_{n=1}^{\infty} a_n X_n(x)$$

Definition of complete:

$$\lim_{N \to +\infty} S_N(x) \overset{L^2[a,b]}{=}f(x)$$

Could we maybe apply the above definition, by picking a specific $X$ ?

EDIT:If we consider the case where $f(0)=0, f(\pi)=0$ on $C^{\infty}[0,\pi]$, how do we deduce that the eigenvalues are $1^2, 2^2, 3^2, \dots$ ? Also, in this case isn't the system $1, \cos{kx}, \sin{kx}$ complete?

Answer 1

The classical sets of orthogonal eigenfunctions come from solving the eigenvalue problem $$ -f'' = \lambda f $$ on $[a,b]$, subject to separated conditions $$ \cos\alpha f(a) + \sin\alpha f'(a) = 0,\\ \cos\beta f(b) + \sin\beta f'(b) = 0, $$ or periodic conditions $$ f(a)= f(b),\\ f'(a)=f'(b). $$ Examples:

1. Consider periodic conditions on $[-\pi,\pi]$, then the eigenvalues $\lambda$ are $0,1^{2},2^{2},3^{3},\cdots$, with eigenfunctions $$ 1,\cos x,\sin x,\cos 2x,\sin 2x,\cdots. $$

2. Consider $0$ endpoint conditions $f(0)=0$, $f(\pi)=0$, on $[0,\pi]$. The eigenvalues $\lambda$ are $1^{2},2^{2},3^{2},\cdots$, with eigenfunctions $$ \sin x,\sin 2x,\sin 3x,\cdots . $$

3. Consider $0$ derivative conditions $f'(0)=0$, $f'(\pi)=0$, on $[0,\pi]$. The eigenvalues $\lambda$ are $0,1^{2},2^{2},3^{2},\cdots$, with eigenfunctions $$ 1,\cos x,\cos 2x,\cos 3x,\cdots . $$

4. Consider mixed conditions $f(0)=0$, $f'(\pi)=0$, on $[0,\pi]$. The eigenvalues $\lambda$ must solve $\cos(\sqrt{\lambda}\pi)=0$, or $\lambda = (n+1/2)^{2}$ for $n=0,1,2,3,\cdots$, with eigenfunctions $$ \sin(x/2),\sin(3x/2),\sin(5x/2),\cdots . $$

For all possible conditions, the resulting eigenfunctions form a complete orthogonal basis of $L^{2}[a,b]$. If you leave out of one the eigenfunctions from the set, then the resulting set cannot be complete because the one you leave out is orthogonal to all the rest and, hence, a series in the remaining functions is orthogonal to the one that was omitted.