Sine Series: ONB How to show that $\{\sin{kx}:k\in\mathbb{N}\}$ for $\{f\in\mathcal{L}^2[0,\pi]:f(0)=f(\pi)=0\}$ is an ONB?
(Clearly they are orthogonal to each other but is their span also dense?)
What general route can one follow when one wants to check total?
 A: NOTE: There is no such thing as a boundary condition $f(0)=0$ or $f(\pi)=0$ on $L^{2}[0,\pi]$. Functions in $L^{2}[0,\pi]$ are equivalence classes of functions which are equal to each other a.e.. That part should be tossed out of your question. What is true is that
$$\left\{ \sqrt{\frac{2}{\pi}}\sin(nx)\right\}_{n=1}^{\infty}$$
is a complete orthonormal basis of $L^{2}[0,\pi]$. There are no conditions required for the $\sin$ series of $f \in L^{2}[0,\pi]$ to converges to $f$ in $L^{2}[0,\pi]$.
To prove this, start with the fact that $\{ \frac{1}{\sqrt{2\pi}} e^{inx}\}_{n=-\infty}^{\infty}$ is a complete orthonormal basis of $L^{2}[-\pi,\pi]$. Every $f \in L^{2}[0,\pi]$ has an odd extension to $\tilde{f}$ on $[-\pi,\pi]$. That is $\tilde{f}(x)=-f(-x)$ for $-\pi \le x < 0$ and $\tilde{f}(x)=f(x)$ for $0 \le x \le \pi$. then
$$
\begin{align}
     \int_{-\pi}^{\pi}\tilde{f}(x)e^{-inx}\,dx 
         & = -\int_{-\pi}^{0}f(-x)e^{-inx}\,dx+\int_{0}^{\pi}f(x)e^{-inx}\,dx \\
         & = -\int_{0}^{\pi}f(x)e^{inx}\,dx+\int_{0}^{\pi}f(x)e^{-inx}\,dx \\
         & = -2i\int_{0}^{\pi}f(x)\sin(nx)\,dx .
\end{align}
$$
Therefore the following converges in $L^{2}[-\pi,\pi]$:
$$
       \tilde{f} = -\sum_{n=-\infty}^{\infty}\frac{2i}{2\pi}
                               \left(\int_{0}^{\pi}f(x')\sin(nx')\,dx'\right)e^{inx}.
$$
Because $\tilde{f}$ is odd, then the following converges in $L^{2}[0,\pi]$:
$$
    f(x)=\frac{1}{2}\left(\tilde{f}(x)-\tilde{f}(-x)\right)=\sum_{n=1}^{\infty}
          \left(\frac{2}{\pi}\int_{0}^{\pi}f(x')\sin(nx')\,dx'\right)\sin(nx).
$$
And that's the result you want.
A: Try the following, if you want to avoid any Fourier series methods: first throw in the function $1$ into your set $\{\sin{kx}\}$. The resulting linear span is a subalgebra of $C(T,\mathbb{R})$, where $T$ is the one-dimensional torus. Then apply Stone-Weierstrass to obtain density in the $C$-norm. Now use the fact that $C(T,\mathbb{R})$ is dense in $L^2(T, \mathbb{R})$ in the $L^2$ norm. Since $T$ is compact, one can control the $L^2$ norm by the $C$-norm, which shows that the subalgebra generated by $1, \{\sin{kx}\}$ is dense in $L^2(T)$. Then remove the constants.
