Base for Fourier series in an interval. INTRODUCTION:
Given the problem $\begin{cases} u_t-u_{xx}=0 & (x,t)\in(0,\pi)\times(0,+\infty) \\ u(x,0)=g(x)&x\in[0,\pi] \\ u_x(0,t)=u(\pi,t) & t>0 \end{cases}$ I was trying to find the solutions with the method of separetion of variables.
After some calculation I find that the solution, trowing the intial condition on the second line, should be $u_k=c_k\cos(\frac{2k+1}{2}x)e^{-(\frac{2k+2}{2})^2t}$ so to satisfy the initial condition a want to find $c_k$ to have a series that express $g(x)$,
I mean solve $\sum_{_{k>0}} c_k\cos(\frac{2k+1}{2}x)=g(x)$ in the interval $[0,\pi]$.
QUESTIONS:
(i) I know that $\{\cos(kx),\sin(kx)\}_k$ is a orthonormal base (respect to the $L^2[-\pi,\pi]$ scalar product) but what can I say about $\cos(\frac{2k+1}{2}x)$ or $\sin(\frac{2k+1}{2}x)$?
(ii) Is it real that $c_k=\frac{2}{\pi}\int g(x)\cos(\frac{2k+1}{2}x)dx$?
(iii) Is there a general rule to find from the standard base other bases?
Here i was thinking that in the theroy of vector spaces, in finite dimension, i know that all the "good" automorphisms are represented by orthogonal matrices, is there any similar thing in infinite dimension using linear operator?
If I miss some details I will add it, if I can, let me know.
Any suggestion, answer, hint.. would be greatful, thanks.
 A: (I)

*

*You can calculate that $\{\cos(\frac{2k+1}{2}x)\}_{k=0}^{\infty}$ is orthogonal system on $L_2[0, \pi]$. This is orthogonal system of eigenfunctions of the Sturm–Liouville problem:
$$X''(x) + \lambda X(x) = 0, X'(0) = X(\pi) = 0.\quad\quad(1)$$
Any function $X$, which satisfy $X'(0) = X(\pi) = 0$, can be represented as $X(x) = \sum_{k=0}^{\infty} A_k \cos(\frac{2k+1}{2}x)$.

*You can calculate that $\{\sin(\frac{2k+1}{2}x)\}_{k=0}^{\infty}$ is orthogonal system on $L_2[0, \pi]$. This is orthogonal system of eigenfunctions of the Sturm–Liouville problem:
$$X''(x) + \lambda X(x) = 0, X(0) = X'(\pi) = 0.\quad\quad(2)$$
Any function $X$, which satisfy $X(0) = X'(\pi) = 0$, can be represented as $X(x) = \sum_{k=0}^{\infty} A_k \sin(\frac{2k+1}{2}x)$.

*These systems do not produce basis of all space $L_2[0, \pi]$, but they produce basis of vector space functions, which satisfy corresponding boundary conditions.

(II) Yes. Because you have $X_k(x) = \cos(\frac{2k+1}{2}x)$ and $T_n(t) = c_k e^{-(\frac{2k+2}{2})^2t}$. $u$ can be represented as $\sum_{k=0}^{\infty} X_k(x) T_n (t)$.
$u(x, 0) = \sum_{k=0}^{\infty} c_k X_k(x) = g(x)$ or $u(x, 0) = \sum_{k=0}^{\infty} c_k \cos(\frac{2k+1}{2}x) = g(x)$. So $c_k = \frac{\langle g, X_k \rangle}{\langle X_k, X_k \rangle}$, where $\langle a, b \rangle = \int_{0}^{\pi} a(x) b(x) dx$.
(III) System of functions $X_k$ can be obtained as eigenfunctions of Sturm–Liouville problem. In your case $u(x, t) = X(x) T(t)$. After substituting into equation $X(x) T'(t)  - X''(X) T(t) = 0$ or $\frac{T'(t)}{T(t)} = \frac{X''(x)}{X(x)} = -\lambda$. Now you have equations: $X''(x) + \lambda X(x) = 0$ and $T'(t) + \lambda T(t) = 0$. Using $u_x(0, t) = X'(0) T(t) = 0$ we get $X'(0) = 0$, and using $u(\pi, t) = X(\pi) T(t) = 0$, we get $X(\pi) = 0$. In these problems we do not need basis of all space, we need basis of functions that satisfy boundary conditions, and we find it as eigenfunctions of the Sturm–Liouville problem.
