How to show if these functions are a complete set? I have the following exercise:
"Consider $H=L^2[0, \pi]$ and the operator given by 
\begin{align} T(f) = gf\end{align} with
\begin{align}g(x) = \begin{cases}
x \quad \text{ for } 0 \leqslant x \leqslant 1\\
1 \quad \text{ for } 1 \leqslant x \leqslant \pi\end{cases}
\end{align}
Let $u(x)=1$ in $L^2[0, \pi]$ and consider the functions $\Phi_n(x) = T^n(u), n=0, 1, ...$ Are they a complete set in $L^2[0, \pi]$?"
How can I show wheter or not they are complete? What does it even mean exactly for a set fo functions to be complete? It's not like saying a space is complete, is it?
 A: A subset $S$ of an inner product space $X$ is said to be complete (or maximal) if for any $x \in X$ we have $$x \perp S \implies x = 0$$ that is $$\langle x, s\rangle = 0, \forall s \in S \implies x = 0$$
Furthermore, if $X$ is a Hilbert space (like in the case of $L^2[0, \pi]$), then it can be shown that an orthonormal set $S$ is complete if and only if $S$ is an orthonormal basis for $X$, i.e. every $x \in X$ can be written as $$x = \sum_{s \in S} \langle x, s\rangle s$$
The term 'maximal' comes from the fact that for a complete orthonormal set $S$ there doesn't exist a vector $e \in X$ such that $S \cup \{e\}$ is also an orthonormal set (namely $e \perp S \implies e = 0$ so $0 = \|e\| \ne 1$).

We'll prove that the set $\{T^nu : n \in \mathbb{N}\} = \{g^n : n \in \mathbb{N}\}$ is not complete. It suffices to find $h \in L^2[0,\pi]$ such that $h \ne 0$, but $\langle h, g^n\rangle = 0, \forall n \in \mathbb{N}$.
As noted in the comments, since $g|_{[1, \pi]} \equiv 1$ for any $f \in L^2[0,\pi]$ we have
$$(Tf)(x) = f(x), \forall x \in [1, \pi]$$
meaning $T$ does not change the function $f$ on the interval $[1, \pi]$.
In particular, $g^n|_{[1, \pi]}\equiv 1$ for all $n \in \mathbb{N}$.
Define $h : [0, \pi] \to \mathbb{R}$ as 
$$h(x) = \begin{cases}
0,  & \text{if $x \in [0,1]$} \\
\sin\left(\frac{2\pi}{\pi - 1}(x-1)\right), & \text{if $x \in [1, \pi]$}
\end{cases}$$
$h \in L^2[0, \pi]$ since it is continuous. We have $h \ne 0$ but:
$$\langle h, g^n\rangle = \int_0^\pi h(x)\overline{g^n(x)}\,dx = \int_1^\pi h(x)\,dx = 0$$
We conclude that $\{g^n : n \in \mathbb{N}\}$ is not complete.
