Prove or disprove: $\{t^{2k}\}_{k=0}^{\infty}$ complete in $L_2[-1,3]$ Is $\{t^{2k}\}_{k=0}^{\infty}$ not complete in $L_2[-1,3]$?(Here, completeness of a system is equivalent to the density of its span) 
Obviously many polynomials in the domain will be irrelevant, but I don't quite understand what I am to do/show. I thought I should look at an odd $n$ for $t^n$, but how can I show it is not a limit of $sp(\{t^{2k}\})$. In addition, how does $[-1,3]$ play a role in it? I would appreciate it if you could address it.
 A: The function $t\cdot \chi_{[-1,1]}(t)$ is orthogonal to each $t^{2k}$ in $L^2[-1,3],$ hence is orthogonal to the the linear span of $\{t^{2k} : k=0,1,\dots \}$ in $L^2[-1,3],$ hence is orthogonal to the closure of this linear span in $L^2[-1,3].$ Therfore this closure cannot be all of $L^2[-1,3].$
A: Let $A$ denote the linear span of $\{t^{2k}\}_{k \in \mathbb{N}}$.
Then, if instead of $[-1,3]$ the domain was $[1,3]$ instead, we could use the Stone-Weierstrass theorem to conclude that $A$ is dense ins $C([1,3])$ and thus in $L^2([1,3])$ since $t^2$ separates the points of $[1,3]$ and $1 \in A$.  
However, for $[-1,3]$, we have $t^{2k}(-1) = t^{2k}(1)$ for all $k \in \mathbb{N}$, and thus $f(-1) = f(1)$ for all $f$ in the linear span of $\{ t^{2k} \}_{k \in \mathbb{N}}$, so in particular the linear span does not separate the points of $[-1,3]$.
Now, you idea of showing $t \not\in \overline{A}$ is the correct way, since $t(-1) \neq t(1)$, and even $t(x) \neq t(-x)$ for all $x \in[-1,0]$ where $f(x) = f(-x)$ for all $f \in A$. We need this, as $[-1,0]$ has non-zero measure where $\{1\}$ has zero measure.
Now, suppose $f_n \in A$ converges in $L^2$ to $t$, then we have, as $f(t) = f(-t)$ for $t \in [0,1]$
\begin{align*}
\| f_n - t\|^2 &= \int |f_n(t) - t|^2 \text{d} t \\
&\ge \int_{-1}^0 |f_n(t) - t|^2 \text{d} t + \int_{0}^1 | f_n(t) -t |^2 \text{d}t\\
&= \int_0^1 |f_n(z)+z|^2 \text{d}z+ \int_0^1 |f_n(t) - t|^2 \text{d}t
\end{align*}
As $f_n$ converges in norm to $t$, we have that $\int_0^1 |f_n(z)+z|^2 \text{d}z$ goes to $0$.
So we can find a subsequence $f_{n_j}$ which converges pointwise on $[0,1]$ to $-z$. However, we also have that $\int_0^1 |f_{n_j}(z) - z |^2 \text{d}z$ goes to zero, so we can find a subsequence of $f_{n_j}$ which converges pointwise to $z$ on $[0,1]$, which is absurd as $f_{n_j}$ converges pointwise to $-z$.
So $t \not\in \overline{A}$, and thus $A$ is not dense.
