Haar condition and double zeros

Question

I wonder if the space $\mathcal{A}=\text{span}\{ x_{0}, x_{2}\}$ where ($x_{i}(t)=t^i$ for all $t\in [0,1]$) is a Haar space? I know that the Haar condition says that the only way an element in $\mathcal{A}$ can have 2 zeros or more is if it's equal to the zero element. In this case $x_{2}$ is in the space but it has a double zero. I'm not sure if it is considered a Haar space or not since the zeros aren't distinct.

Answer 1

By checking a determinantal condition $$\left|\begin{matrix}1&1\\x&y\end{matrix}\right|\ne 0$$ for any $x,y\in[0,1]$ s.t. $x\ne y$, we arrive at $\{1,x^2\}$ is a Chebyshev system on $[0,1]$, because $y^2-x^2=0\iff x=y$. A linear span of a Chebyshev system is a Haar space. Then a double zero of $x^2$ at the endpoint does not violate a Haar property.

Answer 2

a) In $C[0,1]$ as explained earlier (by determinant property) ${A}$ satisfies Haar condition. also note that fn has at most one distict zero b) In $C[-1,1]$, ${A}$ doest not satisfy Haar condition because the determinant is zero for $t_{1}= -1, t_{2}=1$ , also not fn it two distict zeros