Is this function even? I'm working on determining the minimum of a function and in order to do so I would like to simplify my problem a little.
This is the problem:
Provide $C([0,1])$ (C is set of continuous functions, I didn't know how to program the nice C) with the usual $L^2$ inner product.  Determine $min_{a,b,c \in C}\int_{-1}^{1}|t^3-a-bt-ct^2|^2dt$.
The solution I have writes:
$min_{a,b,c \in C}\int_{-1}^{1}|t^3-a-bt-ct^2|^2dt = 2 \cdot min_{a,b,c \in C}\int_{0}^{1}|t^3-a-bt-ct^2|^2dt$
The problem is I don't see why this is.  Does taking the minimum imply that the function has to be even?  Because $|t^3-a-bt-ct^2|^2$ isn't necessarily even.

Hint : You have to compute the $L^2$ projection of $x^3$ on the space $F$ of polynomials with degree less than $2$. Denote $Q(x)=a+bx+cx^2$ the projection of $x^3$. We have $\langle x^3-Q(x), 1 \rangle =0$, $\langle x^3-Q(x), x \rangle =0$ and $\langle x^3-Q(x), x^2 \rangle =0$ since $x^3-Q(x) \in F^\perp$. This is a system with three equations and three unknowns.
