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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.

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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.

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  • $\begingroup$ Thanks but maybe I should have been a bit more clear. I think I'm actually okay with calculating the minimum (although your answer does make the process make more sense in my head) but the inner product I have, the usual $L^2$ inner product, is defined as $(f,g)=\int_{0}^{1}f(t)\bar{g(t)}dt$. So I would like to reduce the problem I have from $\int_{-1}^{1}...$ to $2 \cdot \int_{0}^{1}...$. Is this possible? $\endgroup$ – Daniel Ward Jun 13 '15 at 18:00
  • $\begingroup$ You can easily check that $(f,g)=\int _{-1}^1f(t)\bar{g}(t)dt$ define an inner product on the space of square integrable functions defined on $[-1,1]$, it is exactly the same proof. In particular the considerated polynomials are in this space. It doesn't seem easy to prove $\int _{1}^1\ldots =2\int _0^1$. In addition this method cannot be generalized. $\endgroup$ – Patissot Jun 13 '15 at 20:07

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