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Split the best polynomial into its even and odd parts, $p_n=p_e+p_o$. Both parts belong to ${\Pi }_n$. Since $u$ is odd, $u-p_o$ is odd and $p_e$ is even. Their product is odd, so its integral on the symmetric interval vanishes. The orthogonality givesThe candidate $p_o$ cannot have a strictly smaller error than a minimizer. Consequently $\Vert p_e{\Vert }^2=0$, and continuity gives $p_e=0$. Thus $\boxed{p_n(-x)=-p_n(x)}$. For $n=0$ this means $p_0=0$, which is indeed an odd function. More generally this is parity preservation by polynomial least-squares projection.

By parity preservation by polynomial least-squares projection, $p_2(x)=\beta x$: the constant and quadratic coefficients vanish. The remaining normal equation isThereforeThe residual is automatically orthogonal to $1$ and $x^2$ by parity and to $x$ by construction, so it is orthogonal to every element of ${\Pi }_2$. This checks the full linear least squares condition, rather than only a necessary coefficient equation.