<s>[INST] Find min of $\int_{-1}^1| x^3-a-bx-cx^2|^2 dx$ and max of $\int_{-1}^1 x^3 g(x) dx$ 
1) Apply best approximation theorem to compute min$\int_{-1}^1| x^3-a-bx-cx^2|^2 dx$.
2) Find max $\int_{-1}^1 x^3 g(x) dx$, where $g$ is subject to the constraints 
  $$\int_{-1}^1 g(x)dx = \int_{-1}^1xg(x)dx=\int_{-1}^1 x^2 g(x) dx=0,\quad\text{and}\quad \int_{-1}^1|g(x)|^2 dx=1.$$

I don't know how to start to solve this problem, anyone can help me. 
 [/INST] Hint for 1). Since $\int_{-1}^1x^d dx=0$ when $d$ is odd, and $\int_{-1}^1x^d dx=\frac{2}{d+1}$ when $d$ is even, by expanding the square we get
$$\int_{-1}^1(x^3-a-bx-cx^2)^2 dx
=\frac{2}{3}b^2-\frac{4}{5}b+\frac{2}{7}+2a^2+\frac{4}{3}ac+\frac{2}{5}c^2\\
=\frac{2}{3}\left(b-\frac{3}{5}\right)^2+\frac{8}{175}+2\left(a^2+\frac{2}{3}ac+\frac{1}{5}c^2\right).
$$
Moreover $(1/3)^2-1/5<0$ implies that $\left(a^2+\frac{2}{3}ac+\frac{1}{5}c^2\right)\geq 0$.
Hint for 2) (see Professor Vector's comment). By the given constraints and the Cauchy-Schwarz inequality,
$$\begin{align}
\int_{-1}^1 x^3 g(x) dx&=\int_{-1}^1 (x^3-a-bx-cx^2)g(x) dx\\
&\leq
\left(\int_{-1}^1 (x^3-a-bx-cx^2)^2dx\right)^{1/2}
\left(\int_{-1}^1 (g(x))^2dx\right)^{1/2}\\
&\leq
\left(\int_{-1}^1 (x^3-a-bx-cx^2)^2dx\right)^{1/2}.
\end{align}$$
Hence the maximum of the LHS is less or equal to the minimum of the RHS (see 1)).
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