Maximum of a definite integral I am required to find
$max\int_{-1}^{1}x^3g(x)dx$ 
given that 
$\int_{-1}^{1}x^ng(x)dx=0$ for $n=0,1,2$ and $\int_{-1}^{1}|g(x)|^2dx=1$. 
The hint is to find $\underset{a,b,c}{min}\int_{-1}^{1}|x^3-a-bx-cx^2|^2dx$ first.
Below is what I have found.
$\underset{a,b,c}{min}\int_{-1}^{1}|x^3-a-bx-cx^2|^2dx$ is quite easy.
Let $f(x)=x^3-a-bx-cx^2$, then by Holder's Inequality,
$\int_{-1}^{1}x^{3}g(x)dx\leq \int_{-1}^{1}|f(x)g(x)|dx\leq (\int_{-1}^{1}|f(x)|^2dx)^{\frac{1}{2}}(\int_{-1}^{1}|g(x)|^2dx)^{\frac{1}{2}}=(\int_{-1}^{1}|f(x)|^2dx)^{\frac{1}{2}}$ which has a minimum value from above.
But I am not sure if I am working in the right direction since it seems like I am only able to find an upper bound rather than the maximum value.
Any advice?
 A: Have you tried verifying that $f$ satisfies the orthogonality properties verified by $g$ (which should be the case to satisfy the minimum) and replace $g$ by $f$ to have a case of equality in your inequality ?
A: I suppose you started with
$$f(x)=\frac25P_3(x)+\frac35P_1(x)+c_0P_0(x)+c_1P_1(x)+c_2P_2(x)$$
And found that
$$\int_{-1}^1|f(x)|^2dx=\frac27\left(\frac25\right)^2+\frac25|c_2|^2+\frac23\left|\frac35+c_1\right|^2+\frac21|c_0|^2$$
And so found the minimum value of $\frac8{175}$ then $f(x)=\frac25P_3(x)$. From that point I am assuming that $g(x)$ is also a real function of $x$, else how could you find the maximum of $\int_{-1}^1x^3g(x)dx$?  
Let
$$\begin{align}I(\alpha)&=\int_{-1}^1\left(g(x)-\alpha f(x)\right)^2dx\\
&=\int_{-1}^1(g(x))^2dx-2\alpha\int_{-1}^1g(x)f(x)dx+\alpha^2\int_{-1}^1(f(x))^2dx\\
&\ge0\end{align}$$
Then $I(\alpha)$ has a minimum when
$$I^{\prime}(\alpha)=0=-2\int_{-1}^1g(x)f(x)dx+2\alpha\int_{-1}^1(f(x))^2dx$$
So applying this value of $\alpha$,
$$1-\frac{\left[\int_{-1}^1g(x)f(x)dx\right]^2}{\int_{-1}^1(f(x))^2dx}\ge0$$
So that
$$\begin{align}\left[\int_{-1}^1g(x)f(x)dx\right]^2&=\left[\int_{-1}^1g(x)(x^3-a-bx-cx^2)dx\right]^2\\
&=\left[\int_{-1}^1x^3g(x)dx\right]^2\le\int_{-1}^1(f(x))^2dx\end{align}$$
For any real $f(x)=x^3-a-bx-cx^2$, therefore it can be no bigger than the minimum possible value,
$$\left[\int_{-1}^1x^3g(x)dx\right]^2\le\int_{-1}^1\left(\frac25P_3(x)\right)^2dx=\frac4{25}\frac27=\frac8{175}$$
Of course for $g(x)=\sqrt{\frac72}P_3(x)$,
$$\int_{-1}^1(g(x))^2dx=1$$
and
$$\int_{-1}^11\cdot g(x)dx=\int_{-1}^1xg(x)dx=\int_{-1}^1x^2g(x)dx=0$$
And
$$\int_{-1}^1x^3g(x)dx=\int_{-1}^1\frac25P_3(x)\sqrt{\frac72}P_3(x)dx=\frac25\sqrt{\frac27}$$
So the bound is actually attainable.
