Maximal Value of Integral Calculate the maximal value of $\int_{-1}^1g(x)x^3 \, \mathrm{d}x$, where $g$ is subject to the conditions
$\int_{-1}^1g(x)\, \mathrm{d}x = 0;\;\;\;\;\;\;\;$     $\int_{-1}^1g(x)x^2\, \mathrm{d}x = 0;\;\;\;\;\;\;\;\;\;$     $\int_{-1}^1|g(x)|^2\, \mathrm{d}x = 1.$

I should mention that $g\in C[-1,1]$ with real scalars, and usual inner product. Just looking for the actual answer!
 A: We denote with $$I= \int_{-1}^1\!dx\,g(x)x^3$$ the value of the integral. The constraints, we include via Lagrange multipliers. So we would like to maximize the integral
$$J= I + \lambda \int_{-1}^1\!dx\,g(x) + \mu \int_{-1}^1\!dx\,g(x) x^2 + \nu \int_{-1}^1\!dx\,g(x)^2.$$
At the maximum of $J$, the first variation has to vanish:
$$ \delta J =x^3 + \lambda + \mu x^2 + 2 \nu g(x)=0.$$
Thus we have
$$g(x) = c_0 + c_2 x^2 + c_3 x^3.$$


*

*From $\int_{-1}^1\!dx\,g(x)=0$ it follows $3 c_0 + c_2 =0$.

*From $\int_{-1}^1\!dx\,g(x) x^2=0$ it follows $5 c_0 + 3c_2 =0\quad$ ($\Rightarrow c_0 = c_2 =0$)

*From $\int_{-1}^1\!dx\,g(x)^2$ it follows $c_3 =\sqrt{7/2}$.
Thus, we have $$g(x) = \sqrt{\frac72} x^3$$ and the maximum value of $I$ reads
$$ I_\text{max} = \sqrt{\frac27}.$$
A: Using some Hilbert space theory with the inner product $<f,g> = \int_{-1}^1 f(x)g(x)\,dx$, maximizing $\int_{-1}^1 x^3 g(x)$ with $\int_{-1}^1 g(x)^2\,dx = 1$ is the same as maximizing $<x^3,g>$ with the constraint that $||g|| = 1$. Thus $g(x) = cx^3$ for the constant $c$ giving $\int_{-1}^1 (cx^3)^2\,dx = 1$. The other two constraints are satisfied for such $g(x)$, so this will be your answer. Solving for $c$ you have
$$c^2 \int_{-1}^1 x^6 = 1$$
This gives $c = \sqrt{7 \over 2}$ and $g(x) = \sqrt{7 \over 2}x^3$. In this case $\int_{-1}^1 x^3 g(x) = \sqrt{2 \over 7}$.
A: One solution to this problem would be to use Hilbert space methods. There is a well-known orthogonal basis to $L^2([-1,1])$ given by the Legendre polynmials. Let $P_0,P_1,\ldots$ be the Legendre polynomials, as defined in the Wikipedia link. Your conditions $$\tag{$*$}\int_{-1}^1 g(x)\,dx = 0\mbox{ and }\int_{-1}^1g(x)x^2\,dx = 0$$ are equivalent to the inner product equations $\langle g, P_0\rangle = 0$ and $\langle g, P_2\rangle = 0$. Moreover, it is easy to derive that $x^3 = \frac{2}{5}P_3 + \frac{3}{5}P_1$. If $\hat{P}_1$ and $\hat{P}_3$ denote the normalizations of $P_1$ and $P_2$, then $$\hat{P}_1 = \sqrt{\frac{3}{2}}P_1\mbox{ and }\hat{P}_3 = \sqrt{\frac{7}{2}}P_3,$$ so that $$x^3 = \frac{2}{5}\sqrt{\frac{2}{7}}\hat{P}_3 + \frac{3}{5}\sqrt{\frac{2}{3}}\hat{P}_1.$$ In order for $\langle g, x^3\rangle$ to be maximal, $g$ can therefore only have components in the $P_1$ and $P_3$ directions, i.e., $g = a\hat{P}_1 + b\hat{P}_3$ with $a^2 + b^2 = 1$. We have therefore reduced the problem to maximizing  $$\langle g, x^3\rangle = \frac{2}{5}\sqrt{\frac{2}{7}}b + \frac{3}{5}\sqrt{\frac{2}{3}}a$$ under the constraint $a^2 + b^2 = 1$. Standard Lagrange multipliers techniques give that this quantity is maximized when $$a = \frac{3}{5}\sqrt{\frac{7}{3}}\mbox{ and }b = \frac{2}{5},$$ and that for these values of $a$ and $b$ one has $\langle g, x^3\rangle = \sqrt{2/7}.$ Moreover, $g$ is explicitly computed as $$g = a\hat{P}_1 + b\hat{P}_3 = \sqrt{\frac{7}{2}}x^3.$$
