Finding all measurable functions maximizing an expression given integral conditions

Question

I'd like to know how to answer the question located in this comprehensive exam from the 1990s. The question is:

Find the maximum value of $\int_{-1}^{1} x^3 g(x) dx$ for measurable functions g(x) satisfying

$$ \int_{-1}^{1} g(x) dx = \int_{-1}^{1} x g(x) dx = \int_{-1}^{1} x^2 g(x) dx = 0 . $$

and $\int_{-1}^{1} |g(x)|^2 dx = 1$.

In principle I think I could reduce to the case where $g$ is continuous, and then even smooth/polynomial, and try to formulate things as some kind of optimization problem, but I suspect there's some sort of trick I'm not aware of.

Answer 1

I suppose that it is a linear algebra problem: $g$ needs to belong to the orthogonal complement of $1, x, x^2$ and we want to maximize its inner product with $x^3$. So, what we do is find the projections of $x^3$ onto the (normalized) $1, x,$ and $x^2$ coordinates, the subtract that away to get a function $h$ that is orthogonal to $1, x, x^2$ and will be the closest in direction to $x^3$. Then you normalize $h$ to get $g$.

Answer 2

We have four conditions. Define

$$ g(x) = a_0+a_1 x+ a_2 x^2+ a_3 x^3 $$

and then calculate

$$ \int_{-1}^1 g(x)dx = 2\left(a_0+\frac 13 a_2\right)=0\\ \int_{-1}^1 g(x)x dx = 2\left(\frac 13 a_1+\frac 15 a_3\right) = 0\\ \int_{-1}^1 g(x)x^2 dx =2\left(\frac 13 a_0 +\frac 15 a_2\right) = 0\\ \int_{-1}^1 |g(x)|^2 dx = 2 a_0^2+\frac{4 a_0 a_2}{3}+\frac{2 a_1^2}{3}+\frac{4 a_1a_3}{5}+\frac{2 a_2^2}{5}+\frac{2 a_3^2}{7}=1 $$

and solving we have

$$ \left[ \begin{array}{cccc} a_0 & a_1 & a_2 & a_3 \\ 0 & -\frac{3 \sqrt{\frac{7}{2}}}{2} & 0 & \frac{5 \sqrt{\frac{7}{2}}}{2} \\ 0 & \frac{3 \sqrt{\frac{7}{2}}}{2} & 0 & -\frac{5 \sqrt{\frac{7}{2}}}{2} \\ \end{array} \right] $$

and the maximum is $\frac{5}{\sqrt{14}}-\frac{3 \sqrt{\frac{7}{2}}}{5}$