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.

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    $\begingroup$ Define an inner product $(f,g)=\int_{-1}^{1}fg$ in the space of square integrable functions on $[-1,1]$. Then apply Gram-Smith to the linearly independent system $1,x,x^2,x^3$ to obtain four orthonormal functions $f_0,f_1,f_2,f_3$. This system can be completed to an orthonormal basis $f_0,f_1,...$. Then $g=\sum_n a_nf_n$. Express $x^3=b_0f_1+b_1f_1+b_2f_2+b_3f_3$. Then $\int_{-1}^{1}x^3g=a_0b_0+a_1b_1+a_2b_2+a_3b_3\leq (a_0^2+...+a_3^2)^{1/2}(b_0^2+...+b_3^2)^{1/2}\leq(b_0^2+...+b_3^2)^{1/2}$, by Cauchy's inequality. The equality can be attained when $(a_0,...,a_3)$ and $(b_0,...,b_3)$ are ... $\endgroup$ – user574889 Jul 20 '18 at 0:57
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    $\begingroup$ ... proportional, and $a_0^2+...+a_3^2=1$. The latter ensuring that $\int_{-1}^{1}|g|^2=a_0^2+...+a_3^2=1$. $\endgroup$ – user574889 Jul 20 '18 at 0:57

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 $g$.


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 $\frac{5}{\sqrt{14}}-\frac{3 \sqrt{\frac{7}{2}}}{5}$

  • $\begingroup$ Why do we have to pick $g$ to be a polynomial? $\endgroup$ – Sambo Jul 21 '18 at 16:59
  • $\begingroup$ @Sambo By simplicity. $\endgroup$ – Cesareo Jul 21 '18 at 17:00
  • $\begingroup$ Could you elaborate? $\endgroup$ – Sambo Jul 21 '18 at 17:02

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