Finding the max. of an integral

Question

I have a question which asks:

Let $g\in C[-1,1]$ and the usual inner product $\langle f,g\rangle = \int_{-1}^{1} f(x)g(x)dx$.

Find the max value of $\int_{-1}^{1}g(x)x^3dx$ where $g$ is subject to the restrictions:

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

I'm not really sure how I am supposed to proceed with this question. It is in relation to Hilbert spaces and the orthogonal projection and this question Finding the min of an integral

Please Help.

Answer 1

Here is a hint: Let $P$ be the projection on the span of $1$ and $x^2$, so $Q=I-P$ is the projection on the orthogonal complement of that space. By two of the constraints $g=Qg$, so that $\langle g,x^3\rangle=\langle Qg,x^3\rangle=\langle g,Qx^3\rangle$. So you want to compute $Qx^3$. $g$ is the unit vector with the largest possible inner product with that vector.