I'm an undergraduate student and I've only studied Riemann integrals in detail so far. When I was trying to solve this question, I tried to make use of the boundedness of $h(x)$ but that only works in the interval $[0,1]$, and that helps me find the minimum in that region not the maximum. Then, I tried to think of a function that may satisfy all of these conditions (like a sine wave whose wavelength was adjusted to satisfy the last condition), but I couldn't even find any common continuous function that would satisfy all 4 conditions; one of the zero conditions always got left out. So could someone please help me find such a function, and also give me a direction to proceed to rigorously show what the maximum of this definite integral will be (possibly using elementary techniques)?
Compute $\max \int_{-1} ^1 x^3 h(x) \ dx $ where $h(x)$ is any continuous function on the interval $-1 \leq x \leq 1$ subject to the restrictions \begin{align*} \int_{-1}^1 h(x) \ dx = \int_{-1}^1 xh(x) \ dx = \int_{-1}^1 x^2 h(x) \ dx &= 0; \\ \int_{-1}^1 |h(x)|^2 \ dx &= 1. \end{align*}
We can use the Cauchy-Schwarz inequality for integrals.
The Cauchy-Schwarz inequality for integrals states that: $$\left(\int f(x)g(x)\,dx\right)^2 \leq \int f^2(x)\,dx \cdot \int g^2(x)\, dx $$
If we plug in $f(x) = x^3$ and $g(x) = h(x)$, then we get:
$$ I^2 \le \int_{-1}^1 (x^3)^2 dx \cdot \int_{-1}^1 h^2(x) dx = \frac{2}{7} \cdot 1 = \frac{2}{7}$$
This implies that the integral is bound by the inequality
$$ I \le \sqrt{\frac{2}{7}}$$