In my linear algebra course, we are looking into inner product spaces. The following came up with regards to an inner product on a subspace of the infinitely-differentiable real functions.
Let $f:\mathbb{R}\to\mathbb{R}$ be an infinitely-differentiable function such that $f(x)=0$ for all $|x| \ge 1$. Suppose that $g:\mathbb{R}\to\mathbb{R}$ is also a function satisfying the above properties. Prove that $$\int_{-1}^{1}f''(x)g(x)dx = \int_{-1}^{1}f(x)g''(x)dx.$$
Here are some things that come to mind:
$[f(x)g(x)]''=f''(x)g(x)+2f'(x)g'(x)+f(x)g''(x)$.
If $f$ or $g$ is non-zero on the interval $(-1,1)$, then they need to smoothly leave and return to the x-axis on the interval $[-1,1]$.
Hint: Integrate by parts and use function constraints.