Find all functions such that:$ \int_0^1 g(x)f(x)~dx =\int_0^1f^2(x)~dx$

Question

I am looking to solve this integral problem.

Find all functions such that:$$ \int_0^1 g(x)f(x)~dx =\int_0^1f^2(x)~dx$$

I found a trivial space of solutions and that is $g(x)=f(x).$ Are there any other solutions? Looking for different kinds of examples of $f$ and $g$ for which it's true.

Answer 1

I would use a Fourier type approach.

Let $\Omega$ be a measurable set. Given $f\in L^2(\Omega)$ such that $\|f\|_\Omega \neq 0$ (otherwise the problem is trivial) consider any $g_*\in L^2(\Omega)$ not orthogonal to $f$, i.e., such that $(f,g_*)_\Omega\neq 0$.

Then take $$ g=\frac{\|f\|_\Omega^2}{(f,g_*)_\Omega} g_* $$

Answer 2

There are infinitely many possibilities. Rewrite the equation as $$\int_0^1f(x)[f(x)-g(x)]dx=0$$ Then the following are possible solutions: if $$\int_0^1f(x)dx=0$$any $g(x)=f(x)+C, C\in\mathbb R$ verifies the equation. Here are some examples:

1. $f(x)=x-\frac12$, $g(x)=x+\frac12$
2. $f(x)=\cos(2\pi x)$, $g(x)=\cos(2\pi x)+1=2\cos^2(\pi x)$