$\int_{0}^{1}f(x)\cdot g(x)dx=\int_{0}^{1}f(x)dx\cdot \int_{0}^{1}g(x)dx$ Let $f:[0,1]\rightarrow \mathbb{R} $ be a continuous function so that $\int_{0}^{1}f(x)\cdot g(x)dx=\int_{0}^{1}f(x)dx\cdot \int_{0}^{1}g(x)dx$, for any $g:[0,1]\rightarrow \mathbb{R}$, continuous and not differentiable.
Prove that $f$ is constant.
I have no idea on what to do.

Fro the first mean value theorem for definite integrals follows that the exists $c \in (0, 1)$ such that
$$\int_{0}^{1}f(x)dx = f(c)$$
With the assumption above
$$\int_{0}^{1}(f(x)-f(c))^2dx = \int_{0}^{1}f^2(x)dx - \int_{0}^{1}f^2(c)dx = (\int_{0}^{1}f(x)dx)^2 - \int_{0}^{1}f^2(c)dx = 0$$
follows $f(x)-f(c) = 0$ on $(0,1)$.
