Prove that it is impossible to bound the integral of a product by the product of integrals

Question

How to prove that there is no $c \ge 0$ such that $$ \Big| \int_{0}^{1} f(x)g(x) dx \Big| \le c \int_{0}^{1} |f(x)| dx \cdot \int_{0}^{1} |g(x)| dx $$ where $f,g: [0,1] \to \mathbb{R}$ are continuous.

Answer 1

Take $f(x)=x^n=g(x)$. Then there is no $c$ so that $$ \frac{1}{2n+1} \leqslant \frac{c}{(n+1)^2} $$ for every positive $n$.