TITLE: Is $\operatorname{sinc} x$ the only one?
QUESTION [5 upvotes]: I don't know if this is a very simple question with a very difficult answer, but:
Is $y = \dfrac{\sin x}{x}$ the only function such that
$$\int_{-\infty}^{\infty} f(x) dx =\int_{-\infty}^{\infty} f^2(x) dx \text{ ?}$$
This is to say, if we start with the integral equation  
$$\int_{-\infty}^{\infty} f(x) dx =\int_{-\infty}^{\infty} f^2(x) dx \text{ ?}$$
would we only find $f(x) = \dfrac{\sin x}{x}$, or we can expect other solutions?
Maybe I should have made this clear, but I'm talking about $f(x) \neq 0$ and $f(x)$ continuous in $\mathbb{R}$.

REPLY [10 votes]: Take any $f$  where both integrals exist. Define
$$  A = \int_{-\infty}^{\infty} f(x) dx,   $$ then
$$  B = \int_{-\infty}^{\infty} f^2(x) dx.   $$
Now, define
$$  g(x) = \frac{A}{B} \;  f(x).   $$
Then
$$  \int_{-\infty}^{\infty} g(x) dx = \int_{-\infty}^{\infty} g^2(x) dx =   \frac{A^2}{B}.  $$