TITLE: Number of all positive continuous function $f(x)$ in $\left[0,1\right]$
QUESTION [5 upvotes]: Number of all positive continuous function $f(x)$  in $\left[0,1\right]$ which satisfy $\displaystyle \int^{1}_{0}f(x)dx=1$ and $\displaystyle \int^{1}_{0}xf(x)dx=\alpha$ and $\displaystyle \int^{1}_{0}x^2f(x)dx=\alpha^2$
Where $\alpha$ is a given real numbers.

$\bf{My\; Try::}$ :: Adding $(1)$ and $(3)$ and subtracting $2\times  (2),$  we. Get $$\displaystyle \int^{1}_{0}(x-1)^2f(x)dx=(\alpha-1)^2$$  now how can I solve it after that, Thanks

REPLY [3 votes]: I believe there are no such functions. Indeed, by combining the equations, we get that for every polynomial $P$ of degree $\le2$, we have 
$$\int_0^1  P(x) f(x)  dx = P(\alpha) $$
In particular, for $P=(x-\alpha)^2 $, we get 
$$\int_0^1 (x-\alpha)^2  f(x)  dx = 0$$
The integrand $x \mapsto (x-\alpha)^2  f(x)$ is a non negative continuous function whose integral is zero, so it's zero, and $f=0$, which contradicts the fact that it's integral over $[0,1] $ is $1$.