Number of all positive continuous function $f(x)$ in $\left[0,1\right]$ 
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
 A: One way is to think of $f(x)$ as a probability measure. Then you are seeking a continuous random variable $Y$ such that $E(Y)^2=E(Y^2)$. Since $E(Y^2)-E(Y)^2$ is the variance of $Y$, this value is zero if and only if $Y$ is a constant, and thus $f$ would be a delta function, not a continuous real-valued function.
The standard proof of this is to note:
$$\begin{align}
0<\int_{0}^1 (x-a)^2 f(x)\,dx &= \int_0^1 x^2f(x)\,dx-2a\int_0^1 xf(x)\,dx + a^2\int_0^1 f(x)\,dx\\
&= \int_0^1 x^2f(x)\,dx - 2a^2+a^2\\
&=\int_0^1 x^2f(x)\,dx -a^2
\end{align}$$
So $$\int_0^1 x^2f(x)\,dx >a^2$$
A: By the Cauchy-Schwarz inequality,
$$ \alpha^2 = \bigg( \int_{0}^{1} x f(x) \, dx \bigg)^2 \leq \bigg( \int_{0}^{1} f(x) \, dx \bigg)\bigg( \int_{0}^{1} x^2 f(x) \, dx \bigg) = \alpha^2. $$
Since the inequality is saturated, it reduces to an equality. This implies that $f(x)$ is a constant multiple of $x^2 f(x)$, which is impossible. Therefore no such function $f$ exist.
Remark. Positivity of $f$ is essential in proving non-existence of such $f$. If the positivity of $f$ is dropped off, then there are infinitely many solutions. Even you can find a quadratic solution!
A: 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 $1$.
