How do find the distribution of a R.V. from this functional/differential equation?

Question

Assume that $X$ is some R.V. with domain $[0,1]$.

A function $U:[0,1]\to[0,1]$ is defined as follows: $$U(x)=x\cdot F_X(x)+(1-x)(1-F_X(x)) = 2x\cdot F_X(x)- x - F_X(x) + 1$$

Where $F_X(x)=\int_0^x f_X(x)$ is the CDF function of $X$.

Suppose we are told that $U$ is constant, that is $\forall x,y\in[0,1]:U(x)=U(y)$.

Is there such a distribution?

---

Notice that since $U$ is constant, the requirement is equivalent to: $$U'(x)=2F_X(s)+2x\cdot f_X(x)-1-f_X(x)=0$$

Or $$2(F_X(x)+xf_X(x))=1+f_X(x) $$

Answer 1

$$U(0)=1-p$$

$$U(1)=p$$

Then since $U(0)=U(1)$:

$$p=\frac{1}{2}$$

Replacing:

$$U(x)=\frac{x}{2}F(x)+\frac{(1-x)}{2}(1-F(x))$$

$$U(x)=xF(x)-\frac{F(x)}{2}-\frac{x}{2}+\frac{1}{2}$$

Since $U(x)=\frac{1}{2}$

$$F(x)=\frac{x}{2x-1}$$

Which is not increasing or even positive therefore no distribution exists