Function satisfying $F(a)-F(1)=F(1)-F(\frac{1}{a})$

Question

What continuous monotone increasing function has the property:

$$F(a)-F(1)=F(1)-F(\frac{1}{a}) \space\space\space\space\space\space \forall a>1 $$

The context for this is that I'm looking for a continuous probability distribution on $(0,\infty)$ satisfying:

$$P\left(1\le x\le a\right) = P\left(\frac{1}{a} \le x \le 1\right)$$

Answer 1

Any odd and monotonic increasing function $g(x)$ that has a limit of $\frac12$ when $x\to+\infty$ could work. You define $f(x)=g(\log(x))+\frac12$.

You have the following properties: $$f(1)=\frac12$$ $$f(1/x) + f(x) = 1$$ $$\lim_{x\to0}f(x)=0$$ $$\lim_{x\to+\infty}f(x)=1$$

Answer 2

$$F(a)=\log(a)+C$$ Satisfies for any real $C$.

Using the fundamental theorem of calculus, your equation becomes (I'm assuming a differentiable solution) $$\int_1^a F'(x)dx=\int_{1/a}^1 F'(x)dx$$ Moving everything to the left and differentiating leads to $$F'(a)-F'(1/a)1/a^2=0$$ We are then left with the functional equation $$aF'(a)=F'(1/a)\frac{1}{a}$$ Assume $F'(a)=a^n$ for some $n$. By plugin this into the previous equation we have $n=-1$. So $$F'(a)=1/a$$ Is a candidate. Integration leads to $$F(a)=\log(a)+C.$$ Showing that this is an actual solution is simple.