I am trying to solve $$f(xy)=f\left(\frac{1}{x}\right)+f\left(\frac{1}{y}\right), \text{with }f(0)=0$$
I started by getting $f(1)=2f(1)$ so $f(1)=0$, and then I had $f(x)=f(\frac{1}{x})$. But I do not know how to proceed. How do I solve this?
Functional Equations – Solving Functional Equations Using Inverses
functional-equations
Related Solutions
For every $x$,
$$a[f(x)]^2+bf(x)+c-g(x)=0$$
is an ordinary quadratic equation and
$$f(x)=\frac{-b\pm\sqrt{b^2-4ac+4ag(x)}}{2a}$$ indeed holds.
There are real roots when
$$b^2-4ac+4ag(x)\ge0,$$ a quartic inequation.
If there are no constraints on $f$, except that it must be a function, then for any $x$ such that there are distinct real roots, you can choose either of them. So there are infinitely, non-countably many solutions.
If $f$ is restricted to be continuous, then the choice of the sign must be consistent across the intervals of the domain.
For the case $(1,1,0)$, the roots are
$$\frac{-1\pm\sqrt{1+4(x^4+2x^3+2x^2+x)}}{2}=\frac{-1\pm(2x^2+2x+1)}2=x^2+x,-x^2-x-1.$$
A solution, among many many many others:
Consider any function $g : [1,+\infty) \rightarrow \mathbb{R}$ such that $g(1)=2$.
Then the function $f : \mathbb{R}_+^* \rightarrow \mathbb{R}$ defined by $$f(x)=g(1/x) \quad \text{if } 0<x<1 \quad \quad \quad \text{and} \quad \quad f(x)=g(x) \quad\text{if } x\geq 1 $$
is a solution to your problem. (So there are many solutions)
(If you want continuous solutions, just ask that $g$ is continuous, it will give a continuous $f$. If you want differentiable solutions, then try to see at which condition on $g$ the constructed function will be differentiable - hint : you will have a condition about the derivative of $g$ at $1$)
Best Answer
For $x\neq 0$, let $y=\frac{1}{x}$, then we have $$f(1)=f\left(\frac{1}{x}\right)+f(x) \qquad (\text{where } x \neq 0).$$ Now, we take $y=1$, to obtain $$f(x)=f\left(\frac{1}{x}\right)+f(1).$$ These equations imply that $$f(x)=0 \qquad (\forall x \neq 0).$$ So $f(x)=0$ is the only function.