I'm new to this forum and I don't know how to write mathematical symbols. I have the following functional equation:

  • $f$ defined on $[0, +\infty)$ with values in $[0, +\infty)$
  • $f$ is bijective and the following relation holds (*):

$$f^{-1}(f(x)f(y))=\frac{f(x+y)-f(x)-f(y)}2$$ Find all functions with this property. I found that $f(0)=0$ with standard manipulations. $f$ is also continuous (comments). Thank you!

  • $\begingroup$ I edited you question to typeset the math with Latex. You may find this information useful: MathJax tutorial $\endgroup$ – rubik Feb 2 '15 at 16:29
  • $\begingroup$ Thank you very much! It seems to be very useful and easy to learn. $\endgroup$ – user212396 Feb 2 '15 at 16:30
  • $\begingroup$ Yes! It is definitely not difficult to learn and has many benefits, first of all in readability. $\endgroup$ – rubik Feb 2 '15 at 16:32
  • $\begingroup$ The above shows that $f(x+y) \ge f(x)+f(y)$ so $f$ is monotonically increasing. Since it is bijective, it must be continuous. $\endgroup$ – copper.hat Feb 2 '15 at 17:43

It seems the following.

I shall continue from already obtained results. Since the function $f$ is bijective, there exists a point $x_0$ such that $f(x_0)=1$. Putting in the relation (*) $x=x_0$ and $y=nx_0$ we obtain

$$f^{-1}(f(x_0)f(nx_0))=\frac{f((n+1)x_0)-f(x_0)-f(nx_0)}2$$ $$nx_0=\frac{f((n+1)x_0)-1-f(nx_0)}2$$ $$f((n+1)x_0)= f(nx_0)+2nx_0+1$$

From this recurrence, using initial condition $f(x_0)=1$ we obtain


Putting in the relation (*) $x=2x_0$ and $y=2x_0$ we obtain


$$f^{-1}((2+2x_0)^2)=\frac{4+12x_0-2(2+2x_0)}2$$ $$f^{-1}((2+2x_0)^2)=4x_0$$ $$(2+2x_0)^2=f(4x_0)=4+12x_0$$ $$(1+x_0)^2=1+3x_0$$ $$x_0^2=x_0$$ $$x_0=1$$



Putting in the relation (*) $x=1$ and $y=y$ we obtain

$$f^{-1}(f(y))=\frac{f(y+1)-f(1)-f(y)}2$$ $$2y=f(y+1)-1-f(y)$$ $$f(y+1)=f(y)+2y+1$$

Iterating this recurrence, we obtain (**)


Putting in the relation (*) $x=n$ and $y=y$ we obtain

$$f^{-1}(f(n)f(y))=\frac{f(n+y)-f(n)-f(y)}2$$ $$f^{-1}(n^2f(y))=\frac{f(n+y)-n^2-f(y)}2$$

Using (**), we obtain $$f^{-1}(n^2f(y))=ny$$ $$n^2f(y)=f(ny).$$

Therefore for each positive rational number $\frac pq$ we have

$$f\left(\frac pq\right)=\frac 1{q^2}f(p)= \frac {p^2}{q^2}.$$

Since the function $f$ is continuous, for each $x\ge 0$ we have



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