Solve functional equation:find all strictly monotone functions $f:(0,+\infty)\to(0,+\infty)$ such that $$(x+1)f(\dfrac{y}{f(x)})=f(x+y),\forall x,y>0$$


Take the limit of both sides of the functional equation as $y\to 0$ from the right:

$$\lim_{y\to 0^+}(x+1)f\left(\dfrac{y}{f(x)}\right)=\lim_{y\to 0^+}f(x+y).$$

On the RHS, the limit is clearly $\lim_{y\to 0^+}f(x+y)=f(x)$. The limit on the LHS would reduce to simply $\lim_{y\to 0^+}(x+1)f\left(\dfrac{y}{f(x)}\right)=(x+1)f(0)$ if $0$ was in the domain of $f$, but even though it's not we can still write $\lim_{y\to 0^+}(x+1)f\left(\dfrac{y}{f(x)}\right)=(x+1)\lim_{y\to 0^+}f(y)=(x+1)f_{0^+}$. Thus, $f$ is of the form: $f(x)=a(x+1)$.

Substitute this trial function back into the functional equation to find the allowed values of the parameter $a$:

$$(x+1)a\left(\frac{y}{a(x+1)}+1\right)=a(x+y+1)\\ \iff y+a(x+1)=a(x+1)+ay\\ \iff y=ay\\ \iff a=1.$$

Hence, the only solution is $f(x)=x+1$.

  • $\begingroup$ Why do you have $\lim_{y\to 0^+}f(x+y)=f(x)$? $\endgroup$ – Babymath Apr 12 '14 at 8:10
  • $\begingroup$ @Babymath Because as $y\to 0$, $x+y\to x+0$. $\endgroup$ – David H Apr 12 '14 at 8:19
  • 1
    $\begingroup$ @DavidH So you're assuming $f$ is continuous? $\endgroup$ – Jack M Apr 12 '14 at 9:34
  • $\begingroup$ @JackM Nope.${}$ $\endgroup$ – David H Apr 12 '14 at 9:43
  • $\begingroup$ And what if $f_{0^+}=\infty$? $\endgroup$ – Konstantinos Gaitanas Apr 12 '14 at 15:54

This is just something that is missing from David's answer. (Which i find very clever)
We have to prove first that $f_{0^+}$ is a real number. (Check line 5 at David's answer )

We need to prove first that $f$ is strictly increasing.
Substitute $x=1,y=f(1)$ in the equation. (Both are positive, so we have no problem)

We have $$2\cdot f(1)=f(1+f(1))\Rightarrow f(1)=\frac{f(1+f(1))}{2}<f(1+f(1))$$ since $f(1+f(1))>0$.

But $1<1+f(1)$ and $f(1)<f(1+f(1))$
Knowing that $f$ is strictly monotone, we see that $f$ is strictly increasing and since it's domain is $(0,+\infty)$:
$$\lim_{y \to 0^{+}}f(y)=f_{0^+}\in \mathbb R$$Combining this and David's approach we get the desired result.

  • $\begingroup$ Cuold you prove $\lim_{y\to 0^+}f(x+y)=f(x)$? and $\lim_{y\to 0^+}f(y)$ exists? $\endgroup$ – Babymath Apr 13 '14 at 5:17

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