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#1 h May 1, 2025, 2:59 PM • 3 Y

Y by User141208, MathLuis, cubres

Find all functions $f: \mathbb{R^+} \rightarrow \mathbb{R^+}$ , such that for all $x,y>0$ :
$f(x+f(y))=\dfrac{f(x)}{1+f(xy)}$

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#3 h May 2, 2025, 2:41 AM • 1 Y

Y by cubres

Can anyone check my solution, pls
Let $f(1)=t$
Suppose there exist $a>b$ such that $f(a)=f(b)$ . Let $T=\dfrac{a}{b} >1$
$P(x,a)+P(x,b): f(xa)=f(xb) \rightarrow f(xT)=f(x), \forall x \in \mathbb{R}^+$
By induction, we have: $f\left(xT^k\right)=f(x)$ . Fix $x$ , let $k \to \infty$ , we have $f(x)=0$ (contrdiction)
So that, $f$ is injective $(1)$
Suppose there exist $u>v$ such that $f(u) >f(v)$
$P\left(\dfrac{x}{y},y\right):f\left(\dfrac{x}{y}+f(y)\right)=\dfrac{f\left(\dfrac{x}{y}\right)}{1+f(x)}$
Let $x_0=uv. \dfrac{f(u)-f(v)}{u-v}$
Put $x=x_0,y=u$ in $(2)$ : $f\left(v. \dfrac{f(u)-f(v)}{u-v}+f(u)\right)=\dfrac{f\left(v. \dfrac{f(u)-f(v)}{u-v}\right)}{1+f(x_0)}$
Put $x=x_0,y=v$ in $(2)$ : $f\left(u. \dfrac{f(u)-f(v)}{u-v}+f(v)\right)=\dfrac{f\left(u. \dfrac{f(u)-f(v)}{u-v}\right)}{1+f(x_0)}$
So that: $f\left(v. \dfrac{f(u)-f(v)}{u-v}\right)=f\left(u. \dfrac{f(u)-f(v)}{u-v}\right)$
Since $f$ is injective, we have contradiction.
Therefore, $\forall x>y$ , we have $f(x) \le f(y)$ . From $(1)$ , we have: $\forall x>y$ , we have $f(x) < f(y) (2)$
$(1)$ and $(2)$ give us $f$ is surjective.
$P(1,y): f(1+f(y))=\dfrac{t}{1+f(y)} \rightarrow f(x)= \dfrac{t}{x}, \forall x>1$
Fix $y\le1$ . Then there exist $x>1$ such that $xy>1$
$P(x,y): \dfrac{t}{x+f(y)}=\dfrac{t}{x+\dfrac{t}{y}} \rightarrow f(x) =\dfrac{t}{x}, \forall x \le 1$
Therefore, all solution to this problem are:
$\boxed{f(x)=\dfrac{t}{f(x)}, \forall x \in \mathbb{R}^+}$

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#4 h May 2, 2025, 4:02 AM • 1 Y

Y by cubres

luutrongphuc wrote:

Can anyone check my solution, pls
Let $f(1)=t$
Suppose there exist $a>b$ such that $f(a)=f(b)$ . Let $T=\dfrac{a}{b} >1$
$P(x,a)+P(x,b): f(xa)=f(xb) \rightarrow f(xT)=f(x), \forall x \in \mathbb{R}^+$
By induction, we have: $f\left(xT^k\right)=f(x)$ . Fix $x$ , let $k \to \infty$ , we have $f(x)=0$ (contrdiction)
So that, $f$ is injective $(1)$
Suppose there exist $u>v$ such that $f(u) >f(v)$
$P\left(\dfrac{x}{y},y\right):f\left(\dfrac{x}{y}+f(y)\right)=\dfrac{f\left(\dfrac{x}{y}\right)}{1+f(x)}$
Let $x_0=uv. \dfrac{f(u)-f(v)}{u-v}$
Put $x=x_0,y=u$ in $(2)$ : $f\left(v. \dfrac{f(u)-f(v)}{u-v}+f(u)\right)=\dfrac{f\left(v. \dfrac{f(u)-f(v)}{u-v}\right)}{1+f(x_0)}$
Put $x=x_0,y=v$ in $(2)$ : $f\left(u. \dfrac{f(u)-f(v)}{u-v}+f(v)\right)=\dfrac{f\left(u. \dfrac{f(u)-f(v)}{u-v}\right)}{1+f(x_0)}$
So that: $f\left(v. \dfrac{f(u)-f(v)}{u-v}\right)=f\left(u. \dfrac{f(u)-f(v)}{u-v}\right)$
Since $f$ is injective, we have contradiction.
Therefore, $\forall x>y$ , we have $f(x) \le f(y)$ . From $(1)$ , we have: $\forall x>y$ , we have $f(x) < f(y) (2)$
$(1)$ and $(2)$ give us $f$ is surjective.
$P(1,y): f(1+f(y))=\dfrac{t}{1+f(y)} \rightarrow f(x)= \dfrac{t}{x}, \forall x>1$
Fix $y\le1$ . Then there exist $x>1$ such that $xy>1$
$P(x,y): \dfrac{t}{x+f(y)}=\dfrac{t}{x+\dfrac{t}{y}} \rightarrow f(x) =\dfrac{t}{x}, \forall x \le 1$
Therefore, all solution to this problem are:
$\boxed{f(x)=\dfrac{t}{f(x)}, \forall x \in \mathbb{R}^+}$

why is $f$ surjective? there are injective and strictly increasing functions that aren't surjective (note that strictly increasing implies injective), like x/(x+1)?

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#5 h May 2, 2025, 6:05 AM • 1 Y

Y by cubres

jasperE3 wrote:

luutrongphuc wrote:

Can anyone check my solution, pls
Let $f(1)=t$
Suppose there exist $a>b$ such that $f(a)=f(b)$ . Let $T=\dfrac{a}{b} >1$
$P(x,a)+P(x,b): f(xa)=f(xb) \rightarrow f(xT)=f(x), \forall x \in \mathbb{R}^+$
By induction, we have: $f\left(xT^k\right)=f(x)$ . Fix $x$ , let $k \to \infty$ , we have $f(x)=0$ (contrdiction)
So that, $f$ is injective $(1)$
Suppose there exist $u>v$ such that $f(u) >f(v)$
$P\left(\dfrac{x}{y},y\right):f\left(\dfrac{x}{y}+f(y)\right)=\dfrac{f\left(\dfrac{x}{y}\right)}{1+f(x)}$
Let $x_0=uv. \dfrac{f(u)-f(v)}{u-v}$
Put $x=x_0,y=u$ in $(2)$ : $f\left(v. \dfrac{f(u)-f(v)}{u-v}+f(u)\right)=\dfrac{f\left(v. \dfrac{f(u)-f(v)}{u-v}\right)}{1+f(x_0)}$
Put $x=x_0,y=v$ in $(2)$ : $f\left(u. \dfrac{f(u)-f(v)}{u-v}+f(v)\right)=\dfrac{f\left(u. \dfrac{f(u)-f(v)}{u-v}\right)}{1+f(x_0)}$
So that: $f\left(v. \dfrac{f(u)-f(v)}{u-v}\right)=f\left(u. \dfrac{f(u)-f(v)}{u-v}\right)$
Since $f$ is injective, we have contradiction.
Therefore, $\forall x>y$ , we have $f(x) \le f(y)$ . From $(1)$ , we have: $\forall x>y$ , we have $f(x) < f(y) (2)$
$(1)$ and $(2)$ give us $f$ is surjective.
$P(1,y): f(1+f(y))=\dfrac{t}{1+f(y)} \rightarrow f(x)= \dfrac{t}{x}, \forall x>1$
Fix $y\le1$ . Then there exist $x>1$ such that $xy>1$
$P(x,y): \dfrac{t}{x+f(y)}=\dfrac{t}{x+\dfrac{t}{y}} \rightarrow f(x) =\dfrac{t}{x}, \forall x \le 1$
Therefore, all solution to this problem are:
$\boxed{f(x)=\dfrac{t}{f(x)}, \forall x \in \mathbb{R}^+}$

why is $f$ surjective? there are injective and strictly increasing functions that aren't surjective (note that strictly increasing implies injective), like x/(x+1)?

Oh, tks i will fix it

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#6 h May 2, 2025, 11:48 AM • 2 Y

Y by cubres, navi_09220114

luutrongphuc wrote:

Fix $x$ , let $k \to \infty$ , we have $f(x)=0$ (contrdiction)

Why?

Anyway,
1. You can prove injective by using this sub instead.
2. You can prove surjectivity using the following:
Claim 1: $\lim_{x\to+\infty} f(x) = 0$
Proof

Claim 2: $f$ is continuous
Proof

Claim 3: $\lim_{x\to 0^+} f(x) = +\infty$
Proof
Alternative proof by ja.

These 3 claims imply $f$ is surjective.

This post has been edited 1 time. Last edited by mashumaro, May 2, 2025, 2:05 PM

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#7 h May 2, 2025, 12:09 PM • 1 Y

Y by cubres

mashumaro wrote:

luutrongphuc wrote:

Fix $x$ , let $k \to \infty$ , we have $f(x)=0$ (contrdiction)

Why?

Anyway,
1. You can prove injective by using this sub instead.
2. You can prove surjectivity using the following:
Claim 1: $\lim_{x\to+\infty} f(x) = 0$
Proof

Claim 2: $f$ is continuous
Proof

Claim 3: $\lim_{x\to 0^+} f(x) = +\infty$
Proof

These 3 claims imply $f$ is surjective.

Thank you for spotting my mistake.
I tried to fix it but failed

This post has been edited 2 times. Last edited by luutrongphuc, May 2, 2025, 12:20 PM

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#8 h May 2, 2025, 8:13 PM • 3 Y

Y by cubres, mashumaro, LeDuonggg

Denote $P(x,y)$ the assertion of the given F.E. (New color pog)
Claim 1: $f$ is injective.
Proof: Suppose FTSOC there existed $a>b$ with $f(a)=f(b)$ then from $P(x,a)-P(x,b)$ we get that $P(ax)=f(bx)$ and thus $f(x)=f \left( \left(\frac{a}{b} \right)^k x \right)$ for all $k \in \mathbb Z$ and so $P \left(\frac{bx}{a}, y \right)$ gives $f \left(\frac{bx}{a}+f(y) \right)=\frac{f(x)}{1+f(xy)}$ but here now we can let $x=\frac{af(y)}{a-b}$ to get that $f \left(\frac{ayf(y)}{a-b} \right)=0$ which is of course a contradiction, therefore $f$ is injective as desired.
Claim 2: $f$ is strictly decreasing.
Proof: Suppose FTSOC there existed $m>n$ with $f(m)>f(n)$ then from $P \left(\frac{m(f(m)-f(n)}{m-n}, n \right)-P \left(\frac{n(f(m)-f(n)}{m-n}, m \right)$ we get that $m=n$ are cancelling everything and using Claim 1 and thus a contradiction, meaning $f$ is indeed strictly decreasing as $f(n)=f(m)$ can't happen.
Claim 3: $\lim_{n \to \infty} f(a_n)=0$ for all sequences $a_n>nf(1)$ for all $n>N$ .
Proof: Notice $P(x,1)$ gives $f(x+f(1))=\frac{f(x)}{1+f(x)}$ and inductive here using the jump $x \to x+f(1)$ gives that $f(x+nf(1))=\frac{f(x)}{nf(x)+1}$ for all $n \in \mathbb Z_{>0}$ and obviously taking $n \to \infty$ is sufficient to prove the claim (the more relevant part is to see how it decreases at a slow but sure pace, which we will use next).
Claim 4: $f$ is continuos.
Proof: Fix $x$ and consider $c_n>\max \{nf(1), xnf(1) \}$ for all $n>N$ then by $P(x,c_n)$ and setting $n \to \infty$ makes it clear why this claim holds true, as the inside of LHS becomes closer and closer to $x$ while on the RHS we can see how the whole thing is closer to $f(x)$ so by epsilon-delta definition the claim is true to the right. Now to check its continuos to the left again fix a constant $x$ the same sequence $c_n$ but in addition we instead consider all $n>M$ for which $x-f(c_n)>C$ for some constant $x>C$ and thus $P(x-f(c_n), c_n)$ and the fact that $f(Cc_n)$ will still become arbitrarily small eventually will give from $n \to \infty$ that $f$ is also continuos to the left, finish the claim from epsilon-delta definition.
Claim 5: $f$ is unbounded.
Proof: Suppose $L=\lim_{x \to 0^+} f(x)$ was bounded then on $P(x,y)$ for taking $x \to 0^+$ and a fixed $y$ gives that $f(f(y))=\frac{L}{L+1}$ however $f$ is injective so this can't happen, thus proving the claim.
Claim 6: $f$ is surjective and $f(x)=\frac{f(1)}{x}$ for all $x>1$ .
Proof: $f$ is surjective follows from Claims 2,3,4,5 (mainly continuity carrying lol) and now from $P(1,x)$ and our surjectivity we have that $f(1+t)=\frac{f(1)}{1+t}$ for all $t>0$ as desired.
The finish: Now fix some $x$ then taking the identity on Claim 3 for some large enough $nf(1)>1$ we have that $\frac{f(x)}{nf(x)+1}=\frac{1}{x+nf(1)}$ now if $x>1$ then we have that $\frac{1}{n+x}=\frac{1}{x+nf(1)}$ so $f(1)=1$ and now setting $0<x<1$ would give that $nf(x)+1=xf(x)+nf(1)$ so $xf(x)=1$ and thus combining all this gives $f(x)=\frac{1}{x}$ for all $x \in \mathbb R_{>0}$ thus we are done :cool:

:cool:

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