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314 posts

a2048

#1 h Jun 10, 2020, 6:10 AM • 6 Y

Y by I_am_human, Wisphard, Mango247, Mango247, Mango247, ehuseyinyigit

Determine all functions $f : \mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that $f(a)f(b)=f(a + bf(a))$

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pco

#2 h Jun 10, 2020, 7:02 AM • 7 Y

Y by a2048, Bumblebee60, dangerousliri, Enigma714, vsamc, Abidabi, Math-wiz

a2048 wrote:

Determine all functions $f : \mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that $f(a)f(b)=f(a + bf(a))$

Let $P(x,y)$ be the assertion $f(x)f(y)=f(x+yf(x))$

1) If $f(x)<1$ for some $x$ , then $P(x,\frac x{1-f(x)})$ $\implies$ $f(x)=1$ , impossible
So $f(x)\ge 1$ $\forall x>0$

2) If $f(u)=1$ for some $u>0$
$P(u,x)$ $\implies$ $f(x+u)=f(x)$ $\forall x$
If $f(x)>1$ for some $x\ne u$ , let $k\in\mathbb N$ such that $ku>x$ :
$P(x,\frac{ku-x}{f(x)-1})$ $\implies$ $f(x)f(\frac{ku-x}{f(x)-1})=f(\frac{ku-x}{f(x)-1}+ku)=f(\frac{ku-x}{f(x)-1})$
And so $f(x)=1$ , impossible
So $\boxed{\text{S1 : }f(x)=1\quad\forall x>0}$ which indeed is a solution

3) If $f(x)>1$ $\forall x>0$
Let then $a=f(1)-1>0$
$P(x,\frac y{f(x)})$ $\implies$ $f(x+y)=f(x)f(\frac y{f(x)})>f(x)$ and $f(x)$ is injective
Subtracting $P(x,1)$ from $P(1,x)$ and using injectivity, we get
$\boxed{\text{S2 : }f(x)=ax+1\quad\forall x>0}$ which indeed is a solution, whatever is $a>0$

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#4 h Jun 10, 2020, 1:33 PM • 1 Y

Y by Mango247

$P(a,b): f(a)f(b)=f(a+bf(a))$ If $f(a)<1$ :
$P(a,\frac{a}{1-f(a)}):f(a)f(\frac{a}{1-f(a)})=f(\frac{a}{1-f(a)})\to f(a)=1<1$ $\to f(a)\geq 1$ $\forall$ $a\in \mathbb{R^+}$ .
Let $m>n>0$ then:
$P(n,\frac{m-n}{f(n)}):f(n)f(\frac{m-n}{f(n)})=f(m)\geq f(n)\to f(m)\geq f(n) \text{ if $m>n$}\dots(1)$
Case 1: $\exists$ $c>0$ such that $f(c)=1$
$P(a,c):f(a)=f(a+c)$ Then $f$ is periodic. If $m>n>0$ $\exists$ $k\in\mathbb{Z^+}$ such that $n+kc>m>n$ , then $f(n)=f(n+kc)\geq f(m)\geq f(n)\to f(n)=f(m)$ $\forall$ $m>n$ , then as $f(c)=1\to f(x)=1$ $\forall$ $x\in \mathbb{R^+}$ .

Case 2: $\forall$ $c>0$ $f(c)>1$ :

If $f(m)=f(n)$ for some $m>n$ . Then by $(1)$ we have $f(\frac{m-n}{f(n)})=1$ that is not possible, then $f$ is inyective. Then in $P(a,b),P(b,a)$ we have:
$f(a)f(b)=f(a+bf(a))=f(b+af(b))\to a+bf(a)=b+af(b)\to \frac{f(a)-1}{a}=\frac{f(b)-1}{b}\text{ $\forall$ $a,b>0$}$ Then $f(x)=xc+1$ $\forall$ $x\in \mathbb{R^+}$ , $c>0$ is constant.

Then the solutions are $f(x)=xc+1$ $\forall$ $x\in \mathbb{R^+}$ , $c\geq 0$ is constant.

Now let's see that it works.
$P(a,b):(ac+1)(bc+1)=f(a)f(b)=f(a+bf(a))=f(abc+a+b)=abc^2+ac+bc+1=(ac+1)(bc+1)$

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TuZo

#5 h Jun 10, 2020, 2:06 PM • 2 Y

Y by a2048, Mango247

Solution:
1) We have $P(x,\frac y{f(x)})$ $\implies$ $f(x+y)=f(x)f(\frac y{f(x)})>f(x)$ so $f(x)$ is increasing, thus injective.
2) Swapping the $a,b$ , we get $f(a)f(b)=f(b + af(b))$ , so on the basis of the original equation we get $f(a + bf(a))=f(b + af(b))$ and on the basis of the injectivity we get $a + bf(a)=b + af(b)$ i.e. $\frac{f(a)-1}{a}=\frac{f(b)-1}{b}$ , and fixing the $b$ , we get $f(a)=ka+1$ , and this fit to the equation.
Done.

This post has been edited 2 times. Last edited by TuZo, Jun 12, 2020, 5:13 AM
Reason: typo

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#6 h Mar 31, 2025, 8:42 PM

Y by

a2048 wrote:

Determine all functions $f : \mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that $f(a)f(b)=f(a + bf(a))$

We claim that the only solutions are of the form $\boxed{f(x)=cx+1}$ where $c\ge0$ is any constant. From now on, we look for solutions not of this form. Let $P(x,y)$ be the assertion $f(x)f(y)=f(x+yf(x))$ .

Suppose there were some $u$ with $f(u)<1$ , then $P\left(u,\frac u{1-f(u)}\right)$ gives us $f(u)=1$ , a contradiction. So $f(x)\ge1$ for all $x$ .
This leads to $f(x+yf(x))\ge f(x)$ , so $f(x+y)\ge f(x)$ and so $f$ is nondecreasing.

Suppose there were some $k$ with $f(k)=1$ , then taking $P(k,k)$ gives $f(2k)=1$ and by simple induction we obtain $f(2^nk)=1$ for $n\in\mathbb N$ . Since $f$ is nondecreasing, for any $k\le x\le2^nk$ we have $f(k)\le f(x)\le f(2^nk)$ so $f(x)=1$ . Taking $n\to\infty$ we see that $f(x)=1$ for all $x\ge k$ .
Now fix some $y$ and take some $x\ge k$ . We have $x+yf(x)\ge k$ as well, so $f(x)=f(x+yf(x))=1$ and therefore $f(y)=1$ as well (by $P(x,y)$ ). This gives the solution $f(x)=1$ for all $x$ , which is of the form $cx+1$ , so no such $k$ can exist.

So we have $f(x)>1$ for all $x$ , which leads to $f(x+yf(x))>f(x)$ , so $f(x+y)>f(x)$ and so $f$ is strictly increasing.
Therefore $f$ is injective, and swapping $x,y$ in the equation and comparing, we have finally $f(x+yf(x))=f(y+xf(y))$ which implies $x+yf(x)=y+xf(y)$ . This rearranges to $\frac{f(x)-1}x=\frac{f(y)-1}y$ , so $\frac{f(x)-1}x$ is constant and we will find no additional solutions beyond the ones previously described.

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