Plain functional equation over positive reals

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Plain functional equation over positive reals

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Source: 2021 Thailand MO P7

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#1 h Nov 29, 2021, 2:46 AM • 1 Y

Y by megarnie

Determine all functions $f : \mathbb R^+ \to \mathbb R$ that satisfy the equation
$f(xy) = f(x)f(y)f(x+y)$ for all positive real numbers $x$ and $y$ .

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#2 h Nov 29, 2021, 3:01 AM

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Let $P(x,y)$ denote the assertion.

$P(1,1): f(1)=f(1)^2f(2)\implies f(1)=0\text { or }f(1)f(2)=1$ .

$P(2,2): f(4)=f(2)^2f(4)\implies f(4)=0\text { or }f(2)=\pm 1$ .

Suppose $f(k)=0$ for some positive real $k$ .
$P(k,x): f(kx)=0\implies \boxed{f\equiv0}$ , which works.

Henceforth assume no $k$ satisfies $f(k)=0$ .

Then $f(1)f(2)=1$ and $f(2)=\pm 1$ .

Case 1: $f(2)=1$ .
Then $f(1)=1$ .

$P(1,x): f(x)=f(x)f(x+1)\implies f(x+1)=1$ .

So the only solution here is $\boxed{f\equiv1}$ , which works.

Case 2: $f(2)=-1$ .
Then $f(1)=-1$ .

$P(1,x): f(x)=-f(x)f(x+1)\implies f(x+1)=-1$ .

So the only solution here is $\boxed{f\equiv-1}$ , which works.

Edit: I made a little mistake, it’s only true for x>1, but

$P(x,\frac{1}{x})$ gives f is constant

Edit: Thanks Jupiter123

This post has been edited 6 times. Last edited by megarnie, Jun 18, 2022, 8:15 PM

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#3 h Nov 29, 2021, 3:07 AM • 1 Y

Y by megarnie

Case 1: $f(k)=0$ for some $k$
$P\left(\frac xk,k\right)\Rightarrow\boxed{f(x)=0}$ which is a solution.

Case 2: $f(x)\ne0\forall x$
Let $c=\frac1{f(1)}$ which is defined.
$P(x,1)\Rightarrow f(x+1)=c\Rightarrow f(x)=c\forall x>1$
Let $x\le1$ now. Then $x+\frac1x+1\ge\frac1x+1>1$ , so:
$P\left(x,\frac1x+1\right)\Rightarrow f(x)=c$
Testing solutions of the form $f(x)=c$ , we have the two new solutions $\boxed{f(x)=-1}$ and $\boxed{f(x)=1}$ .

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#4 h Nov 29, 2021, 4:26 AM • 1 Y

Y by megarnie

megarnie wrote:

Case 1: $f(2)=1$ .
Then $f(1)=1$ .

$P(1,x): f(x)=f(x)f(x+1)\implies f(x+1)=1$ .

So the only solution here is $\boxed{f\equiv1}$ , which works.

Case 2: $f(2)=-1$ .
Then $f(1)=-1$ .

$P(1,x): f(x)=-f(x)f(x+1)\implies f(x+1)=-1$ .

So the only solution here is $\boxed{f\equiv-1}$ , which works.

True for all $x > 1$ . Shouldn't you also prove it for $0 < x \leq 1$ ?

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#5 h Nov 29, 2021, 10:05 PM

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Oh shoot I’ll edit when I get home

P(x,1/x) should work

This post has been edited 1 time. Last edited by megarnie, Nov 29, 2021, 10:07 PM

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#6 h Nov 29, 2021, 11:05 PM • 1 Y

Y by megarnie

MarkBcc168 wrote:

Determine all functions $f : \mathbb R^+ \to \mathbb R$ that satisfy the equation
$f(xy) = f(x)f(y)f(x+y)$ for all positive real numbers $x$ and $y$ .

Let $P (x, y)$ be the statement $f (xy) = f (x) f (y) f (x + y)$
we do $P (1, y)$ :
$f (x) = f (1) f (x) f (x + 1)$
Now we have $2$ cases:
Case 1: $f (x) = 0$
then $P (x, y)$ holds for all x and y, therefore $f (x) = 0$ is a solution.
Case 2: $f (x) \not 0$
then $1 = f (1) f (x + 1)$ , and we do $P (x + 1, 1)$ :
$f (x + 1) = f (1) f (x + 1) f (x + 2)$
$f (x + 1) = f (x + 2)$ , we replace and we have that $1 = f (1) f (x + 2)$ , then by induction we obtain that $1 = f (1) f (n) \forall n$
we make $n = 1$ , and then we have $f (1) ^ 2 = 1$ , so $f (1) = 1$ or $f (1) = - 1$ , this implies that $f (x) = - 1$ or $f (x) = 1$ , so in this case the solutions are $f (x) = 1$ or $f (x)= -1$

This post has been edited 1 time. Last edited by e61442289, Nov 29, 2021, 11:10 PM

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puntre

#7 h Dec 2, 2021, 2:17 AM

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The answers are $f\equiv0$ , $f\equiv1$ and $f\equiv-1$ which is not hard to see that they work.

Let $P(x,y)$ be the functional equation.
If $f(x_0)=0$ for some $x_0>0$ , then $P(\frac{x}{x_0},x_0)$ gives $f(x)=0$ for each $x\in\mathbb{R^+}$ .
So, the rest is to prove that if $f(x)\neq0$ for each $x$ , then $f\equiv1$ or $f\equiv-1$ .
$P(2,2)$ , gives $f(2)^2=1$ . Plug $y=1$ yields $f(x)=\frac{1}{f(1)}$ for all $x>1$ .
So $f(x)$ is constant for each $x\geq 1$ . assume $x_1\geq 1$ , $P(x_1,\frac{1}{x_1})$ gives $f(1)=f(\frac{1}{x_1})$ .
So $f$ is a constant function implies $f(x)^2=1\implies f\equiv 1$ or $f\equiv-1$ as desired.

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#8 h Dec 4, 2021, 5:40 PM

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By setting y = 1 we get f(x) = f(x)f(1)f(x+1). If there exists such x that f(x) = 0 then for any possitive real y f(xy) = 0, thus f(x) = 0 for any positive real x is our first solution. If there does not exist such x we can cancel out f(x) on both sides and divide our equation by f(1). We get f(x+1) = 1/f(1). Thus f(x) = c, where c satisfies c = c^3. Polynomial c^3 - c has exactly 3 roots which are -1,0,1, hence f(x) = -1 or 0 or 1 for any positive real x. Sorry for not using latex, I'm new, so I can't

This post has been edited 2 times. Last edited by MR_D33R, Dec 4, 2021, 5:42 PM

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#9 h Jun 18, 2022, 7:41 AM

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If $f$ is zero at any point then it is zero everywhere. If not then setting $y=1$ implies $f(x)=1/f(1)=k$ for all $x>1.$ But $y=1/x$ implies $f$ is constant everywhere whence testing we get $k\in \{1,-1\}.$ All of them work.

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#10 h Jun 18, 2022, 8:14 PM • 1 Y

Y by Taco12

Solved with the AoPS user @Taco12.

We claim the only functions that satisfy the given equation are $f(x) = 0$ , $f(x) = 1$ , and $f(x) = -1$ .

Let $P(x,y)$ denote the assertion.

First, notice that $f(x) = 0$ is a solution.

$P(x, 1)$ : $f(x) = f(x)f(1)f(x+1) \Rightarrow f(1)f(x+1) = 1$ . (We can do this since we already acknowledged $f(x) = 0$ is a solution.)
$P(1, 1)$ : $f(1) = f(1)^2 \cdot f(2)$ . Notice that we can divide $f(1)$ from both sides as we know it is not equal to $0$ because then $f(1)f(x+1) = 1$ would not hold true. So, $f(1)f(2) = 1$ .

Substituting, we have $f(x+1) = f(2) = c$ . Hence, $f(x) = c$ , where $c$ is a fixed real number.

So, we see that $c = c^3 \Rightarrow c^3 - c = 0 \Rightarrow c(c^2-1) = 0 \Rightarrow c = 0, \pm 1$ , and we are done. $\blacksquare$

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#11 h Jun 19, 2022, 2:26 AM

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@above I believe you miss solutions in the interval $(0,1]$

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