Hong Kong 2020 TST 2 Q4

Art of Problem Solving

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32 posts

#1 h Oct 26, 2019, 10:41 AM • 6 Y

Y by itslumi, HWenslawski, Hopeooooo, tiendung2006, Adventure10, ItsBesi

Find all real-valued functions $f$ defined on the set of real numbers such that $f(f(x)+y)+f(x+f(y))=2f(xf(y))$ for any real numbers $x$ and $y$ .

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#3 h Oct 26, 2019, 1:36 PM • 5 Y

Y by pco, SMSGodslayer, HWenslawski, tiendung2006, Adventure10

Notice that any constant function satisfies the condition. Let's look for non-constant solutions.

Changing $x$ with $y$ we get that $f(f(y)+x)+f(f(x)+y)=2f(yf(x))$ and by the condition we get that $2f(xf(y))=2f(yf(x))\Leftrightarrow f(xf(y))=f(yf(x))$ (*), for every real $x$ and $y$ .

Setting $x=0$ to (*) we get that $f(yf(0))=f(0)$ , implying that if $f(0)\ne 0$ , then $f(x)=f(0)=c$ , for every real $x$ , that is to say $f$ is constant.

So we focus on $f(0)=0$ .

Setting to the original $x=0$ and we obtain that $f(y)+f(f(y))=0$ (**), for every real $y$ .

Working on the above we get that $f(f(y))=-f(y)\Rightarrow f(f(f(y)))=f(-f(y))$ and using the above again $f(y)=-f(f(y))=f(f(f(y)))=f(-f(y))\Rightarrow f(-f(y))=f(y)$ (***), for every real $y$ .

Now set to (*) $y=1$ to get that $f(xf(1))=f(f(x))=-f(x)$ (1), by (**), for every real $x$ .

Note that (1) for $x=-1$ is $f(-f(1))=-f(-1)$ and using (***) $f(1)=-f(-1)$

Setting to (*) $y=-1$ and using the above and (***) we obtain that $f(xf(-1))=f(-f(x))=f(x)$ or $f(-xf(1))=f(x)\Leftrightarrow f(xf(1))=f(-x)$ , for every real $x$ (2).

Using (1) and (2) we get that $f(-x)=-f(x)$ for every real $x$ , that is to say $f$ is odd.

Setting to the original $y=x$ we get that $2f(x+f(x))=2f(xf(x))$ or $f(x+f(x))=f(xf(x))$ . Now replacing $x$ with $-x$ we get that $f(-x+f(-x))=f(-xf(-x))$ and using that $f$ is odd we eventually get that $-f(x+f(x))=f(-x-f(x))=f(-x+f(-x))=f(-xf(-x))=f(xf(x))=f(x+f(x))$ .

Consequently $f(x+f(x))=f(xf(x))=0$ , for every real $x$ .

Setting $x=1$ we get that $f(f(1))=0$ or by (**) $-f(1)=0\Rightarrow f(1)=0$ . Now (1) implies that $f(x)=0$ , for every real $x$ .

To sum up the only solution is $f(x)=c$ , for every real $x$ .

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23439 posts

pco

#4 h Oct 26, 2019, 3:56 PM • 5 Y

Y by skrublord420, abdirasulov, Wizard0001, Adventure10, Mango247

Blastzit wrote:

Find all real-valued functions $f$ defined on the set of real numbers such that $f(f(x)+y)+f(x+f(y))=2f(xf(y))$ for any real numbers $x$ and $y$ .

The same proof than above can be shortened (skipping proof that $f(x)$ is odd, for example) a bit ::
Let $P(x,y)$ be the assertion $f(f(x)+y)+f(x+f(y))=2f(xf(y))$
Let $a=f(0)$ and $c=f(1)$

Subtracting $P(x,y)$ from $P(y,x)$ , we get
new assertion $Q(x,y)$ : $f(xf(y)=f(yf(x))$

1) If $a\ne 0$ : $Q(\frac xa,0)$ $\implies$
$\boxed{\text{S1 : }f(x)=a\quad\forall x}$ which indeed is a solution, whatever is $a\in\mathbb R$

2) If $a=0$
$P(x,0)$ $\implies$ $f(f(x)=-f(x)$ and so :
a) $f(c)=f(f(1))=-f(1)=-c$
b) $f(f(c))=-f(c)=c$
c) $f(-c)=f(f(c))=c$

Subtracting $P(c,c)$ from $P(-c,1)$ , we get $f(c+1)=0$
Then $P(1,1)$ $\implies$ $c=0$
And $Q(x,1)$ $\implies$ $f(f(x))=0$ and so
$f(x)=-f(f(x))=0\quad\forall x$ , already included in solution found in 1) above

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#5 h Jan 1, 2020, 2:34 PM • 1 Y

Y by Adventure10

is this right f(x+f(x))=f(xf(x))
x=f(y) , f(f(y)+ff(y))=0=f(-f(y)^2)
Also we know if f(u)=0 then u can only be 0
So f(y)=0 for all reals( This is not the complete solution I what to know whether it's correct pls help)

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23439 posts

pco

#6 h Jan 1, 2020, 2:56 PM • 1 Y

Y by Adventure10

NOOB1729 wrote:

Also we know if f(u)=0 then u can only be 0

How ?

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#8 h Jan 1, 2020, 3:48 PM • 2 Y

Y by Adventure10, Mango247

Oh ok I got my mistake

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1107 posts

#9 h Jan 1, 2020, 5:29 PM • 2 Y

Y by Adventure10, Mango247

Blastzit wrote:

Find all real-valued functions $f$ defined on the set of real numbers such that $f(f(x)+y)+f(x+f(y))=2f(xf(y))$ for any real numbers $x$ and $y$ .

Clearly any $f\equiv \text{const}$ satisfies the equation.So now assume $f$ is non-constant.
$P(x,y)\implies f(f(x)+y)+f(x+f(y))=2f(xf(y))$ $P(y,x)\implies f(f(x)+y)+f(x+f(y))=2f(yf(x))$ Thus $f(xf(y))=f(yf(x))$ .Plugging $x=0\implies f(yf(0))=f(0)\implies f(0)=0$ since $f$ is non-constant.

$P(0,y)\implies f(f(y))=-f(y)$ .Now $P(0,f(y))\implies f(f(f(y)))=-f(f(y))$ ,but $f(f(f(y)))=f(-f(y))$ .Thus $f(-f(y))=-f(f(y))$ .But $-f(y)=f(f(y))=f(yf(1))$ .Plug $y=-f(1)$ to get $-f(1)=f(f(1))=-f(-f(1))=f(-f(1)^2)=0\implies f(1)=0$ .Now $f(f(y))=f(yf(1))=f(0)=0\implies -f(y)=0\implies f(y)=0$ as desired. $\square$
Whence we conclude that the only solution is $f\equiv \text{const}$

This post has been edited 2 times. Last edited by Pluto1708, Jan 1, 2020, 5:46 PM
Reason: Fixed @below

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pco

#10 h Jan 1, 2020, 5:41 PM • 1 Y

Y by Adventure10

Pluto1708 wrote:

Clearly if $f\equiv \text{const} \implies f\equiv 0$ .

Wrong.
Any constant function fits.

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114 posts

#11 h Apr 28, 2020, 1:10 PM

Y by

Redacted

This post has been edited 1 time. Last edited by stayhomedomath, Jul 15, 2021, 6:23 AM

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Functional_equation

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Functional_equation

#12 h Apr 28, 2020, 2:43 PM • 1 Y

Y by SMSGodslayer

$f(xf(y))=f(yf(x))$

Case 1:
$f(0)\not=0,x=\frac{t}{f(0)},y=0$
$f(\frac{t}{f(0)}*f(0))=f(0)=f(t)=c,c\not=0$

Case 2:
$f(0)=0$
$P(0,x)\implies f(f(x))+f(x)=0$

$P(f(x),y)\implies f(f(f(x))+y)+f(f(x)+f(y))=2f(f(x)f(y))$
$P(f(y),x)\implies f(f(f(y))+x)+f(f(y)+f(x))=2f(f(y)f(x))$
$Then$
$f(y+f(f(x)))=f(x+f(f(y)))$
$f(f(x))=-f(x)$

$f(y-f(x))=f(x-f(y))$
$y=f(x)$
$0=f(f(x)-f(x))=f(x-f(f(x)))=f(x+f(x))$
$Then$

$f(x+f(x))=0$

$P(x,x)\implies f(x+f(x))+f(x+f(x))=2f(xf(x))\implies f(xf(x))=0$

$f(xf(x))=0,x=1\implies f(f(1))=0=-f(1)$
$f(1)=0$

$f(xf(y))=f(yf(x)),y=1\implies 0=f(0)=f(f(x))$
$Then$
$0=f(f(x))=-f(x)\implies f(x)=0$

Solution

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464 posts

#13 h Mar 22, 2021, 6:25 AM • 3 Y

Y by Mango247, Mango247, Mango247

$\clubsuit \color{red}{\textit{\textbf{ANS:}}}$ $f(x)\equiv \textrm{const.} \quad \forall x\in \mathbb{R}$ .
$\spadesuit \color{blue}{\textit{\textbf{Proof:}}}$ It is easy to see that the above constant function is indeed a solution to the given FE. Let $P(x,y)$ be the given assertion, if $f(0)\ne 0$ , then we have $P\left(\frac{x}{f(0)},0\right) \quad \textrm{and} \quad P\left(0, \frac{x}{f(0)}\right): \boxed{f(x)=f(0)=c \quad \forall x\in \mathbb{R}, \textrm{where} \ c\in \mathbb{R}}.$ If $f(0)=0$ , then $P(x,0): f(f(x))=-f(x),$ $P(f(x),0): f(-f(x))=f(x)$ and $P(x,x): f(x+f(x))=f(xf(x)).$ Therefore, $0=f(0)=f(f(x)-f(x))=f(f(x)+f(f(x)))=f(f(x)f(f(x)))=f(-f(x)^2).$ This leads us to $P(-f(x),x): f(x+f(x))=f(f(-f(x))+x)=2f(-f(x)^2)=0.$ This means $f(1+f(1))=0$ and so $P(1,1): f(f(1))=0$ together with $0=f(f(1))=-f(1) \implies f(1)=0$ . Thus, $P(1,x) \quad \textrm{and} \quad P(x,1): f(f(x))=0$ which implies $0=f(f(x))=-f(x) \implies \boxed{f(x)=0}$ . $\quad \blacksquare$

This post has been edited 2 times. Last edited by Keith50, Mar 22, 2021, 6:31 AM

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#14 h Mar 24, 2021, 11:16 PM

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The only answer is $f \equiv c$ , where $c$ is an arbitrary real number.

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

From $P(x,x)$ , we have that $f(x+f(x))=f(xf(x))$ .
Thus giving us $f(f(0))=f(0)$ .
Now notice that $P(0,x)=P(x,0)$ , thus giving us $f(xf(0))=f(0)$ .

Now a case bash ensues.

The first case is when $f(0) \neq 0$ , this implies that we have that $xf(0)$ is surjective in the real numbers, thus giving us $f(t)=c$ , for every real $t$ .

The second case is when $f(0)=0$ .
From $P(0,x)$ , we get that $f(f(x))=-f(x)$ .
Thus the logical thing to do is to look at $P(x,f(t))$ and $P(f(t),x)$ .
Since they are both equal we have that $f(f(x)f(t))=f(-xf(t))=f(-tf(x))$ .
Thus from $f(-xf(t))=f(-tf(x))$ we must have that $f(t-f(x))=f(x-f(t))$ . for every $x$ and $t$ .
This implies that $f(t+f(t))=0$ , implying that $f(tf(t))=0$ .
This implies that $f(f(1))=0$ , which implies that $f(1)=0$ .
Since we hav ethat $P(x,1)=P(1,x)$ , then we must have that $f(f(x))=0$ , which impluies that $-f(x)=0$ , which gives us that $f(x)=0$ .

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#15 h Apr 22, 2021, 3:31 AM

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Let $P(x,y)$ be the assertion $f(x+f(y))+f(y+f(x))=2f(xf(y))$ .
All solutions of the form $\boxed{f(x)=c},c\in\mathbb R$ work. Assume now that $f$ is nonconstant.

$\textbf{1. }Q(x,y):f(xf(y))=f(yf(x))$
$P(x,y)-P(y,x)$ . $\blacksquare$

$\textbf{2. }f(0)=0$
If $f(0)\ne0$ then $Q\left(\frac x{f(0)},0\right)\Rightarrow f(x)=f(0)$ , contradiction

$\textbf{3. }f(\pm f(x))=\mp f(x)$
$P(x,0)\Rightarrow f(f(x))=-f(x)$
In addition, $f(-f(x))=f(f(f(x)))=-f(f(x))=f(x)$ .

$\textbf{4. }f(1)=0$
$P(x,-f(x))\Rightarrow f(x+f(x))=2f(xf(x))$
$P(x,x)\Rightarrow 2f(x+f(x))=2f(xf(x))$
So $f(x+f(x))=0$ , thus $f(xf(x))=0$ . In particular, $x=1$ yields $0=f(f(1))=-f(1)$ , so $f(1)=0$ .

$Q(x,1)\Rightarrow f(f(x))=0$
So $-f(x)=f(f(x))=0$ , thus $f(x)=0$ , which is constant.

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#16 h Jul 29, 2021, 2:02 PM

Y by

I found another, somewhat miraculous solution.

Denote by $P(x,y)$ the assertion $f(f(x)+y)+f(x+f(y))=2f(xf(y))$ , and by $Q(x,y)$ the assertion $f(xf(y))=f(yf(x))$ , which follows from comparing $P(x,y)$ and $P(y,x)$ . As in the other solutions, $Q(x,0)$ shows that either $f$ is constant (indeed a solution), or $f(0)=0$ . So WLOG $f(0)=0$ .
Then
$P(x,0): f(f(x))+f(x)=0\implies f(f(x))=-f(x)$ and thus using $Q(x,1)$ also $f(f(1)x)=f(f(x))=-f(x)$ for all $x$ . Therefore
$P(f(x)f(1),1): f(\underbrace{f(f(x)f(1))}_{=-f(f(x))=f(x)}+1)+\underbrace{f(f(x)f(1)+f(1))}_{=-f(f(x)+1)}=\underbrace{2f(f(1)^2f(x))}_{=-2f(f(1)f(x))=2f(f(x))=-2f(x)}$ $\implies 0=f(f(x)+1)-f(f(x)+1)=-2f(x).$ So we obtain $f\equiv 0$ , which is indeed a solution.

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gghx

#17 h Oct 19, 2021, 1:18 AM

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$P(x,y): f(f(x)+y)+f(x+f(y))=2f(xf(y))$ . All constant solutions work, we assume $f$ is non-constant.
Exchanging $x$ and $y$ , we get $f(xf(y))=f(yf(x))$ .
Thus, $f(f(x))=f(xf(1))$ and $f(-f(x))=f(xf(-1))$ .
Furthermore, $f(xf(0))=f(0)$ . Since non-constant, $f(0)=0$ .
$P(x,0): f(f(x))+f(x)=0$ .
Hence $-f(x)=f(f(x))=f(xf(1))$ .
$P(f(1),f(1)): f(f(1)f(f(1))=0$ .
But $f(f(1)f(f(1))=-f(f(f(1))=-f(-f(1))=f(-1)$ , hence $f(-1)=0$ .
Since $f(-f(x))=f(xf(-1)$ , $f(-f(x))=0$ .
$P(x,-f(x)): f(x)=0$ .

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#18 h May 20, 2022, 1:08 PM

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I claim that all constant functions work.
Let $P(x,y)$ be our assertion
$P(0,0)\rightarrow f(f(0)=f(0)$
$P(x,0)\rightarrow f(f(x))+f(0)=2f(xf(0))$
$P(0,x)\rightarrow f(f(0)+x)+f(f(x))=2f(0)$
$P(y,x)-P(x,y) \implies f(xf(y))=f(yf(x))$
Now plug $(\frac{x}{f(0)},f(0))$ in the last equation and assume $f(0) \neq 0$
Then we have $f(x)=f(0)=c$
If $f(0)=0$
Then plug this value in the 2nd equation
we get $f(f(x))=-f(x)$ and from the first equation we have $f(x)=0 \ \forall x \in \mathbb{R}$ and thus we conclude our proof.

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ZETA_in_olympiad

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ZETA_in_olympiad

#19 h Sep 23, 2022, 1:04 PM

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Let $P(x,y)$ denote the assertion $f(f(x)+y)+f(x+f(y))=2f(xf(y)).$ Comparing $P(x,0)$ and $P(0,x)$ implies $f(xf(0))=f(0).$ So we have solution as constant function, which works. Or we have $f(0)=0.$

$P(x,0)$ implies $-f(x)=f(f(x))=f(xf(1))$ and also $f(-f(x))=f(x).$ $P(f(1),f(1))$ gives $f(-f(1)^2)=0.$ Now, $-f(-f(1))=-f(1)=f(-f(1)\cdot f(1))=0,$ so $f(1)=0.$ Comparing $P(x,1)$ and $P(1,x)$ implies $0=f(f(x))=-f(x).$ So $f$ is zero function, which works.

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ZETA_in_olympiad

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ZETA_in_olympiad

#20 h Sep 23, 2022, 1:10 PM

Y by

Fakesolver19 wrote:

$P(x,0)\rightarrow f(f(x))+f(0)=2f(xf(0))$

No, it should be $f(f(x))+f(x+f(0))=2f(xf(0)).$

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