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Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ so that:

a) $[f(x)+f(y)][f(x+2y)+f(y)]=[f(x+y)]^2+f(2y)f(y)$

b) for every real $a>b\ge 0$ we have $f(a)>f(b)$

As much as I know:

  • putting $x=y=0$ we get $f(0)=0$.

  • let $x+y=0$, the we get $f(y)+f(-y)=\frac{f(2y)}{2}$

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Step 1: Prove : $f(y) = 0$ iff. $y=0$

It's easy to see $f(y)>0$ for $y>0$ (from porperty (b) with value $a=y,b=0$)

Following proof is given by @Kyson

To prove that $f(y)≠0$ for all$ y<0.$

Assume there is a $y<0$ such that $f(y)=0.$

Choose $x=−2y>0$ and hence $f(x)>0$ and $f(x+y)=f(−y)>0 $Then $[f(x)+f(y)][f(x+2y)+f(y)]=0$ while $f(x+y)2+f(y)f(2y)>0.$

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Step 2: Prove $f(y)$ is even.

Put $x+y=0$ in equation given to get:$f(y)+f(-y)=\frac{f(2y)}{2}$ for all y. $f(y)+f(-y)=\frac{f(2y)}{2} = f(-y) +f(y) = \frac{f(-2y)}{2} $ Thus $f(x) $is a even function, so $f(y)+f(-y)=2f(y)=\frac{f(2y)}{2} \Rightarrow f(2y)=4f(y) $ (Since $f(x)$ is even ,property (b) actually shows that $f(y) > 0$ on $(-\infty,0)\cup(0,\infty)$)

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Step 3: Prove:$ f(y)=y^2f(1)$

By induction , find $f(ny)=n^2f(y)$ where $n \in \mathbb N$

Details of induction: Suppose $m\in \mathbb N ,m\ge 2 ,\ f(ny)=n^2f(y)$ holds for all $n\le m$
Put $x=(m-1)y$ in property (a).$\\ \Rightarrow ((m-1)^2+1)f(y)f((m+1)y)=(m^4+4-(m-1)^2-1)f^2(y) \\ \Rightarrow f((m+1)y)=(m+1)^2f(y) \ for\ y\not = 0 \ and \ it\ still\ holds\ for\ y=0$

Hint: $(m+1)^2 ((m-1)^2 +1) = (m^2-1)^2+(m+1)^2=m^4 -m^2+2 =m^4+4-(m-1)^2-1$

Remark: Result induces $f(zy)=z^2f(y)$ where $z\in \mathbb Z$ , since $f$ is even.

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$\forall r\in \mathbb Q , \ r = \frac ab \ , \ a,b\in \mathbb Z$ . Then $f(ry)=f(\frac ab y)=a^2f(\frac 1b y)$ and $f(y)=b^2f(\frac 1b y)$

Hence $f(ry)=f(\frac ab y)=(\frac ab)^2f(y)=r^2f(y)\Rightarrow f(r)=r^2f(1)$

As @MarkBennet points out:

the function is strictly increasing for $r≥0$ and the rationals are dense in the reals.

And $f(r)$ is even, so it's strictly decreasing for $r\le 0$.

Conclude $f(x)=x^2f(1)$ where $x\in \mathbb R $

Thus $f(x)$ can only be the form such as $ax^2$ where $a\in \mathbb R^+$

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  • $\begingroup$ You can get $f(r)=r^2f(1)$ for rationals, and then use the fact that the function is strictly increasing for $r\ge 0$ and the rationals are dense in the reals to conclude. But the final equation in the original post involves dividing by $f(y)$ and this could be zero for some negative $y$, in which case that equation need not hold (without more work). $\endgroup$ – Mark Bennet Jan 24 '15 at 15:10
  • $\begingroup$ @MarkBennet thank you! I will complete proof later. $\endgroup$ – Syuizen Jan 24 '15 at 15:15
  • $\begingroup$ Can you please show the induction part and why it also holds for rational beacuse it is a key part in solving the problem. $\endgroup$ – HeatTheIce Jan 24 '15 at 19:58
  • $\begingroup$ @SoulEater I edit my answer. Show all details of the proof and the ideas. $\endgroup$ – Syuizen Jan 25 '15 at 2:29
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This may help to get you to the next step.

From the last equation, put $z=-y$ to obtain $\frac {f(-2z)}2=f(z)+f(-z)=\frac {f(2z)}2$. This means that generally $f(x)=f(-x)$ so that, again from the last equation, $f(2y)=4f(y)$.

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  • $\begingroup$ Let $x+y=0$, we get $f(y)=0$ or $f(y)+f(-y)=\frac{f(2y)}{2}$. Since $f(y)>0$ for all $y>0$, we have $f(y)+f(-y)=\frac{f(2y)}{2}$ for all $y\ge0$. I wonder whether we can substitute $z=-y$. $\endgroup$ – velut luna Jan 24 '15 at 14:41
  • $\begingroup$ $f(x)=ax^2$, where a is a real parameter, is a solution. Maybe the only one. $\endgroup$ – HeatTheIce Jan 24 '15 at 14:45
  • $\begingroup$ @Kyson Yes, there is more work needed to establish the equation for $x+y=0$ because it may involve dividing through by zero if $f(y)=0$ for some $y\lt 0$. It holds when $f(y)\neq 0$ even if $y$ is negative. $\endgroup$ – Mark Bennet Jan 24 '15 at 15:00
  • $\begingroup$ It cant be zero bcs f is not a constant function (it is a strictly incrising one). $\endgroup$ – HeatTheIce Jan 24 '15 at 15:02
  • $\begingroup$ Putting $x=y$ you get $f(3x)=9f(x)$. $\endgroup$ – HeatTheIce Jan 24 '15 at 15:18
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This may help Mark Bennet's proof a little bit.

To prove that $f(y)\ne0$ for all $y<0$.

Assume there is a $y<0$ such that $f(y)=0$.

Choose $x=-2y>0$ and hence $f(x)>0$ and $f(x+y)=f(-y)>0$

Then $[f(x)+f(y)][f(x+2y)+f(y)]=0$ while $f(x+y)^2+f(y)f(2y)>0$.

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  • $\begingroup$ Nicely reasoned! $\endgroup$ – Mark Bennet Jan 24 '15 at 15:25

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