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Denote the set of non-negative real numbers by $\mathbb R^+_0$. Find all functions $f:\mathbb R \rightarrow \mathbb R_0^+$ s.t. $\forall a,b,c,d\in\mathbb R$ satisfying $ab+bc+cd=0$ we have $$f(a-b)+f(c-d)=f(a)+f(b+c)+f(d)$$

My attempt:

set $b=d=0$ and get $f(0)=0$.

Now set $a=b=c=0$ to get $f(-d)=f(d)$.

Now we pretty much reduced this fe's domain to $\mathbb R^+_0\rightarrow\mathbb R^+_0$.

That's where I got stuck. I tried a few things but none of them gave me any progress. Any help/hints appreciated.

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  • $\begingroup$ @AhmedS.Attaalla If $c=0$, $ab=0$. Thus $a=0$ or $b=0$. $\endgroup$ – induction601 Mar 1 at 7:11
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    $\begingroup$ When $a=b$, we have $a^2 = -(a+d)c$. By letting $c=a$ and $d=-2a$, we have $\frac{f(3a)-f(a)}{2} = f(2a)$. $\endgroup$ – induction601 Mar 1 at 7:12
  • $\begingroup$ When $b=d$, we have $b(a+2c)=0$. By letting $a=-2c$, we have $f(2c+b) + f(c-b) = f(2c) + f(c+b) + f(b)$. Let $c=2b$. Then this gives us $f(5b) + f(b) = f(4b) + f(3b) + f(b)$. Thus we have $f(5b) = f(4b) + f(3b)$. $\endgroup$ – induction601 Mar 1 at 7:36
  • $\begingroup$ $\sf{f(a-b)+f(a+b)=f(2a)+f(2b)}$ $\endgroup$ – TheSimpliFire Mar 1 at 8:05
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    $\begingroup$ $f(x) = kx^2$ satisfies the given conditions for any $k\ge 0$. $\endgroup$ – FredH Mar 1 at 11:32
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Let $c \neq 0$. To remove our condition, we write $d$ in terms of the rest of the variables : $$d=\frac{-ab-bc}{c}=-b-\frac{ab}{c}$$ Now, we can substitute this in our functional equation : $$f(a-b)+f(b+c+\frac{ab}{c})=f(a)+f(b+c)+f(-b-\frac{ab}{c})$$ For $c=-b$, we have : $$f(a-b)+f(-a)=f(a)+f(0)+f(a-b) \implies f(0)=f(-a)-f(a)$$ Replacing $a$ by $-a$ : $$f(0)=f(a)-f(-a) \implies f(0)=0 \implies f(a)=f(-a) \space \forall \space a \in \mathbb{R}$$ Next, for $a=b=c$ : $$f(0)+f(3a)=f(2a)+f(2a)+f(a) \implies f(3a)=2f(2a)+f(a) \space \forall \space a \in \mathbb{R}$$ For $b=-nc=-na$ : $$f((n+1)a)+f((2n-1)a)=f(a)+f((n-1)a)+f(2na)$$ At $n=2$, this yields: $$f(4a)=2f(3a)-2f(a)=4f(2a) \implies f(2a)=4f(a) \space; f(3a)=9f(a)\space; f(4a)=16f(a)$$ For all naturals $n \leqslant 4$, we see that $f(na)=n^2 \cdot f(a)$. We continue with induction hypothesis. Let $f(na)=n^2 \cdot f(a) \space \forall \space n<2n-1$. We can see $f(2na)=4f(na)$. Observe the equation: $$f((n+1)a)+f((2n-1)a)=f(a)+f((n-1)a)+4f(na)$$ All coefficients of $a$ inside the function are less than or equal to $2n-1$, and we can solve this equation for $f((2n-1)a)$ to yield one solution in terms of $f(a)$. We can see that in our functional equation, $f(ta)=t^2 \cdot f(a)$ satisfies our condition. Since there is only one solution in terms of $f(a)$, it must be $f(2n-1)=(2n-1)^2 \cdot f(a)$. Thus, induction hypothesis yields us the equation $f(na)=n^2 \cdot f(a) \space \forall \space n \in \mathbb{N}$

If $f$ is continuous, we conclude that the only solution to this functional equation is $f(x)=kx^2$ for some real number $x \geqslant 0$. If not, $k$ may vary (or not) for linearly independent $n$. I am unaware of how to tackle this part of the problem.

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  • $\begingroup$ Don't you need some sort of continuity to rule out examples like $f(x) = \begin{cases} k_1x^2 : x \in \mathbb Q \\ k_2x^2 : \text{else} \end{cases}, k_1 \neq k_2$? $\endgroup$ – eccheng Mar 2 at 8:02

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