# Find $f$ such that $f(a-b)+f(c-d)=f(a)+f(b+c)+f(d)$

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.

• @AhmedS.Attaalla If $c=0$, $ab=0$. Thus $a=0$ or $b=0$. – induction601 Mar 1 at 7:11
• When $a=b$, we have $a^2 = -(a+d)c$. By letting $c=a$ and $d=-2a$, we have $\frac{f(3a)-f(a)}{2} = f(2a)$. – induction601 Mar 1 at 7:12
• 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 $f(5b) = f(4b) + f(3b)$. – induction601 Mar 1 at 7:36
• $\sf{f(a-b)+f(a+b)=f(2a)+f(2b)}$ – TheSimpliFire Mar 1 at 8:05
• $f(x) = kx^2$ satisfies the given conditions for any $k\ge 0$. – FredH Mar 1 at 11:32

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