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Find all the functions in $\mathbb{R}$, satisfying the given equation $(x+y)(f(x)-f(y)) = (x-y)f(x+y)$ for all $x,y$ in $\mathbb{R}$.

I tried to find something like a pattern that would become a constant. If I simplify, I get $xf(x)-xf(x+y)-xf(y)=yf(y)-yf(x+y)-yf(x)$ What should I do now? Is this going to help? Thank you.

Sangchul Lee
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Avijit Mukherjee
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    $\frac{f(x+y)}{(x+y)} = \frac{f(x)-f(y)}{x-y}$ now if $f$ is continuous $f'(x) = \frac{f(2x)}{2x}$ – Cesareo Oct 06 '18 at 11:22
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    All polynomial solutions take the form $f(x)=ax^2+bx$, where $a$ and $b$ are fixed real constants. I do not know if there are other solutions. – Batominovski Oct 06 '18 at 16:12
  • From Cesareo's observation, one could compare series expansions of the left and right hand sides to get that non-zero coefficients appear only in degrees $1$ and $2$. Thus, all analytical solutions are of the form $f(x) = ax^2 + bx$. – Ennar Oct 07 '18 at 11:37

2 Answers2

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For $x,y\in\mathbb{R}$, let $P(x,y)$ denote the condition $$(x+y)\,\big(f(x)-f(y)\big)=(x-y)\,f(x+y)\,.$$ From $P(1,0)$, we get $$f(1)-f(0)=f(1)\text{ so that }f(0)=0\,.$$ With $P\left(\dfrac{x+y}{2},\dfrac{x-y}{2}\right)$, we obtain $$x\,\Biggl(f\left(\frac{x+y}{2}\right)-f\left(\frac{x-y}{2}\right)\Biggr)=y\,f(x)\,.\tag{*}$$ This shows that $$f\left(\frac{1+y}{2}\right)-f\left(\frac{1-y}{2}\right)=ay\,,$$ where $a:=f(1)$. Plugging in $2$ for $x$ and $y+1$ for $y$ in (*) yields $$f\left(\frac{3+y}{2}\right)-f\left(\frac{1-y}{2}\right)=b\left(\frac{1+y}{2}\right)\,,$$ where $b:=f(2)$. Thus, subtracting the two equations above gives us $$f\left(\frac{3+y}{2}\right)-f\left(\frac{1+y}{2}\right)=b\left(\frac{1+y}{2}\right)-ay$$ for all $y\in\mathbb{R}$. In other words, $$f(t)-f(t-1)=b(t-1)-a(2t-3)=(b-2a)t+(3a-b)\,.\tag{#}$$

Now, we look at $P(x,x-1)$. This gives $$(2x-1)\,\big(f(x)-f(x-1)\big)=f(2x-1)\,.$$ Thus, from (#), we obtain $$f(2x-1)=(2x-1)\,\big((b-2a)x+(3a-b)\big)\,.$$ That is, $$f(t)=t\,\Biggl((b-2a)\left(\frac{t+1}{2}\right)+(3a-b)\Biggr)=\frac{b-2a}{2}t^2+\frac{4a-b}{2}t$$ for every $t\in\mathbb{R}$. It is easy to verify that every quadratic function in the form above satisfies the functional equation.

Batominovski
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The equation from Cesaro requires "differentiable".

A class of solutions is f(x) = ax

Sam Ginrich
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    I do not understand how you made this conclusion. There are other differentiable solutions. Every quadratic function $f(x)=ax^2+bx$ for all $x\in\mathbb{R}$ (where $a$ and $b$ are fixed constants) is a solution. – Batominovski Oct 06 '18 at 16:10