# $xf(x) - yf(y) = (x-y)f(x+y)$

Determine all $$f : \mathbb R \rightarrow \mathbb R$$ that satisfies

$$xf(x) - yf(y) = (x-y)f(x+y)$$ ; $$\forall x , y \in \mathbb R$$

I tried reform the equation to

$$f(x) + f(-x) = 2f(0)$$ and $$f(x) + f(3x) = 2f(2x)$$. Also, $$f$$ is injective if $$f$$ isn’t constant.

Can anyone give me some hints please. Thank you very much!

• One observation is that if $f$ has these properties, then so does $f+c$ for any constant $c$. So you can safely assume $f(0)=0$. – Qiyu Wen May 17 '20 at 5:43
• take x = y+1, then try to find derivative of f – Sagar Chand May 17 '20 at 5:48

Write the given functional equation for the three pairs $$(x,y)$$, $$(y,z)$$, $$(z,x)$$, and add up. You then obtain $$0=(x-y)f(x+y)+(y-z)f(y+z)+(z-x)f(z+x)\ .$$ We now have a functional equation with three free variables. Put $$x:={t+1\over2},\quad y:={t-1\over2},\quad z:={1-t\over2}\ ,$$ and you get $$0=1\cdot f(t)+(t-1)\cdot f(0)-t\cdot f(1)\ ,$$ or $$f(t)=f(0)+t\bigl(f(1)-f(0)\bigr)\qquad\forall t\in{\mathbb R}\ .$$ This shows that $$f$$ has to be of the form $$f(t)=at+b$$ with arbitrary constants $$a$$, $$b$$.
Here are a few hints : use $$\frac{xf(x)-(y+z)\bigg(\frac{yf(y)-zf(z)}{y-z}\bigg)}{x-(y+z)}=\frac{(x+y)\bigg(\frac{xf(x)-yf(y)}{x-y}\bigg)-zf(z)}{(x+y)-z}$$ to show that $$f(z)=\frac{xf(y)-yf(x)+z(f(x)-f(y))}{x-y}$$. Deduce that there are two constants $$a,b$$ such $$f(z)=az+b$$, then plug into the original equation.
Here's another interesting solution: $$xf(x)−yf(y)=(x−y)f(x+y)\iff x(f(x+y)-f(x))=y(f(x+y)-f(y))$$ Let, $$g(x)=\frac{f(x)-f(0)}{f(1)-f(0)} \iff g(x)(f(1)-f(0))+f(0)=f(x)$$ So, now, $$g(1)=1$$ and $$g(0)=0$$ and $$x(g(x+y)-g(x))=y(g(x+y)-g(y))$$ $$y=1 \implies x(g(x+1)-g(x))=g(x+1)-1 \tag{1}$$ $$y=-x \implies-g(x)=g(-x) \tag{since g(0)=0}$$ $$y=1-x \implies x(1-g(x))=(1-x)(1-g(1-x))$$ Replacing $$x$$ with $$-x$$ in the previous equation yields (knowing the fact that $$g$$ is odd) $$-x(1+g(x))=(x+1)(1-g(x+1))\iff 2x+1=x(g(x+1)-g(x))+g(x+1)$$ $$\implies2g(x+1)=2(x+1) \tag{by (1)}$$ $$\implies g(y)=y \ \ \ \ \forall \ y \in \mathbb{R}$$ $$\implies f(x)=ax+b \ \ \ \ \forall \ a,b,x \in \mathbb{R}$$