Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $$xf(x)-yf(y)=(x-y)f(x+y)$$ for all $x,y\in\mathbb{R}$.

Problem by me.

Most elegant solution gets the checkmark!


2 Answers 2


If a function $f(x)$ satisfies the given functional equation, then so does $f(x)+c$ for any fixed constant $c$. So we can assume without loss of generality that $f(0)=0$.

  • Putting $y=-x$, we find

    $f(-x)=-f(x)$ for all $x\neq0$, i.e. the function $f$ must be odd.

  • Putting $y=-2x$, we find

    $f(-2x)=-2f(x)$ for all $x\neq0$. Combining this with the above result gives $f(2x)=2f(x)$ for all $x\neq0$.

  • Then by induction, putting $y=2x$ then $y=3x$ and so on, we find

    $f(nx)=nf(x)$ for all $n\in\mathbb{Z}$ and $x\in\mathbb{R}$. Therefore, the same is true for all $n\in\mathbb{Q}$ and $x\in\mathbb{R}$.

So far this shows that $f$ must be

linear on every "rationally linear subset" $r\mathbb{Q}$ for $r\in\mathbb{R}$, i.e. that the function $\frac{f(x)}{x}$ must be constant on any such subset.

Can these constants be different for different such subsets? Let's use $(x,y)$ and $(x,-y)$ to get

$(x-y)f(x+y)=xf(x)-yf(y)=xf(x)+yf(-y)=(x+y)f(x-y)$, which means $\frac{f(x+y)}{x+y}=\frac{f(x-y)}{x-y}$ for any $x,y$ not equal to $\pm$ each other.

But then for any two real numbers which are not rational multiples of each other,

we can express them as $x+y$ and $x-y$ for some $x$ and $y$, thus showing that the constants corresponding to these rational-multiple subsets must be the same.

In conclusion,

$f$ must be linear, and every linear function satisfies the given constraint.

  • 1
    $\begingroup$ Tip: After the prove of odd, you can put $(x,y)$ for $(x,-y)$ in the original equation and observe what happens. $\endgroup$ Jun 12, 2020 at 12:46
  • $\begingroup$ Correct! You did well! $\endgroup$ Jun 13, 2020 at 5:15

I think that $f$ satisfies the equation iff

$f(x) = mx + c$, i.e, $f$ is linear, for $m,c \in \mathbb{R}$


If $x=y$ then the equation is trivially satisfied so let $x = y+a$ with $a \neq 0$.
Then the functional equation becomes $$(y+a)f(y+a) - yf(y) = a f(2y+a)$$ Furthermore, if we let $x=a$ in the original equation and rearrange we get $$(y-a)f(y+a) - yf(y) = -af(a) $$ Subtracting the second equation from the first gives $$2af(y+a) = af(2y+a) + af(a)$$ and dividing by $a$ and rearranging gives $$f(2y + a) - f(y+a) = f(y+a) - f(a)$$ and this holds for all values of $y$ and $a \neq 0$.
In particular, setting $y=1$ gives $f(a+2) - f(a+1) = f(a+1) - f(a)$ and it follows that $$f(a+N) - f(a) = N(f(a+1) - f(a)) = m(a+N) + c$$ for all integers $N$ where $m = f(a+1)-f(a)$ and $c = f(a) - a(f(a+1)-f(a))$.

Now let $\mathbb{Z}_z = \{x \in \mathbb{R} | x-z \in \mathbb{Z}\}$.
The above equation tells us that for all $x$ in $\mathbb{Z}_z$, $f(x) = m_z x + c_z$ for some $m_z, c_z \in \mathbb{R}$.
In particular, for all $x$ in $\mathbb{Z}$, $f(x) = m_0x + c_0$.
For arbitrary $z$, it's not necessarily true that the values of $m_z$ and $c_z$ are equal to $m_0$ and $c_0$, respectively. To show this, let $x$ be and integer and $y \in \mathbb{Z}_z$.
Then $x+y \in \mathbb{Z}_z$ and using our original equation, we have $$xf(x) - yf(y) = (x-y)f(x+y)$$ $$\Rightarrow x(m_0x+c_0) - y(m_zy+c_z) = (x-y)(m_z(x+y) + c_z)$$ $$ \Rightarrow m_0 x^2 + c_0 x = m_z x^2 + c_z $$ Since this must hold for all integers $x$ it follows that $m_0 = m_z$ and $c_0 = c_z$.
Therefore $f(x) = mx + c$ for all $x \in \mathbb{R}$.

The final part is checking that this works for the original equation, indeed $$xf(x) - yf(y) = m(x^2 - y^2) + c(x-y) = (x-y)(m(x+y) + c) = (x-y)f(x+y) $$

  • $\begingroup$ This seems like it could use some elaboration on the arguments indicated by "by varying a" and "by varying y". $\endgroup$
    – Gareth McCaughan
    Jun 10, 2020 at 14:03
  • $\begingroup$ I'm worried that you might be implicitly assuming continuity or something like that. $\endgroup$
    – Gareth McCaughan
    Jun 10, 2020 at 15:04
  • $\begingroup$ @GarethMcCaughan I've updated my answer which I think works this time. I think you were right, I was trying to fudge a continuity earlier. $\endgroup$
    – hexomino
    Jun 10, 2020 at 16:14
  • $\begingroup$ There is a more elegant solution. Try find it. $\endgroup$ Jun 11, 2020 at 12:59

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