What should you substitute?

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!

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.

• Tip: After the prove of odd, you can put $(x,y)$ for $(x,-y)$ in the original equation and observe what happens. – Culver Kwan Jun 12 '20 at 12:46
• @CulverKwan Thanks, that cracked it! – Rand al'Thor Jun 12 '20 at 13:31
• Correct! You did well! – Culver Kwan Jun 13 '20 at 5:15

I think that $$f$$ satisfies the equation iff

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

Proof

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 $$xf(x) - yf(y) = m(x^2 - y^2) + c(x-y) = (x-y)(m(x+y) + c) = (x-y)f(x+y)$$

• This seems like it could use some elaboration on the arguments indicated by "by varying a" and "by varying y". – Gareth McCaughan Jun 10 '20 at 14:03
• I'm worried that you might be implicitly assuming continuity or something like that. – Gareth McCaughan Jun 10 '20 at 15:04
• @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. – hexomino Jun 10 '20 at 16:14
• There is a more elegant solution. Try find it. – Culver Kwan Jun 11 '20 at 12:59