Functional Equation: Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $(x+y)(f(x)-f(y))=(x-y)f(x+y)$

Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $$(x+y)(f(x)-f(y))=(x-y)f(x+y)$$

My attempt:

If $x=-y \not = 0$ then $0= 2x f(0)$ so $f(0)=0$.

Suppose for the sake of contradiction that $f(x)=f(x+\epsilon)$ for some $x$ and $\epsilon>0$. Let $y=x+\epsilon$. Then $$0=\epsilon \cdot f(2x+\epsilon)$$ therefore as $2x+ \epsilon$ can take any real value $f$ is either strictly increasing or strictly decreasing or $f(x)=0 \; \; \forall \; \;x \in \mathbb{R}$.

Note that $f(x)=ax$ is a solution $\forall \; \;a \in \mathbb{R}$. Thanks so much for any help!

• Your "as $2x+\epsilon$ can take any real value" does not appear to be warranted, since $x$ and $\epsilon$ were chosen specifically to make $f(x)=f(x+\epsilon)$. Nov 19, 2014 at 0:00
• Also, even if you do manage to prove that $f$ must be injective, it doesn't follow that it is strictly increasing of strictly decreasing, at least not unless you also prove that $f$ is necessarily continuous. Nov 19, 2014 at 0:01
• note tha if $2<n\in\mathbb{N}$ $$f(n)=\frac{n}{n-2}(f(n-1)-f(n-2))$$ Nov 19, 2014 at 0:17
• Yeah, I see. Thanks. I will cross out the wrong stuff. Nov 19, 2014 at 0:17
• How did you come up with this question? Jul 9, 2015 at 9:52

$f(x)$ must be of the form $ax^2 + bx$.

Letting $x = 1$ and $y = 0$ in the equation, we find $f(0) = 0$.

Now define $g(x) = f(x)/x$ for $x \ne 0$. Taking $f(0) = 0$ for granted, the functional equation can be rewritten as $$(x-y)g(x + y) = xg(x) - yg(y), \qquad x, y, x+y \ne 0.$$

Substituting $1$ for $y$, we find $$(x-1)g(x+1) = xg(x) - g(1), \qquad x \ne -1, 0.$$

Now substituting $x + 1$ for $x$ and $-1$ for $y$, we find \begin{align*} (x+2)g(x) &= (x+1)g(x+1) + g(-1) &\quad \text{(x \ne -1,0)}\\ (x-1)(x+2)g(x) &= (x+1)(x-1)g(x+1) + g(-1)(x-1) \\ (x^2 +x - 2)g(x) &= (x+1)[xg(x) - g(1)] + g(-1)(x-1) \\ -2g(x) &=-g(1)(x+1) +g(-1)(x-1). \end{align*} The last relation is in fact true for all $x \ne 0$, including $x = -1$.

This proves that the function $g(x)$ is linear, say $g(x) = ax + b$. Thus $f(x) = xg(x) = ax^2 + bx$. The relation $f(x) = ax^2 + bx$ is valid for all $x$, including $x = 0$.

Note that $f(x) = a x^2 + b x$ is a solution. I believe these are all the analytic solutions.

EDIT:

Yes, in fact they are all the differentiable solutions.

Taking $x=0$ we get $y f(0) = 0$, so $f(0) = 0$. Now suppose $f$ is differentiable. Taking the derivative of the equation with respect to $x$ and substituting $x=0$ we get $$- 2 f(y) + y f'(0) + y f'(y) = 0$$ Letting $f'(0) = b$, the solutions of the differential equation $-2 f(y) + b y + y f'(y) = 0$ are $f(y) = b y + a y^2$ where $a$ is arbitrary.

EDIT: Since $\dfrac{f(x) - f(y)}{x-y} = \dfrac{f(x+y)}{x+y}$, any solution that is continuous will be differentiable except possibly at $0$. I am not at all convinced that every solution must be differentiable at $0$, or indeed that every solution must be continuous.

• Could we prove it has to be differentiable? I am not familiar with how to approach those sort of real analysis proofs. Nov 19, 2014 at 0:40

Hint:

First take $y=-x/2$ then

$$f(x) = 3f(x/2) + f(-x/2)$$

Taking $x\to -x$ we get

$$f(-x) = 3f(-x/2) + f(x/2)$$

This motivates us to define $g(x) = f(x) + f(-x)$ which by adding the equations above is found to satisfy

$$g(x) = 4g(x/2)$$

which is much easier to work with. We can for example by a simple inductive argument show that $g(2^n x) = 4^n g(x)$ for all $x$ and $n\in\mathbb{Z}$. A simple consequence of this is that if we know $g$ on any interval on the form $(2^{-n-1},2^{-n}]$ with $n\in\mathbb{Z}$ then we know it for all other values.

• But how do you go from $g$ to $f$? Not all $f$'s with the same symmetrization will satisfy the equation. Nov 19, 2014 at 0:52
• @RobertIsrael That is true, haven't considered that. If we follow the same line as you did and assume differentiabillity then this is no problem (which was the thought I had when writing this). Though I think one might get away with only assuming differentiabillity at $x=0$ (or one might need to assume $g$ to be two times differentiable at $x=0$ to get the argument through?!). Nov 19, 2014 at 0:58

$$f$$ can be only of the form $$ax^2+bx$$.

To see this, we observe that for any $$a$$ and $$b$$, if we define $$g(x):=f(x)-\big(ax^2+bx\big)$$, then $$g$$ satisfies the same functional equation $$(x+y)\big(g(x)-g(y)\big)=(x-y)g(x+y)$$. Letting $$a=\frac{f(1)+f(-1)}{2}$$ and $$b=\frac{f(1)-f(-1)}{2}$$ we get $$g(1)=g(-1)=0$$. now putting $$y=\pm1$$ in the functional equation, we'll have: $$(x+1)g(x)=(x-1)g(x+1)$$ $$(x-1)g(x)=(x+1)g(x-1)$$ Substituting $$x+1$$ for $$x$$ in the last equation, we'll have: $$(x+2)g(x)=xg(x+1)$$ By using both equations above containing $$g(x)$$ and $$g(x+1)$$, we get: $$x(x-1)g(x+1)=x(x+1)g(x)=(x-1)(x+2)g(x)$$ $$\therefore\big(x^2+x\big)g(x)=\big(x^2+x-2\big)g(x)$$ $$\therefore g(x)=0$$ Hence for every real number $$x$$, $$f(x)$$ is equal to $$ax^2+bx$$.