6
$\begingroup$

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!

$\endgroup$
7
  • 1
    $\begingroup$ Your "as $2x+\epsilon$ can take any real value" does not appear to be warranted, since $x$ and $\epsilon$ were chosen specifically to make $f(x)=f(x+\epsilon)$. $\endgroup$ Nov 19, 2014 at 0:00
  • $\begingroup$ 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. $\endgroup$ Nov 19, 2014 at 0:01
  • $\begingroup$ note tha if $2<n\in\mathbb{N}$ $$f(n)=\frac{n}{n-2}(f(n-1)-f(n-2))$$ $\endgroup$ Nov 19, 2014 at 0:17
  • $\begingroup$ Yeah, I see. Thanks. I will cross out the wrong stuff. $\endgroup$
    – John Marty
    Nov 19, 2014 at 0:17
  • 1
    $\begingroup$ How did you come up with this question? $\endgroup$
    – Nikolaj-K
    Jul 9, 2015 at 9:52

4 Answers 4

7
$\begingroup$

$f(x)$ 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 $x = 0$.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Could we prove it has to be differentiable? I am not familiar with how to approach those sort of real analysis proofs. $\endgroup$
    – John Marty
    Nov 19, 2014 at 0:40
2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ But how do you go from $g$ to $f$? Not all $f$'s with the same symmetrization will satisfy the equation. $\endgroup$ Nov 19, 2014 at 0:52
  • $\begingroup$ @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?!). $\endgroup$
    – Winther
    Nov 19, 2014 at 0:58
1
$\begingroup$

$f$ 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 $ax^2+bx$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .