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Find all functions $ f: \mathbb R \to \mathbb R $ such that for all reals $ x $ and $ y $, $$ ( x + y ) \big( f ( x ) - f ( y ) \big) = ( x - y ) f ( x + y ) \text . $$

I actually got the answer by guessing and checking, $ f ( x ) = a x ^ 2 + b x $, but I want to see the solution. My friend suggested surjectivity but I don't see how to continue.

$ f ( x ) = f \left( \frac x { f ( x ) ^ 2 } \right) $; this is what I just got.

The answer is correct, just need a solution. Thanks!

Parcly Taxel
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2 Answers2

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For differentiable functions, the only solution is as you have suggested (I can’t tell if differentiability can be dropped). To see this, first let $y=0$ and $x\ne 0$, and you obtain $$x(f(x)-f(0))=xf(x)\implies f(0)=0\,.$$ Now, assume $xy\ne0$ and rewrite your identity as follows: $$\frac{f(x+y)-f(y)}{x}+\frac{f(x)}{x}=\frac{f(x+y)-f(x)}{y}+ \frac{f(y)}{y}\,$$ which upon taking the limit as $y\to 0$ on the both sides gives us $$\frac{f(x)}{x}+\frac{f(x)}{x}= f’(x) +f’(0)\implies f’(x)-\frac{2}{x}f(x)+f’(0)=0\,.$$ Using the integrating factor method, the solution to this becomes $$f(x)=ax^2+f’(0)x\,,$$ for some constant $a$.

Jack LeGrüß
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  • Welcome; my pleasure. (You may consider accepting this answer if you are satisfied; otherwise, I am attempting to see if I could determine existence of wild solutions in the absence of differentiability/continuity and add that later.) – Jack LeGrüß Oct 29 '20 at 18:39
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For any real numbers $ a $ and $ b $, if we define the function $ g : \mathbb R \to \mathbb R $ with $ g ( x ) = f ( x ) - a x ^ 2 - b x $, then using the original functional equation, we'll have $$ ( x + y ) \big( g ( x ) - g ( y ) \big) = ( x - y ) g ( x + y ) \text . \tag 0 \label 0 $$ In particular, setting $ a = \frac { f ( 1 ) + f ( - 1 ) } 2 $ and $ b = \frac { f ( 1 ) - f ( - 1 ) } 2 $, we will also have $ g ( 1 ) = g ( - 1 ) = 0 $. Letting $ y = 1 $ in \eqref{0}, we get $$ ( x + 1 ) g ( x ) = ( x - 1 ) g ( x + 1 ) \text , \tag 1 \label 1 $$ while substituting $ x + 1 $ for $ x $ and $ - 1 $ for $ y $ in \eqref{0} we have $$ x g ( x + 1 ) = ( x + 2 ) g ( x ) \text . \tag 2 \label 2 $$ \eqref{1} and \eqref{2} together show that $$ 2 g ( x ) = x ( x + 1 ) g ( x ) - ( x - 1 ) ( x + 2 ) g ( x ) = x ( x - 1 ) g ( x + 1 ) - ( x - 1 ) x g ( x + 1 ) = 0 \text . $$ Therefore $ g $ is the constant zero function, and hence $ f ( x ) = a x ^ 2 + b x $ for all $ x \in \mathbb R $. It's straightforward to see that any function of this form satisfies the original functional equation, and thus they form the class of all the solutions.

Mohsen Shahriari
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