# Variant of Cauchy Functional Equation

Consider the equation $$f(kx-f(x))=x=kf(x)-f(f(x))$$ for montonic $f$.

What can we say about the solutions to this equation. Comparing with Cauchy equation $f(x+y)=f(x)+f(y)$, I think the solution must be somewhat close to being linear. Any hints. Thanks beforehand.

• The $= x =$ in between, is that meant to be there? – Rutger Moody Jan 31 '17 at 21:36
• @RutgerMoody yes, that is meant to be there. I think that makes the function bijective. – vidyarthi Feb 3 '17 at 11:18

$f(kx-f(x))=x \tag{1}$
$kf(x)-f(f(x)) =x \tag{2}$

From $(1) \implies$ $f(x)$ is surjective.

$(2) \implies$ Let $f(a)=f(b)\implies a=kf(a)-f(f(a))=kf(b)-f(f(b)) = b \implies$

$f(x)$ is injective.

This means $f(x)$ is bijective. So $f^{-1}(x)$ exists.

$(1)\implies f(kf^{-1}(x)-f(f^{-1}(x)))=f^{-1}(x) \implies f^{-1}(f(kf^{-1}(x)-x))=f^{-1}(f^{-1}(x)) \implies x =kf^{-1}(x) -f^{-1}(f^{-1}(x))$
$(2) \implies kx-f^{-1}(x) = f(x) \implies f^{-1}( kx-f^{-1}(x)) = x$ $\implies$

If $f(x)$ is a solution to $(1)$ and $(2)$ then so is $f^{-1}(x)=kx-f(x)$

Let : $g(x)=-f(-x)$.
$(1)\implies -f(-kx-f(-x))=x \implies g(kx+f(-x))=x \implies g(kx-g(x))=x$
$(2)\implies -kf(-x)+f(f(-x)) =x \implies kg(x)+f(-g(x))=x\implies kg(x)-g(g(x))=x$

$\implies$ If $f(x)$ is a solution to $(1)$ and $(2)$ then so is $g(x):=-f(-x)$.

Fixed points :
From $(2)$ we see that if there exists an $a$ for which $f(a)=a$ then: $kf(a)-f(f(a)) =a \implies ka =2a \implies a=0 \lor k=2$
$f(ka-f(a))=a \implies ka-f(a) =a \implies ka =2a \implies a=0 \lor k=2$

Also we see that if there exists an $a$ for which $f(a)=-a$ then: $kf(a)-f(f(a)) =a \implies -ka =2a \implies a=0 \lor k=-2$
$f(ka-f(a))=a \implies -ka-a =a \implies -ka =2a \implies a=0 \lor k=-2$

From the above we see that when $k=2$ , then $f(x)=x$ is a solution. And when $k=-2$ , then $f(x)=-x$ is a solution.

General solution :

(Note: I screwed up a couple of times before here. Apologies to everyone who read it, if anyone did..I do think this must be the correct argument.)

We transform $f(x)$ into a new function $g(x)$ like this : $g(x)+\frac{kx}{2}=f(x)$
Which gives :
$g(\frac{kx}{2}-g(x))-\frac{kg(x)}{2} =(1-\frac{k^2}{4})x \tag{1a}$
$-g(\frac{kx}{2}+g(x))+\frac{kg(x)}{2} =(1-\frac{k^2}{4})x \tag{2a}$
There are two cases to consider :

Case 1 when $k^2 \neq 4$ :

Let $g(a)=0 \implies (1a) \implies g(\frac{ka}{2}) =(1-\frac{k^2}{4})a$
$(2a) \implies -g(\frac{ka}{2} ) =(1-\frac{k^2}{4})a$

$\implies a=0 \implies g(0)=f(0)=0 \enspace ( k^2 \neq 4 )$

From bijectivity and monotonicity I think we can conclude that $f(x)$ and $g(x)$ must be continuous, (and maybe even differentiable).
Here I'll assume $g(x)$ can be written as a Taylor series around $x=0$ :
We calculate the first and second derivative :

$g'(\frac{kx}{2}-g(x))(\frac{k}{2}-g'(x))-\frac{kg'(x)}{2} =(1-\frac{k^2}{4})$
$-g'(\frac{kx}{2}+g(x))(\frac{k}{2}+g'(x))+\frac{kg'(x)}{2} =(1-\frac{k^2}{4})$

$g'(0)(\frac{k}{2}-g'(0))-\frac{kg'(0)}{2} =(1-\frac{k^2}{4})$
$-g'(0)(\frac{k}{2}+g'(0))+\frac{kg'(0)}{2} =(1-\frac{k^2}{4})$

$g'(0) = \pm \sqrt{\frac{k^2}{4}-1}$

$g''(\frac{kx}{2}-g(x))(\frac{k}{2}-g'(x))^2+g'(\frac{kx}{2}-g(x))(-g''(x))-\frac{kg''(x)}{2} =0$
$-g''(\frac{kx}{2}+g(x))(\frac{k}{2}+g'(x))^2 -g'(\frac{kx}{2}+g(x))(g''(x))+\frac{kg''(x)}{2} =0$

$g''(0)(\frac{k}{2}-g'(0))^2+g'(0)(-g''(0))-\frac{kg''(0)}{2} =0$
$-g''(0)(\frac{k}{2}+g'(0))^2 -g'(0)(g''(0))+\frac{kg''(0)}{2} =0$

$g'(0)g''(0)( k + 1) =0 \implies g''(0)=0 \enspace \lor \enspace k=-1$
$-g''(0)(\frac{k^2}{4}+(k+1)g'(0)+g'(0)^2 -\frac{k}{2}) =0 \implies \text{(with$k=-1$and$g'(0)^2=\frac{k^2}{4}-1$)} \implies -g''(0)(\frac{1}{4} +\frac{1}{4}-1 -\frac{1}{2}) =0 \implies g''(0)=0$

$\implies g''(0) =0$ . Also the higher derivatives in $x=0$ are $0$ (I think).

So for $k^2 \neq 4$ we have : $g(x)=\pm \sqrt{\frac{k^2}{4}-1} \cdot x$. Or : $f(x)=( \frac{k}{2} \pm \sqrt{\frac{k^2}{4}-1} ) \cdot x$

Fill this in in the original equations to check :

$g(x)= \sqrt{\frac{k^2}{4}-1} \cdot x$
$\sqrt{\frac{k^2}{4}-1} (\frac{k}{2}-\sqrt{\frac{k^2}{4}-1} )x-\frac{k}{2}\sqrt{\frac{k^2}{4}-1} x =(1-\frac{k^2}{4})x$
$-\sqrt{\frac{k^2}{4}-1} (\frac{k}{2}+\sqrt{\frac{k^2}{4}-1} )x+\frac{k}{2}\sqrt{\frac{k^2}{4}-1} x =(1-\frac{k^2}{4})x$

$g(x)= -\sqrt{\frac{k^2}{4}-1} \cdot x$
$-\sqrt{\frac{k^2}{4}-1}(\frac{k}{2}+\sqrt{\frac{k^2}{4}-1} )x+\frac{k}{2}\sqrt{\frac{k^2}{4}-1} x =(1-\frac{k^2}{4})x$
$+\sqrt{\frac{k^2}{4}-1} (\frac{k}{2}-\sqrt{\frac{k^2}{4}-1} )x-\frac{k}{2}\sqrt{\frac{k^2}{4}-1} x =(1-\frac{k^2}{4})x$

$\square$

Case 2 when $k^2 = 4$ :

$g(\frac{kx}{2}-g(x))-\frac{kg(x)}{2} =0 \tag{1b}$
$-g(\frac{kx}{2}+g(x))+\frac{kg(x)}{2} =0 \tag{2b}$

We see that $g(x)$ must be constant (see here for example : Functions $f$ satisfying $f\circ f(x)=2f(x)-x,\forall x\in\mathbb{R}$.) and fill in as a general solution :

For $k = 2$ we have : $g(x)= c$. Or : $f(x)= c + x$.

For $k = -2$ we have : $g(x)= 0$. Or : $f(x)= - x$.

$\square$

Below some solutions to modified versions :

Below some things that can be proved about the more general version of $(1)$ and $(2)$ : equation $(3)$ :

More general case without '$=x=$' in between :

$$f(kx-f(x))=kf(x)-f(f(x)) \tag{3}$$

$f(x)=0$ and $f(x)=kx$ and $f(x)=\frac{kx}{2}$ are all solutions to $(3)$ .

$f(x)=0$ is trivial.
Let: $f(x)=kx \implies \\ f(kx-f(x))=kf(x)-f(f(x)) \implies f(kx-kx)=k^2x-f(kx) \implies 0=0$
Let: $f(x)=\frac{kx}{2} \implies f(kx-\frac{kx}{2})=k\frac{kx}{2}-f(\frac{kx}{2}) \implies \frac{k^2x}{2}= \frac{k^2x}{2} \enspace \square$

There are many more solutions, including the ones above for $(1)$ and $(2)$.

Below a solution to a modified version of $(1)$ and $(2)$ : equation $(4)$ and $(5)$:

Modified case with '$=\frac{k^2x}{4}=$' in between :

$f(kx-f(x))=\frac{k^2x}{4} \tag{4}$
$kf(x)-f(f(x)) =\frac{k^2x}{4} \tag{5}$

$f(x)=\frac{kx}{2}$ is a solution to $(4)$ and $(5)$.

Let: $f(x)=\frac{kx}{2} \implies f(kx-\frac{kx}{2})=\frac{k^2x}{4} \implies \frac{k^2x}{4}=\frac{k^2x}{4}$ and :
$k\frac{kx}{2}-f(\frac{kx}{2}) =\frac{k^2x}{4} \implies \frac{k^2x}{2}- \frac{k^2x}{4} =\frac{k^2x}{4}$ $\enspace \square$

• great answer, the answer would be complete if you provide proofs for expanding into taylor series – vidyarthi Feb 16 '17 at 11:09
• @vidyarthi Thanks! I realize this was an assumption, it might be that this can be proved. But then again it might be that there are other exotic functions (eg fractal-like) so weird in their definition that they are non-differentiable solutions to the above. Allways hard to prove a negative. I'll get back to it if I can find an answer. – Rutger Moody Feb 16 '17 at 12:24