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Find all $a\in\mathbb{R} $ for which there exists a function $f : \mathbb{R}\to\mathbb{R} $ , such that

(i) $f(f(x))=f(x)+x$, for all $x\in\mathbb{R} $,

(ii) $f(f(x)–x)=f(x)+ax$, for all $x\in\mathbb{R} $

Normally in such functional equations, I'd put different values of $x$ (like $x+1$, $x+2$ etc) and try to get an equation for $f(x)$. But here I'm not getting anywhere by that method. I can't use the coefficient comparison method either, because $f(x)$ may not be a polynomial.

Cameron Williams
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Righter
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1 Answers1

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NOTE: $$f^{-1}f(x) = f(f^{-1}(x)) =x$$ GIVEN:

(i) $f(f(x))=f(x)+x$, for all $x$ $\in$ $ℝ$,

(ii) $f(f(x)–x)=f(x)+ax$, for all $x$ $\in$ $ℝ$

Substitute f(x) as x in equation (i). you can verify the equation below is equivalent to equation (i) if you put $x = f(x)$ in the equation below. $$f(x)= x+ f^{-1}(x)$$

Then in RHS of equation 2, substitute what we have above: $x = f(x)- f^{-1}(x)$:

$f(f(x)-(f(x)-f^{-1}(x)))=f(x)+ax \implies$ $x=f(x)+ax \implies f(x)=(1-a)x$

Substitute and solve... I got a = golden ratio and its conjugate... Ask doubt if any below

Righter
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Sid
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  • You are assuming here that the inverse function $f^{-1}$ is well defined (or even defined at all for all $x$). It is not clear why that is necessarily true. – JimT Feb 14 '22 at 20:39
  • is there a better way to solve this? @JimT – Sid Mar 28 '22 at 13:34
  • Proving that $f$ is injective is easy; if $f(x) = f(y)$ then it follows that $f(f(x)) = f(f(y))$ and then from the first recursive equation we obtain $x = y$. Thus, $f^{-1}$ is well defined but possibly not on the entire $\mathbb R$. Proving that $f$ is surjective does not seem easy,... of course, if that is true at all. But perhaps I am wrong and there exists a simple solution. – JimT Mar 29 '22 at 16:59