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Let a given real $A$ satisfy $0 < A < 2$.

Conjecture :

For any real entire nonconstant $f(x)$ there exist real entire $g(x)$ and $h(x)$ such that

$A f(x) = f(g(x)) + f(h(x))$

or

$ A f(x) = f(g(x)) - f(h(x))$

is satisfied.

Is this true ?

If its true , then how to compute $g(x)$ and $h(x)$ efficiently ?

edit

I changed the question due to HansEnglers comment. Basically I assumed the equations to hold on a real interval but because of analytic continuation the equation must hold everywhere.

edit

edit 2

I added the condition that $f(x)$ needs to be a nonconstant function. ( I am tempted to wonder if I need to say nonpolynomial )

This edit is thanks to the comment of user43208.

edit 2

mick
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    The tag complex analysis was removed. No big deal but I wonder : Does asking questions about real-entire functions not justify the tag of complex analysis ? Afterall entire functions are a subject of complex analysis. – mick Dec 10 '14 at 22:13
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    Wouldn't this then follow on all $\mathbb{R}$, by analytic continuation? Or do you mean to say $\forall x , Af(x) = f(g(x)) + f(h(x)) , \vee , Af(x) = f(g(x)) - f(h(x))$? – Hans Engler Dec 10 '14 at 22:48
  • @HansEngler well suppose $g$ is something like $x^2 + 1$ and $h$ is something like $x^6 + 1$. If $x>0$ implies $f(x) > 1$ then I think it does not follow from analytic continuation because of problems with range and domain ... Im not sure though. – mick Dec 10 '14 at 23:12
  • Hmm I think Hans is correct. by analytic continuation it must follow on all of R. – mick Dec 10 '14 at 23:16
  • When you use "entire" to describe a function on reals, do you mean "real analytic"? – Mohsen Shahriari Nov 15 '21 at 14:41

1 Answers1

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The conjecture is false. Consider $ A = 1 $ and $ f ( x ) = \frac 1 2 \left ( 1 + \frac 1 { 1 + x ^ 2 } \right ) $, both satisfying the requirements of the problem. for any $ x \in \mathbb R $ we have $ \frac 1 2 < f ( x ) \le 1 $. Therefore for any $ y , z \in \mathbb R $ we have $ f ( y ) + f ( z ) > 1 $ and $ f ( y ) - f ( z ) < \frac 1 2 $. This means that $ A f ( x ) = f ( y ) + f ( z ) $ and $ A f ( x ) = f ( y ) - f ( z ) $ never happen for any $ x , y , z \in \mathbb R $.

Mohsen Shahriari
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