I want to show that the set of all the solutions to this functional equation $f(x-t) = f(x) -h$ where $f$ is strictly monotone continuous function is the set of affine functions $ax +b$. Update: The constants $t$ can be arbitrarily chosen, and $h$ is only dependent on $t$ and $f$.

Is my statement correct? And what would be a good approach to this problem? I am not sure how to tackle this problem without adding other assumptions such as $f$ is differentiable everywhere.

  • $\begingroup$ Are $t,h$ given constants? $\endgroup$ – YoTengoUnLCD Sep 17 '16 at 21:09
  • $\begingroup$ are they both positive or both negative ? To be strictly increasing we must have $a>0$ $\endgroup$ – WW1 Sep 17 '16 at 21:13
  • $\begingroup$ ah, I see. I changed my question to strictly monotone. $\endgroup$ – Xiao Sep 17 '16 at 21:14

Those are not the only solutions. For instance, for $f(x+1)=f(x)+1$, you have the solution $x+\frac{\sin(\pi x)}{2}$. Basically, given your affine function $ax+b$ that solves the equation, you can add any periodic, continuous, differentiable function with a period that divides $t$ and with derivative bounded above and below by $\pm a$. (You can relax the differentiable criterion somewhat. For instance, $\frac{|\sin(\pi x)|}2$ also works in the example above. But I don't know a more general, sufficient criterion.)

  • $\begingroup$ Thank you for your solution, I have updated my question. I was thinking about where $t$ is arbitrarily chosen, and there exists $h$ depends on $t$ and $f$. I wanted to find all $g$ such that $g\circ(\cdot-t) \circ g^{-1} (x) = x-h$ which makes translation closed under conjugation. $\endgroup$ – Xiao Sep 19 '16 at 22:08

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