Let $\;f:\mathbb R \to \mathbb R^m\;$ be a function such that $\;\vert f(x)-l_{\pm} \vert \le Ce^{-\vert x \vert}\;$. Assume $\;g:\mathbb R \supset I \to \mathbb R\;$ and consider the shift $\;f(x-g(y))\;$. I want to prove (if possible) that $\; \vert f(x-g(y)) -l_{\pm} \vert \le C_0e^{-\vert x \vert}\;$ if and only if $\;g\;$ is a bounded function.(NOTE: $\;C_0:=C_0(C)\gt 0\;$ and $\; \vert \cdot \vert\;$ stands for the euclidean norm)
My attempt:
($\; \Leftarrow\;$) If $\;g\;$ is a bounded function then $\;\vert g(y) \vert \le M\;\;\forall y \in I\;$ where $\;M \gt 0\;$. Hence:
- if $\;x \gt M\;$ then $\;\vert f(x)-l_{+} \vert \le Ce^{-(x-g(y))} \le Ce^Me^{-x}\;$
- if $\;x \lt -M \lt 0\;$ then $\;\vert f(x) -l_{-} \vert \le Ce^(x-g(y))\le Ce^Me^x\;$
For $\;C_0=Ce^M \gt 0\;$ the proof is complete.
However I've been stuck to the other direction and I don't know how to proceed. I don't ask for a complete proof rather than some hints. The above is something I need to prove in order to solve a bigger problem of mine, thus it might be wrong the statement "if and only if".
Any help would be valuable. Thanks in advance!