I have the following question: Let $h \colon \mathbb{R}^n \rightarrow \mathbb{R}^n$ such that $h(x)=\Psi(x)+\rho(x)$ with $\Psi$ isometry and $\rho$ a bounded function. Are $\Psi$ and $\rho$ unique in such a way that $h(x)=\Psi(x)+\rho(x)$ ?
My attempt is the following:
Let $\Psi,\varphi$ isometries and $\rho,\phi$ bounded functions such that
$h(x)=\Psi(x)+\rho(x)=\varphi(x)+\phi(x)$
We know we can write any isometry as a descomposition of a rotation and translation (uniques), so
$h(x)=A(x)+t+\rho(x)=B(x)+r+\phi(x)$ with $A, B$ rotations
$A(x)-B(x)=\phi(x)-\rho(x)+r-t$
We know that rotation fix $0$, then
$0=\phi(0)-\rho(0)+r-t$ but I do not know how to finish, also I do not if its true, any suggestion?
Any isometry $\Psi$ is generated by rotation, reflections and translations, namely $\Psi(x)=Ax+b$ where $A\in \mathbb O_n(\mathbb R), b\in \mathbb R^n$. Thus if $h=\Psi+\rho$, replacing $\rho$ by a translate of $\rho$, namely $\rho+b$, one can assume that $\Psi$ is linear.
Suppose $$\Psi_1 +\rho_1= \Psi_2+\rho_2$$ Then we get $$\Psi_1 - \Psi_2=\rho_2-\rho_1$$ The $RHS$ is the difference of $2$ bounded functions and hence bounded. The $LHS$ is a linear map. So we get $\Psi_1-\Psi_2$ is a linear map which is bounded as a function (please do not confuse this with bounded linear map). This forces $\Psi_1-\Psi_2=0$ and hence $\rho_1=\rho_2$
Thus you do get uniqueness upto a translation.