Nonexistence of UMVUE for non-constant function? I tried to prove the problem:

Suppose X $\sim \ U(\theta-1,\theta+1)$, $\theta \in \mathbb{R}$. Then there is no UMVUE for $g(\theta)$ unless $g$ is a constant function.

Here is my attempt:
Suppose $T(X)$ is the UMVUE for $g(\theta)$, then it must be uncorrelated with any unbiased estimator of zero $U(X)$, that is $E[T(X)U(X)]=0.$
$$\mathbb{E}[U(X)]=0 \Rightarrow \int_{\theta-1}^{\theta+1}\frac{1}{2}U(x)dx=0$$
Take the derivative with respect to $\theta$, I get $U(\theta+1)=U(\theta-1) \ \forall \theta \in \mathbb{R}$.
$$\mathbb{E}[U(X)T(X)]=0 \Rightarrow \int_{\theta-1}^{\theta+1}\frac{1}{2}U(x)T(x)dx=0$$
Agian, take the derivative with respect to $\theta$, I get $T(\theta+1)U(\theta+1)=T(\theta-1)U(\theta-1)$, thus $T(\theta+1)=T(\theta-1) \ \forall \theta \in \mathbb{R}$.
$$\frac{d}{d\theta}\mathbb{E}[T(X)]=g'(\theta) \Rightarrow\frac{1}{2}[T(\theta+1)-T(\theta-1)]=g'(\theta)$$
Since $T(\theta+1)-T(\theta-1)=0$, so this implies $g$ is a constant function $\forall \theta \in \mathbb{R}$.
Is this proof correct or not?

$\bullet$ From $\mathbb  EU =0,$ one yields
$$U(x+1)= U(x-1)~~\textrm{a.s.}~~\forall x\in \mathbb R.\tag 1\label 1$$
$\bullet$ $T(x)U(x)$ is unbiased for $0,$ so from $\eqref 1,$
$$T(x+1)= T(x-1)~~\textrm{a.s.}~~\forall x\in \mathbb R.\tag 2\label 2$$
$\bullet$ Differentiating $\displaystyle \int_{\theta-1}^{\theta+1} T(x)~\mathrm dx = g(\theta)$ w.r.t. $\theta$ and using $\eqref 2,$
$$g^\prime(\theta) = 0~~\textrm{a.s.};\tag 3$$
OP's approach and derivation seem legit.
