Completeness of $X$ with normal distribution $N(\theta,\theta^2)$ with equal mean and standard deviation: Integral of scale of a function

Question

Suppose that $f:\mathbb{R} \to \mathbb{R}$ is a Borel function such that, for every $a > 0$, we always have:

$$\begin{equation*} \int_{\mathbb{R}}f(ax)\exp \left(-\frac{1}{2}(x-1)^2\right)dx =0 \end{equation*}$$

Does it follow that $f = 0$ almost surely ? Does the answer change if we limit the attention only to $f \in L^{\infty}$ ?

The background of the question is that, suppose $X$ obeys the distribution $N(\theta, \theta^2)$ for some $\theta> 0$ and we wish to determine if $X$ is a (boundly) completes statistic for $\theta$. Expanding the definition out with a change of variable will give the statement of the problem above. I am looking for an analysis solution .

Intuitively, the exponential centers at $1$ and the scale is not centered at $1$. A counterexample would be surprising but I cannot prove it.

Answer 1

Noting that the integral equation can be written as

$$\mathbb E \left (f(aZ+a)\right )=0, \, Z \sim \mathcal N(0,1),$$

a counterexample can be designed as follows:

$$\color{blue}{f(x)=\frac{\max(x,0)}{|x|\Phi(1)}+\frac{\min(x,0)}{|x|(1-\Phi(1))}=\begin{cases} \frac{1}{\Phi(1)} & x>0\\ \frac{-1}{1-\Phi(1)} & x<0 \end{cases}}$$

for $x\neq0$ where $$\Phi(x)=\int_{-\infty}^x\frac{1}{\sqrt{2\pi}}\exp \left(-\frac{1}{2}x^2\right)\text{d}x$$ denotes the cdf of $Z$, and $f(0)$ can be set to any number.

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PS: From the section "Relation to sufficient statistics" in the Wikipedia page on complete statistics, it can be indirectly proven that there is no complete statistic for $\mathcal N(\theta,\theta^2)$.