I am given a function $h(t)$ such that for $\forall\theta\in\mathbb R$,$$f(\theta)=\int_{-\infty}^\infty h(t)\exp\left\{-\frac1{2n}(t-n\theta)^2\right\}dt=0,~n\in\mathbb N\text{ (constant)}$$Is this sufficient to conclude $h(t)$ is identically $0$ in $\mathbb R$?
Context: I am required to prove (or disprove) completeness of the statistic $T=\sum_{i=1}^n x_i$ where $\{x_i\}_{i=1}^n$ is a simple random sample in $N(\theta,1)$ population, using the definition of complete statistic.
I have differentiated and integrated $f(\theta)$ to obtain that $$\int_{-\infty}^\infty t^mh(t)\exp\left\{-\frac1{2n}(t-n\theta)^2\right\}dt=0\\\int_{-\infty}^\infty t^mh(t)dt=0$$for all $m\in\mathbb Z_{\ge0}$. Now I am stuck.
You have
$$E_{\theta}\left[h(T)\right]=\frac1{\sqrt{2n\pi}}\int_{-\infty}^\infty h(t)\exp\left\{-\frac1{2n}(t-n\theta)^2\right\}\,dt=0\quad,\forall\,\theta\in \mathbb R$$
Expanding the square, this implies
$$\int_{-\infty}^\infty e^{\theta t}\,h(t)\exp\left(-\frac{t^2}{2n}\right)\,dt=0\quad,\forall\,\theta \tag{$\star$}$$
The left hand side of $(\star)$ is a two-sided Laplace transform of $h(t)\exp\left(-\frac{t^2}{2n}\right)$.
By property of integral transforms, this implies $$h(t)\exp\left(-\frac{t^2}{2n}\right)=0\,,\,\text{a.e.}$$
Hence, $$h(t)=0\,,\,\text{a.e.}$$