Hermite Polynomials are orthonormal basis for $L(\mathbb{R})$ In the process of showing Hermite polynomials are dense in  $L^2(\Bbb{R})$, I asked to claim for $g \in L^2(R)$ (We may assume $g$ to be continuous)


*

*$\int\limits_{-\infty}^{\infty} g(t) e^{\frac{-t^2}{2}} \frac{\sin
   a(t-y)}{t-y} dt =0,\forall a,y\in \Bbb{R}$
then $g$ is identically zero on $\Bbb{R}$. 
I have no idea what the integral means and how to conclude the claim. Please help me with this. 
 A: Disclaimer: I'm not sure this answer works since I never had to really use the $e^{-\frac{t^2}{2}}$. If my argument is correct, I'm sure it's lacking rigor. I leave it to you to fill in the details since this seems like a homework question. 
I'm going to take $a > 0$. Define 
$$f(t) = g(t)e^{-\frac{t^2}{2}}$$ 
and 
$$h(t) = \frac{\sin at}{t} = a \, \text{sinc}(at)$$
Then, by definition, the integral is $(f * h)(y)$, where $f * h$ denotes the convolution of $f$ and $h$. By the convolution theorem, $\mathcal{F}\{f*h\} = \mathcal{F}\{f\} \mathcal{F}\{h\}$.
We have
$$
\mathcal{F}\{h\}(x) = a \mathcal{F}\{t \mapsto \text{sinc}(at)\}(x) = \text{rect}\left(\frac{x}{ka} \right)
$$
for some constant $k$ (I don't know what convention you're using for the fourier transform). 
On the other hand, we are given that $f * h = 0$. This is just the statement that the integral is $0$ for all choices of $y$. Thus $\mathcal{F}\{f * h\} = \mathcal{F}\{0\} = 0$. Therefore
$$
0 = \left(\mathcal{F}\{f\}(x)\right) \cdot \text{rect} \left(\frac{x}{ka}\right)
$$
for all choices of $a > 0$. This shows that $\mathcal{F}\{f\}$ is $0$ on the interval $\left[-\frac{ka}{2}, \frac{ka}{2} \right]$. But since we can take $a$ as large as we like, this shows that $\mathcal{F}\{f\}$ itself is identically $0$. But every function in $L^2(\mathbb{R})$ with a Fourier transform of $0$ is equivalent to $0$. Thus, we conclude that $f = 0$, which shows that $g = 0$ since $e^{-\frac{t^2}{2}}$ is positive on $\mathbb{R}$.
