Suppose that the following convolution equation holds: for all $y\in \mathbb{R}^n$ and every $i =1 ..n$ $$\int f(x_i) \exp \left(- \frac{\| x \|^2}{2} \right) g(x-y) d x =0$$
where
- $f(x)$ is an odd increasing function.
- $g$ is multilinear polynomial for some degree $N$
Question: Can we show that $g(x)=c$ (i.e., it is a constant)?
Here is the proof for $n=1$ case:
We can write $g(x) = \sum_{i=1}^N a_i x^i$ and $g(x -y) = \sum_{i=1} c_{i,j} x^i y^j$ .
Now expanding the integral, we get that for all $y$ \begin{align} 0= \sum_{i+j \le N } c_{i,j} y^j \int f(x) \exp(-x^2/2) x^i d x = \sum_{i+j \le N } c_{i,j} y^j b_i \end{align} where we define $b_i = \int f(x) \exp(-x^2/2) x^i d x $. We also not that $b_i=0$ for even $i$ and $b_i>0$ otherwise. This results in \begin{align} 0= \sum_{i+j \le N : i \text{ is odd}} c_{i,j} b_i y^j \end{align}
Now by the identity theorem a polynomial is zero iff all of the coefficients are zero. The rest is very tedious and follows by playing with coefficients and leads to $a_i =0$ forall $i \ge 1$, which implies that $g(x) =a_0$.
Question 2: I wonder if there is a simpler proof too even for $n=1$?
One can also write the above in terms of expecations \begin{align} E[ f(Z_i) g(Z-y)] =0 \end{align} where $Z$ is standard normal vector.