Non-trivial examples of non-discriminatory functions In a 1989 paper by Cybenko, he defines a discriminatory function as follows:
For a fixed $n \in \mathbb{Z}^+$, let $I_n = [0,1]^n$. A function $\sigma : \mathbb{R} \to \mathbb{R}$ is said to be discriminatory if the only signed regular Borel measure $\mu$ that satisfies $\int_{I_n} \sigma (y^T x + \theta ) d \mu (x)= 0$ for all $y^T \in \mathbb{R}^n$ and $\theta \in \mathbb{R}$ is $\mu = 0$.
I'm interested in finding examples of non-discriminatory functions. Naturally, a function that is almost everywhere zero with respect to the $d$-dimensional Lebesgue measure is non-discriminatory, but this example is fairly trivial. Are there non-trivial examples of non-discriminatory functions?
 A: If $\sigma$ is a polynomial, then it is not discriminatory.
Assume that $\sigma$ is a polynomial of degree $m$ and $f$ be a function with $m+1$ vanishing moments (i.e., its inner product with polynomials of degree $m$ vanishes). Further assume that $f$ is supported in $[0,1]$. Then, setting $\mu = f d\lambda$, where $\lambda$ is the Lebesgue measure on $\mathbb R$ yields that for all $y, \theta \in \mathbb R$
$$
\int_{[0,1]} \sigma(y x +\theta) d \mu = \int_{\mathbb R} \sigma(y x +\theta) f d \lambda = 0,
$$
since $x \mapsto \sigma(y x +\theta)$ is a polynomial of degree at most $m$.
Hence, to complete the example, we only need to convince ourselves that functions $f$ with $m+1$ vanishing moments exist that are also supported in $[0,1]$. If you accept that this is possible then there is no need to read further. Otherwise, I give a construction below.
Take any non-zero, smooth function $g$ supported in $(0,1/2)$. Then 
$h(x) := g(x) - g(x-1/2)$ is supported in $(0,1)$ and we have that 
$$
\int_{\mathbb R}h(x) dx = 0.
$$
This implies that $h$ has one vanishing moment. Now the $m$'th derivative of $h$, which we denote by $h^{(m)}$, satisfies
$$
\int_{\mathbb R}h^{(m)}(x) x^{m} dx = (-1)^m \int_{\mathbb R}h(x) dx = 0,
$$ 
by partial integration. You can proceed similarly for all monomials of order less than $m$. This shows that $h^{(m)}$ has $m+1$-vanishing moments.
