# 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?

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.