Proposition: Unrestricted smooth weights

The real vector space

is infinite-dimensional, even after imposing first–third exchange symmetry.

Proof. The exchange-symmetric weights $a_n=(x^0+x^{\prime 0}{)}^n$, $n\ge 0$, give linearly independent distributions. If $P(x^0+x^{\prime 0})r=0$ for a polynomial $P$, the positive pushforward density implies $P(t+t^{\prime })=0$ throughout the parameter space: a nonzero value would persist on an open set and be detected by a compact test. Since $t+t^{\prime }$ ranges over $(0,\infty )$, $P$ is identically zero.