Theorem: Infinite observation-derivative orbit

The distributions

are linearly independent over $\mathbb{C}$. No finite-dimensional subspace of ${\mathcal{D}}^{\prime }(X)$ containing $r$ is invariant under $-{\partial }_{x^0}$.

Proof. Let

Then $F=0$ on $\mathrm{Rel}$, so $Fr=0$, whereas

is nowhere zero on $\mathrm{Rel}$.

For an integer $N\ge 1$ and $j<N$, every term in ${{\partial }_{x^0}}^j(F^N\Phi )$ retains a factor $F$. Consequently

At derivative order $N$, restriction to $\mathrm{Rel}$ gives

Because the transpose of $(-{\partial }_{x^0}{)}^N$ is ${{\partial }_{x^0}}^N$, this proves

The right-hand side is nonzero. Indeed, $r$ is the positive parameter measure (eq:r), and the multiplier has constant nonzero sign; a nonnegative test positive near any point of $\mathrm{Rel}$ detects it.

If ${{\sum }_{j=0}}^N{c}_j{r}_j=0$ with $c_N\ne 0$ and $N\ge 1$, multiplication by $F^N$ leaves precisely $c_{N}N!({\partial }_{x^0}F{)}^{N}r=0$, a contradiction. For $N=0$, nonvanishing of $r$ gives the conclusion directly. Thus every finite relation is trivial. An invariant finite-dimensional space containing $r$ would contain all $r_n$, which is impossible.