If a distribution and its gradient have the same finite order then the order must be zero

Question

As the title suggests, I would like to know whether it is possibile for a distribution and its (distributional) gradient to have the same finite and greater or equal than one order.

Answer 1

(There are some complications due to "support", which I think are irrelevant to the question... so I consider compact_support distributions...)

A compact-support distribution is a continuous linear functional on some $D_E$, meaning smooth functions on a compact set $E$. The topology on $D_E$ is the Frechet space topology from the countable collection of seminorms $|f|_k=\sum_{j\le k} \sup_{x\in E} |f^{(j)}(x)|$.

The projective limit topology (which is correct in every way...) indicates that a continuous linear functional (such as a distribution) factors through the topology of some "limitand", meaning that it is continuous with respect to some finite collection of these seminorms.

In particular, the "order/degree/whatever" of a distribution on $D_E$ is the smallest $k$ such that it is continuous with respect to the $|\cdot|_k$ topology.

It is a reasonable calculus exercise to find explicit, fairly elementary, functionals which are continuous in the $k$-th topology, but not the $k-1$...th... and so on.

EDIT: in response to a followup question... and to be more precise/explicit: if a distribution is continuous in at_best a particular seminorm/topology, that is exactly to declare that it is not continuous in any further.

Whatever that may exactly mean/entail, it is effectively definitional that a functional continuous in a particular topology but no finer... is not continuous in any finer.

The/any derivative operator naturally maps from one TVS to another, and the topology "loses" one degree of differentiability. Not surprisingly, considering the definition.

So, really, it's definitional (unless one somehow cannot believe that there are functions that are this-much differentiable, but not that-much...)