# Recognizing Schwartz regular distributions

Are there characterizations of Schwartz regular distributions other than being locally integrable (which does not lend itself to easy manipulations)?

To be more detailed: if I want to show that some distribution is in fact a smooth function, I can try to use the concepts of singular support or wavefront, together with hypoellipticity. What can I do if I want much less, namely to show that a distribution is only a locally-integrable function (not necessarily smooth)?

The question is rather vague but one characterisation is the following: for each compact interval $I$ (I am working on the real line for simplicity) and each uniformly bounded sequence $(x_n)$ of smooth functions with support in $I$ which converges in $L^1$ to $0$ we have $T(x_n)\to 0$. This can easily be reformulated in terms of inequalities if that is more to your taste.
• Did you mean to say that $x_n$ converges to $0$? Otherwise, I can just take a constant sequence of functions, and I do not think that this is what you meant. – Alex M. Jun 23 '15 at 18:40
• Thank you, this seems to be exactly what I needed. 1) Could you direct me to a proof, please? 2)Would your result still be valid if $T$ were just linear, but not continuous? I.e. showing that a linear form is in fact a regular distribution. 3) If I wanted "continuous" instead of "locally integrable", could I just replace $L^1$ by the space of continuous functions? – Alex M. Jul 6 '15 at 8:27
This is a response to the above comment but is too long for a comment. ad 1. This is a corollary of the fact that $L^1$ is the dual of $L^\infty$ with the so-called strict topology, i.e., the finest locally convex topology which agrees with the $L^1$ topology on the unit ball. This essentially goes back to Saks (T.A.M.S. 35 (1933) 549-556) and a modern approach in the secondary literature can be found in the book "Saks spaces and applications to functional analysis". ad 2. yes. ad 3. no, but a variant is true. You use sequences of smooth functios which tend to zero for the weak topology induced by the continuous functions. This follows from the fact that the space of continuous functions, say on a compact interval, is the dual of the space of measures thereon with the so-called bounded weak star topology.