4
$\begingroup$

In the book "Index Transforms" by S.B. Yakubovich (World Scientific, 1996), I encountered the notion of a "smooth function of the order two" with compact support on $\mathbb R_+$, denoted $C^{(2)}(\mathbb R_+)$, cf. here (I hope the link works).

What does that mean?

At first I thought it refers to the space of functions with continuous first and second derivative, but then it is the term "smooth" which puzzles me in this context. So maybe it means something different.

Annoyingly, the notation is not explained in the book, and it also is not obvious for me from the context. Yakubovich cites the result that $C^{(2)}(\mathbb R_+)$ is dense in $L_{1,2}(\mathbb R_+):=L^2 (\mathbb R_+, x\, dx)$ and refers to the book of Titchmarsh, "Introduction to the Theory of Fourier Integrals" (1937), but I could not find the result in there. So maybe someone of you knows more about this. Thank you in advance!

$\endgroup$

1 Answer 1

2
$\begingroup$

The symbols $C^k$, with $k=0,\dots,\infty$ are today universally acknowledged and can be used with no need of other specification; yet sometimes (and more often $20$ years ago) for the sake of clarity, or style, one would still insert them in expressions like regular/regularity or smooth/smoothness of order/class $C^k$ and equivalently smooth/smoothness, using smooth/smoothness as a generic term. Note that this author has infinitely smooth at page $72$. In conclusion, I would bet for $C^2$.

PS: I remember seminars where the lecturer would state hypotheses like "Let $\Omega$ be a bounded open set with smooth boundary...", and as soon as somebody asked "What order of smoothness" the answer was "Smooth enough for the needed Sobolev theorems to hold true", followed by a not convinced, respectful silence of the audience. "OK, let's say infinitely smooth". Maybe this is the origin of the meaning of smooth as $C^\infty$?

$\endgroup$
2
  • $\begingroup$ Thanks, also for the little anecdote! That sounds totally plausible, in particular considering the fact that the term "infinitely smooth" appears too (I have missed that). It's interesting however that Yakubovich's book is the only source I know of, where "smooth of order $k$" appears. So at least in written documents this does not seem to be a common expression. $\endgroup$
    – kolaka
    Commented Dec 7, 2019 at 17:39
  • 1
    $\begingroup$ It's less common than "regular of order k", yet I've seen several times. Also, a google search of "smooth of order " gives tens of thousands of results... $\endgroup$ Commented Dec 7, 2019 at 19:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .