0
$\begingroup$

I have posted a similar question in the past but let me make a final try in a simpler framework.

Let $g \in C_0 ^\infty (\mathbb{R})$ be smooth and compactly supported. Define $$ f(x) = \int \big ((x - y)^2 - 1 \big )^{1/2}(x-y) g (y) \,dy $$ where integration is performed over the set where $|y - x|>1$ and $y\in \operatorname {supp}g$.

If $g$ fails to be real analytic at some point $x_0$ can we deduce that also $f$ fails to be real analytic at some point depending on $x_0$, like perhaps $x_0 \pm 1$?

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ I think it seems unlikely, without a lot of extra assumptions, that there will be any useful general statement. Knowing that g is not analytic at some point doesn't really give you much information, since there are many possible ways for analyticity to fail; and furthermore, one single value of g (maybe even all its derivatives) doesn't change f in any way (whereas by contrast, having g analytic at a point tells you something about g on a whole interval). Of course if you could invert the integral operator and express g directly in terms of f, you would get a lot of information. $\endgroup$
    – Zen Harper
    Commented Sep 2, 2012 at 22:38
  • 1
    $\begingroup$ Have you tried using a Taylor series expansion in $(z^2-1)^{1/2} = z(1-1/z^2)^{1/2}$, for $z = |x-y|>1$? Then, at least formally, you can get a decomposition into simpler operators, using a formal infinite sum. $\endgroup$
    – Zen Harper
    Commented Sep 2, 2012 at 22:42
  • $\begingroup$ Thanks a lot for your response. The idea of inversion was my first attempt. I do know the Fourier transform of $(x^2 - 1)^{1/2}x 1_{|x| > 1}$ but I don't get much further after that. Actually I'm looking at this in the framework of distribution theory where $g\in \mathcal{E}'(\mathbb{R})$. $\endgroup$
    – flavio
    Commented Sep 3, 2012 at 7:33

1 Answer 1

Reset to default
3
$\begingroup$

Your function $g$ in $C_0^\infty$, if not identicaly equal to zero, certainly fails to be real analytic at some point. Because a real analytic function cannot be in $C_0^\infty$. Your convolution is irrelevant for this conclusion.

Edit: your comment indicates that you are really asking about the relation of the singular sets of $f$ and $g$. The singular set $S(f)$ (singular support) is the set where the function is not analytic. The general fact here is that $S(f)$ is contained in $(S(g)+1)\cup (S(g)-1)$, the union of shifts of $S(g)$ by one unit left and right. Perhaps this answers your question. This can be found in the second volume of Hormander's Analysis of linear differential operators, sect 16.3, together with a discussion when the equality happens.

Improve this answer
$\endgroup$
3
  • $\begingroup$ But if instead we have $f(x) = \int (x-y)^\alpha _+ g(y) dy$ then, since the Fourier transform of $x_+^\alpha $ and $1/(x_+^\alpha \hat{)}$ are known one can deduce that $g$ extends to a holomorphic function near $x_0$ if and only if $f$ extends near $x_0$. $\endgroup$
    – flavio
    Commented Aug 31, 2012 at 14:13
  • $\begingroup$ Of course such a $g$ will not be everywhere real analytic. But what I'm asking is if those points where it isn't will induce points where $f$ is not real analytic, if we define $f$ by the above convolution. $\endgroup$
    – flavio
    Commented Sep 2, 2012 at 13:18
  • $\begingroup$ Thank you for your edited comment. I'm familiar with what you are saying. But what I want is more or less the opposite inclusion, i.e. a conclusion of the form; if we start with a point $x_0\in S(g)$, then the point $\varphi (x_0)$ belongs to $S(f)$. I want to know what $\varphi $ is, if there is such a $\varphi $. $\endgroup$
    – flavio
    Commented Sep 2, 2012 at 15:10

You must log in to answer this question.

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