1
$\begingroup$

Let $f_n$ and $g_n$ two generalized functions such that :

  • the product $f_n g_n$ is well defined for all $n\in \mathbb{N}$, which means $WF(f_n)+WF(g_n)$ does not contain an element of the form $(x,0)$.
    ( where $WF(u)$ is the wavefront of generalized function $u$)
  • $f_n$ converges to $f$ and $g_n$ converges to $g$ (Weak* topologie).
  • the product $f g$ is well defined.
    My question is : Can we deduce that $f_n g_n$ converge to $f g$ (weak* topologie) ?
Improve this question
$\endgroup$
4
  • $\begingroup$ The standard example for this kind of question is the sequence of Rademacher functions--converge to zero in various weak topoloy but the squares converge to $1$. $\endgroup$
    – bathalf15320
    Commented Dec 1, 2021 at 8:44
  • $\begingroup$ Thank you @bathalf15320 , but I am not sur if the product r_n^2 is well defined in the sens of generalized functions, for exemple the Heaviside function verifies that H.H = H but the product H.H is not well defined becuse does not satisfy the leibniz rule. $\endgroup$
    – Didine Zaidni
    Commented Dec 1, 2021 at 12:42
  • $\begingroup$ It depends on how you define the product, i.e., under which conditions on the distributions. In my book, it is well-defined but if this bothers you, you can smooth them out to very good $C^\infty$ approximations by the usual methods. $\endgroup$
    – hordubal
    Commented Dec 1, 2021 at 14:33
  • $\begingroup$ Let u and v be distributions in D′(U). Assume that there is no point (x, k) in WF(u) such that (x, −k) belongs to WF(v), then the product uv can be defined. (Lars Hörmander) $\endgroup$
    – Didine Zaidni
    Commented Dec 1, 2021 at 18:44

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .