Timeline for Approximate identities: a converse question

8 events

when toggle format	what		by	license	comment
S May 10, 2020 at 20:02	history	suggested	CommunityBot	CC BY-SA 4.0	I added an assumption that may make the problem easier to answer.
May 10, 2020 at 19:58	review	Suggested edits
S May 10, 2020 at 20:02
May 9, 2020 at 23:09	comment	added	Mateusz Kwaśnicki		If $f$ corresponds to, say, a bounded distribution (a continuous linear functional on the class of test functions $\phi$ such that all derivatives of $\phi$ are absolutely integrable), then $K_\lambda * f$ converges to $f$ in the space of bounded distributions, and to $g$ in $L^2$. Thus, $f = g$ almost everywhere. That said, I do not know if it is sufficient to assume that the integrals in the definition of $K_\lambda * f$ are absolutely convergent.
May 9, 2020 at 22:39	history	edited	nickkatz2018	CC BY-SA 4.0	added 106 characters in body
May 9, 2020 at 22:37	comment	added	nickkatz2018		If $f$ is measurable, then the convolution function $K_{\lambda}*f$ is well-defined. I have added the condition that it is finite valued at every $x$.
May 9, 2020 at 22:09	comment	added	Christian Remling		How do we define $K_{\lambda}*f$ if we only know that $f$ is measurable?
May 9, 2020 at 20:19	history	edited	nickkatz2018	CC BY-SA 4.0	added 6 characters in body
May 9, 2020 at 19:53	history	asked	nickkatz2018	CC BY-SA 4.0