Timeline for Are all continuous linear operators on the space of entire functions "simple"?

11 events

when toggle format	what		by	license	comment
May 7, 2011 at 11:06	vote	accept	CommunityBot
May 6, 2011 at 9:38	answer	added	Neil Strickland		timeline score: 0
May 6, 2011 at 9:18	comment	added	Marc Palm		What about restricting your functions to a compact subset of the reals and Fourier transforming it?
May 6, 2011 at 9:15	comment	added	user5810		@Denis, that cuts it down to $L_1,...,L_4$. (as I just edited to reflect)
May 6, 2011 at 9:12	history	edited	user5810	CC BY-SA 3.0	subscripted function correctly
May 6, 2011 at 9:11	history	rollback	user5810		Rollback to Revision 2
May 6, 2011 at 9:06	history	edited	user5810	CC BY-SA 3.0	added composition
May 6, 2011 at 9:02	comment	added	Marc Palm		Why would you expect this? You mean probably $S$ being the algebra generated by the operators? $S$ will seperate points using a Taylor expansion in $0$, but how would you get $f(g(z))$ or $g(f(z))$, e.g. start $f( \alpha z)$ for some $\alpha \in \mathbb{C}$.
May 6, 2011 at 9:02	comment	added	Denis Serre		It seems to me that you forget the following operators: $$(L_6(f))(z)=f\circ g(z),$$ where $g\in$Ent is given.
May 6, 2011 at 8:45	history	edited	user5810	CC BY-SA 3.0	changed "constant" to "complex numbers"
May 6, 2011 at 8:39	history	asked	user5810	CC BY-SA 3.0