Timeline for A weakly sequentially continuous operator which is not weakly continuous

8 events

when toggle format	what		by	license	comment
Sep 23, 2020 at 14:00	comment	added	fedja		Then just use some fancy space $X$ like $\ell^1$ in which the weak convergence (of sequences) is equivalent to the norm convergence but the weak topology is substantially different from the norm one. For the operator, just take $Tx=\\|x\\|e$ where $e$ is any non-zero vector in $X$.
Sep 23, 2020 at 13:45	history	edited	Motaka	CC BY-SA 4.0	added 25 characters in body
Sep 23, 2020 at 13:45	comment	added	Motaka		$T$ is not linear
Sep 23, 2020 at 12:21	comment	added	Jochen Glueck		I added the "Banach spaces" tag. By the way, what does the symbol "$)^*$" mean at the end of assertion 1.?
Sep 23, 2020 at 12:20	history	edited	Jochen Glueck		Added the tag "Banach spaces".
Sep 23, 2020 at 12:19	comment	added	Jochen Glueck		Both assertions are equivalent, and they are also equivalent to "3. $T$ is norm continuous". Proof: "3. => 2." This follows from the existence of the dual operator. "2. => 1." As mentioned in the question, this is obvious. "1. => 3": Let $(x_n)$ be a sequence in $X$ which converges to $0$ in norm. Then $(x_n)$ converges also weakly to $0$. Hence, $(Tx_n)$ converges weakly to $0$ by 1. Thus, $(Tx_n)$ is bounded by the uniform boundedness principle. So $T$ maps sequencs that norm converge to $0$ to bounded sequences; this is sufficient for norm continuity of $T$.
Sep 23, 2020 at 11:27	history	edited	Martin Sleziak	CC BY-SA 4.0	a minor typo
Sep 23, 2020 at 11:25	history	asked	Motaka	CC BY-SA 4.0