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I'm reading some papers where the condition of weak sequential continuity is crucial instead of the weak continuity.

So, let

  • $T$ an operator between a Banach space $X$ and itself.
  1. $T$ is weakly sequentially continuous if for any $(x_n)_{n\in \mathbb N}$, $x_n \stackrel{w}{\rightharpoonup}x$ in $X$ $\Rightarrow$ $T(x_n) \stackrel{w}{\rightharpoonup } T( x )$ in $X$)*.
  2. $T$ is weakly continuous iff $T$ is continuous with respect to the weak topologies on $X$

While it is clear that 2 implies 1, I want to find a counter-example of an operator which verifies 1 but not 2.

$T$ is not linear

Improve this question
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    $\begingroup$ 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 $T$. $\endgroup$
    – Jochen Glueck
    Commented Sep 23, 2020 at 12:19
  • $\begingroup$ I added the "Banach spaces" tag. By the way, what does the symbol "$)^*$" mean at the end of assertion 1.? $\endgroup$
    – Jochen Glueck
    Commented Sep 23, 2020 at 12:21
  • $\begingroup$ $T$ is not linear $\endgroup$
    – Motaka
    Commented Sep 23, 2020 at 13:45
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    $\begingroup$ 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 $X$. $\endgroup$
    – fedja
    Commented Sep 23, 2020 at 14:00

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