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Let $E$ be a Banach space, $T:E\rightarrow E$ a non-linear operator.

$T$ is said to be Weakly Sequentially Continuous (shortly W.S.C) on $E,$ if for every $\left(x_{n}\right)_{n}\subseteq E$ with $x_{n} \rightharpoonup x,$ we have $T x_{n} \rightharpoonup T x$.

In papers they use this type of continuity because, "although it is not always possible to show that a given mapping between functional Banach spaces is weakly continuous, quite often its weak sequential continuity can be checked easily. This follows, among other things, from the fact that Lebesgue’s dominated convergence theorem is valid for sequences but it fails for nets".

I have some questions here about strong continuity (in the strong topology) and W.S.C,

  1. In a Banach space, which one implies the other one (if there is an implication between them)? If there are some examples, I will be very grateful
  2. Let's suppose that I am working under the weak topology and I have a result that works for both strong continuity or W.S.C, which one is more interesting to choose?
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    $\begingroup$ I assume that $T$ is allowed to be non-linear again, as in your previous question? If yes, then I suggest to replace "an operator" in your first sentence either with "an (in general, non-linear) operator" or with "a mapping". $\endgroup$
    – Jochen Glueck
    Commented Oct 26, 2020 at 12:10
  • $\begingroup$ Thank you for pointing out this. $\endgroup$
    – Motaka
    Commented Oct 26, 2020 at 13:50
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    $\begingroup$ It is easy to show that a Banach space is a Shur space (every weakly convergent sequence is norm convergent) if and only if the norm is weakly sequentially continuous. Hint: Show that in every non Shur space there is a sequence of unit vectors that converges weakly to zero. $\endgroup$
    – Bill Johnson
    Commented Oct 26, 2020 at 22:26

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