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Im reading some papers where the condition of weak sequential continuity is crucial, and this question come to my mind:

Why it's more interesting to use this condition instead of the classical weak continuity?

Context:

  • $T$ between a Banach space X and itself.
  • $T$ 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 $X$).
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  • $\begingroup$ All of these conditions boil down to norm continuity of $T$. But of course some might be easier to verify in concrete situations. $\endgroup$
    – Ruy
    Sep 21, 2020 at 20:25
  • $\begingroup$ Can you give more explication, I will appreciate it! $\endgroup$
    – Motaka
    Sep 22, 2020 at 9:47

1 Answer 1

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Theorem. Let $X$ and $Y$ be normed vector spaces and let $T:X\to Y$ be a linear map. Then the following are equivalent:

  1. $T$ is continuous,
  2. $T$ is weakly continuous,
  3. $T$ is sequentially weakly continuous,
  4. For every sequence $\{x_n\}_{n\in {\bf N}}$ in $X$ which is weakly convergent to $0$, one can find a weakly converging subsequence $\{T(x_{n_k})\}_{k\in {\bf N}}$.

Proof. The implications (1) $\Rightarrow$ (2) $\Rightarrow$ (3) $\Rightarrow$ (4) are trivial, so let us prove that (4) $\Rightarrow$ (1). Arguing by contradiction assume that $T$ is not continuous, hence unbounded on the unit ball of $X$. So we can find a sequence $\{y_n\}_{n\in {\bf N}}$ in that ball such that $\Vert T(y_n)\Vert >n^2$, for every $n$. Setting $x_n=y_n/n$, we have that $x_n\to 0$ in norm, so also weakly, while $$ \Vert T(x_n)\Vert = {1\over n} \Vert T(y_n)\Vert > {1\over n}n^2 = n, $$ so no subsequence of $\{T(x_n)\}_{n\in {\bf N}}$ is weakly convergent because weakly convergent sequences are bounded (by a well known Corollary of the uniform boundedness principle). This contradicts (4), so the proof is complete. QED

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  • $\begingroup$ Unfortunately, $T$ is not supposed to be linear, but thanks for your answer anyway.. $\endgroup$
    – Motaka
    Sep 22, 2020 at 13:40
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    $\begingroup$ Oh! I am sorry. But since the majority of facts about functions between Banach spaces involve linear maps, it is perhaps wise to emphasize it when you are considering non-linear maps! $\endgroup$
    – Ruy
    Sep 22, 2020 at 13:44

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