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Show that every closed linear subspace $ \ Y \ $ of a Banach space $ \ X \ $ is weakly sequentially closed ,

that is , $ \ Y \ $ contains weak limits of all weakly convergent sequences $ \ \{y_n \} \ $ in $ \ Y \ $.

Answer:

We know closed subspace of a Banach space is also a Banach space.

Thus $ \ Y \ $ is a Banach space or Completely normed linear space.

That means every cauchy sequence in $ \ Y \ $ converges in it.

Now let $ \ \ y_n \to \ y \ $ weakly.

To finish the proof , we have to show that the weak limit $ \ y \ \in Y $ .

But I got no way to show it.

Help me doing this.

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  • $\begingroup$ Why $\quad$ the we$\ $ird $\qquad$ spacin$\,$g? $\endgroup$ – Asaf Karagila Jun 24 '18 at 15:53
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$Y$ is weakly closed (strongly closed linear subspaces are weakly closed, this follows from Hahn-Banach, essentially; see the argument here e.g.).

In any topology (so also the weak topology on a Banach space), being closed implies being sequentially closed: if $A$ is closed and $a_n \to a$, then let $O$ be an open neighbourhood of $a$. By convergence, $O$ contains a tail of $(a_n)$, so in particular, $O \cap A \neq \emptyset$. As $O$ was arbitary, $a \in \overline{A} = A$, so $A$ is sequentially closed.

So $Y$ is weakly sequentially closed.

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Let $Y$ be a closed subspace of $X$. Then $Y$ is weakly closed (and hence weakly sequentially closed).

To see this, consider its annihilator $Y^0 = \{f \in X^* : f|_Y = 0\} \subseteq X^*$. We have

$$Y = \bigcap_{f \in Y^0} \ker f$$

Namely, clearly $Y \subseteq \ker f$ for every $f \in Y^0$ so $Y \subseteq \bigcap_{f \in Y^0} \ker f$.

Conversely, let $x \in X \setminus Y$. A corollary of Hahn-Banach states that there exists $f \in X^*$ such that $f(x) = 1$ and $f|_Y = 0$. Then $x \notin \ker f$ so we conclude $\bigcap_{f \in Y^0} \ker f \subseteq Y$.

Now, for $f \in X^*$ notice that $\ker f$ is a weakly closed set as a preimage of $\{0\}$ by a weakly continuous function $f$. Hence $Y$ is weakly closed as an intersection of weakly closed sets.

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