2
$\begingroup$

Let $l^{\infty}(\mathbb{N})$ denotes the set of all bounded sequences, which has weak* topology as it is the dual of $l^{1}(N).$ Let $c_0$ denotes the subspace consisting of sequences which converges to zero. I have the following question:

Is the norm closed unit ball of $c_0$ is weak* dense in the norm closed unit ball of $l^{\infty}(\mathbb{N})?$

$\endgroup$
2
  • $\begingroup$ Notice that the dual of $c_{0}$ is $\mathscr{l}^{1}$ if that is to any help $\endgroup$ Nov 23, 2015 at 2:16
  • $\begingroup$ Let $a=(a_i)_{i=1}^\infty\in\ell_\infty$ and let $a|_n$ be the sequence consisting of the first $n$ terms in $a$ followed by zeros. Does $a|_n$ go weak* to $a$? Let $x=(x_i)_{i=1}^\infty\in\ell_1$, and notice that $|(a|_n-a)(x)|=|\sum_{i=n+1}^\infty a_ix_i|\leq\|a\|_\infty\sum_{i=n+1}^\infty|x_i|\to 0$ as $n\to\infty$. $\endgroup$
    – Ben W
    Nov 23, 2015 at 6:53

1 Answer 1

3
$\begingroup$

As explained by anonymous, this can be achieved by a cut-off argument.

However, it can be proved in a more general way. Let $X$ be a Banach space, $J : X \to X^{**}$ the canonical embedding into the bidual.

For a Banach space $Y$, we denote by $B_Y$ the unit ball of $Y$.

Now, we will show that $J(B_X)$ is weak-$*$ dense in $B_{X^{**}}$. We will use the bipolar theorem in the dual pairing of $X^{**}$ (with the weak-$*$ topology) and $X^*$ (with the weak topology). It is easy to show $$J(B_X)^\circ := \{x^* \in X^* : J(x)(x^*) \le 1 \; \forall x \in B_X\} = B_{X^*}$$ and $$(B_{X^*})^\circ := \{x^{**} \in X^{**} : x^{**}(x^*) \le 1 \; \forall x^* \in B_{X^*}\} = B_{X^{**}}.$$ Hence, $$J(B_X)^{\circ\circ} = B_{X^{**}}$$ and the bipolar theorem tells you that the weak-$*$ closure of $J(B_X)$ is $B_{X^{**}}$.

(Note that $J(B_X)$ is always closed and, if $X$ is not reflexive, we have $J(B_X) \ne B_{X^{**}}$ and $J(B_X)$ is not weak-$*$ closed.)

To apply this result to your situation, one recalls that the dual of $c_0$ is isometric to $\ell^1$ and the dual of $\ell^1$ is isometric to $\ell^\infty$. Moreover, under these identifications, the canonical embedding $J : c_0 \to \ell^\infty$ is the identity.

$\endgroup$
1
  • 1
    $\begingroup$ This is a straightforward application. Yes, the weak* density also follows. $\endgroup$
    – gerw
    Apr 11 at 19:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.