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 $l^{\infty}(\mathbb{N})?$

  • $\begingroup$ 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


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.

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

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