Let $c_0 = \{x \in l^\infty \ | \ x_i \ \to 0 \}$. Show that $c_0$ is not weakly dense in $l^\infty$.

This means that the closure of $c_0$ w.r.t the weak topology is not $l^\infty$, right?

Does that mean that $\exists x\in l^\infty$ such that $\not\exists (x_1, x_2, ....), x_i \in c_0: x_i \to^w x $?

  • 1
    $\begingroup$ Weak topology is not metrizable so you cannot use sequences to characterize denseness. $\endgroup$ Jan 9 at 11:31
  • 1
    $\begingroup$ Hint: The (geometric) Hahn-Banach theorem might come handy. $\endgroup$
    – MaoWao
    Jan 9 at 11:41
  • $\begingroup$ This may help. $\endgroup$ Jan 9 at 12:13

1 Answer 1


Here is a sketch as to how you can proceed:

Claim: If $C$ is a convex subset of a normed space, then the weak closure and the norm closure of $C$ are equal.

Proof: Apply the Hahn-Banach separation theorem. Details are left for you. $\quad \square$

Corollary: The weak closure of $c_0$ is equal to the norm closure of $c_0$ in $\ell^\infty$. Since $c_0$ is norm-closed in $\ell^\infty$, we deduce that $c_0$ is weakly closed in $\ell^\infty$ as well. So basically you are asked to show that the inclusion $c_0 \subseteq \ell^\infty$ is strict, which is trivial (consider the function $n \mapsto 1$).


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