As the title indicates I'm trying to show that $\ell_{\infty}^{*}$ is not $\ell_1$. I've shown that for p, q conjugate and finite we do indeed have $\ell_{p}^{*} = \ell_q$, with the correspondence between $y_i = \phi(e_i)\in \ell_{q}$ and $\phi$ where $e_i$ is the standard (0,....,0,1,0,....) with a 1 in the 'i'th place and $\phi \in \ell_{p}^{*}$.

However, I gather the reason this falls apart in the infinite case is that the dual $\ell_{\infty}^{*}$ is far larger than (but contains) $\ell_1$, being in fact the set of bounded sequences eventually tending to 0. However, I'm having some trouble seeing this. In theory, I'm working over $\mathbb{C}$, but I don't think that's where my issue is arising. It's certainly obvious that $\ell_1$ is contained in $\ell_{\infty}^{*}$: $\sum_i |y_i| < \infty \Rightarrow |\phi(x)| = |\sum_i y_i x_i| \leq (\sum_i |y_i|)(\|x\|_{\infty})$ so $\|\phi\| \le \sum_i |y_i| < \infty$: but then this is my problem, how do we get anything which corresponds to a sequence of $y_i$ -not- in $\ell_1$? I don't think working in the reals is an issue but perhaps I'm overlooking something, so suppose (e.g.) $y_i = \frac{1}{i}$: unless I've misunderstood, this should correspond to a functional since the $y_i$ are bounded and tend to 0, i.e. $\phi(e_i) = \frac{1}{i}$ and extend linearly. But if this is the case, then surely x=(1,1,1,...) gives us an x with norm 1 ($\|x\|_{\infty} = 1$) and for which $\phi$ is unbounded? So in that case this sequence of $y_i$ is a bounded sequence tending to 0 which does not in this way correspond to a bounded linear operator: or is the correspondence no longer in the obvious way? Surely no matter how you look at it each functional gives rise to a sequence of $y_i$, and unless i've confused myself these $y_i$ must be in the space $\ell_1$, which is wrong. So where have I gone wrong?


You assume that any element of the dual must come from a sequence of $y_i$. Equivalently, you assume that the space spanned by the indicator functions $e_i$ is dense in $\ell_{\infty}$. This is false: the closure of the span of the indicator functions $e_i$ is the subspace $c_0$ of $\ell_{\infty}$ of sequences which tend to $0$, and then the existence of an element of $\ell_{\infty}^{\ast}$ not in $\ell_1$ follows by the Hahn-Banach theorem. (But there is an easier way to see that the dual of $\ell_{\infty}^{\ast}$ is not $\ell_1$ which does not produce an explicit element not in $\ell_1$, which I think was the point of your exercise.)

It is known that examples cannot be explicitly written down in the sense that their existence is independent of ZF.

  • $\begingroup$ Yes, if $x$ is in $\ell_\infty\setminus c_0$, the functional on $c_0+\mathbb C x$ defined by $y+\lambda x\mapsto \lambda$ is bounded (with norm $\frac{1}{d(x,c_0)}$) and so extends by Hahn-Banach to a nonzero element of $\ell_\infty^*$ that vanishes on $c_0$. And yes, the independence from ZF came up in the following MathOverflow question: mathoverflow.net/questions/5351/…. Edit: I only commented because of the "I think" parts, but I see those were gone before I posted. $\endgroup$ – Jonas Meyer Apr 27 '11 at 0:50
  • $\begingroup$ Ah I see, thank you for the clarification, that makes sense. What was the easier way of seeing the dual is not $\ell_1$, looking at separability of dual spaces? Interesting that you can not write down an explicit example, that could have saved me 10 minutes of head-scratching, hah. $\endgroup$ – Warner B. Apr 27 '11 at 1:00
  • 1
    $\begingroup$ @Warner: yes, separability is the argument I was thinking of. The argument actually proves that the dual of $\ell_{\infty}$ cannot even be isomorphic to $\ell_1$, whereas the argument I describe above only shows that the obvious inclusion of $\ell_1$ into the dual is not surjective, which is a weaker statement. $\endgroup$ – Qiaochu Yuan Apr 27 '11 at 1:01
  • $\begingroup$ Brilliant, many thanks to both of you. $\endgroup$ – Warner B. Apr 27 '11 at 1:19

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