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Show that $(\mathcal{l}^\infty)^*$ is not homeomorphic to $\mathcal{l}^1$, by showing that the dual of $\mathcal{l}^\infty$ contains a proper subspace which is homemorphic to $\mathcal{l}^1$.

L is homemorphic if it is linear, one-to-one, continuous, and has a continuous inverse. I have also proved in class that L is homeomorphism if and only if it is surjective and $\exists A, B>0$ so that $B||x||_x \leq ||L(x)||_Y \leq A||x||_x$.

I don't know how to get start with the proof can I have some help starting it off?

Username Unknown
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2 Answers2

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It is not true that "if $X$ contains an isomorphic copy of $Y$, then $X$ is not isomorphic to $Y$". For example $X=\ell_2$ is isomorphic to its subspace $Y=\{x\in\ell_2:x_1=0\}$.

Now, $\ell_\infty^*$ indeed contains the copy of $\ell_1$. I leave it to you to show that $$ L:\ell_1\to\ell_\infty^*:x\mapsto f_x $$ is an isometric embedding. Here $f_x$ is a bounded linear functional defined as $$ f_x:\ell_\infty\to\mathbb{C}:y\mapsto \sum_{i=1}^\infty x_iy_i $$

But, $\ell_\infty^*$ is not isomorphic to $\ell_1$. Since $\ell_\infty$ is not separable, so does its dual $\ell_\infty^*$. Since $\ell_1$ is separable it can not be isomorphic to non-separable space $\ell_\infty^*$.

Community
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Norbert
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  • Is it possible to use another fact instead of separable? – Username Unknown Mar 19 '14 at 05:44
  • Non-existence-of-isomorphisms-proofs are always difficult and require knowledge of properties of both spaces which are preserved under isomorphisms. The separability is the simplest one, I think. – Norbert Mar 19 '14 at 05:50
  • Separability is not a hard proof here, though. I can write it up if you want it, @UsernameUnknown. –  Mar 19 '14 at 06:02
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You can find a bounded linear functional in $(\ell^\infty)^* $ that does not correspond to an element of $\ell^1$ in the following way:

Consider the following linear subspace of $\ell^\infty$: $$W=\{a=(a_1,a_2,\dots)\in \ell^\infty : \lim_{n\to\infty }a_n\textrm{ exists}\}$$ and define the linear functional $\lambda:W\to \mathbb R$ by $$\lambda (a)=\lim_{n\to\infty } a_n$$ it is easy to see that this is bounded and $\|\lambda\|\leq1$. Now, using the Hahn- Banach theorem extend this functional to a bounded linear functional on $\ell^\infty$, such that $$\lambda(a)=\lim_{n\to\infty} a_n$$ for $a\in W$. This extended functional does not correspond to any element $b\in \ell^1$. If it did, then we would be able to write $$\lambda(a)= \sum_{n=1}^\infty a_nb_n$$ for all $a\in \ell^\infty$. However, if you test $e_n=(0,\dots,0,1,0,\dots)\in \ell^\infty$ you get $\lambda(e_n)=0$ using the first definition, and $\lambda(e_n)=b_n$, using the second. This implies that $b_n=0$ for all $n$, therefore $\lambda=0$. That is a contradiction because $\lambda$ is a non-trivial functional. For example, if $a=(1,1,\dots)$ then $\lambda(a)=\lim_{n\to\infty}a_n=1$.

Dimitris
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  • But this doesn't proof that $\ell_1$ is not isomorphic to $\ell_\infty^$, it proves only that the copy of $\ell_1$ in $\ell_\infty^$ is a proper subspace! – Norbert Mar 19 '14 at 07:51