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Let $X$ be a Banach space and $X^*$ its dual

Does anyone have an example where $X$ does not embed isomorphically/isometrically into $X^*$.

Does $C([0,1])$ embed isomorphically/isometrically into its dual?

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  • $\begingroup$ You forgot to mention that $X$ is infinite dimensional. $\endgroup$ – Kavi Rama Murthy Aug 5 '18 at 23:28
  • $\begingroup$ Yes, $X$ is infinite dimensional $\endgroup$ – Tom Chalmer Aug 6 '18 at 0:12
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    $\begingroup$ Isn't $L^p$ a counterexample for suitable $p$? $\endgroup$ – David C. Ullrich Aug 6 '18 at 3:31
  • $\begingroup$ I think you have to make clear what you mean by "embed isomorphically". To me, an isomorphism is bounded, injective and surjective. Is this what you mean? Why do you use "embed" then? $\endgroup$ – amsmath Aug 9 '18 at 11:48
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If $X=l^{p}$ with $1<p<\infty $ then $X^{*}$ does not contain an isomorphic copy of $C[0,1]$. To prove this we need two well known theorems from FA:

Theorem 1

(Pitt) and bounded linear map from $l^{r}$ to $l^{s}$ is compact if $1<s<r<\infty$.

Theorem 2

$C[0,1]$ is universal for separable Banach spaces (in the sense any separable Banach space can be embedded isometrically in $C[0,1]$). Ref.: Geometric FA by Holmes.

Now let $q>p^{*}$ where $p^{*}$ is the conjugate of $p$. Suppose there is an isomorphism $T: C[0,1] \to X^{*}=l^{p^{*}}$. Let $S:l^{q} \to C[0,1]$ be an isometric isomorphism (into). Then $T\circ S$ is compact by Pitt's Theorem. But this map is also an isomorphism. It is easy to see that there cannot be an isomorphism on an infinite dimensional space which is compact.

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  • $\begingroup$ For the last part note that $\|T(S(x))\| \geq C \|x\|$ for some $c$. If $\{x_n\}$ is any sequence in the unit ball of $l^{q}$ then $T(S(x_n))$ has a convergent subsequence and the inequality implies that the corresponding subsequence of $\{x_n\}$ is Cauchy. This makes the unit ball of $l^{q}$ compact, a contradictioin. $\endgroup$ – Kavi Rama Murthy Aug 6 '18 at 4:57
  • $\begingroup$ Hi Kavi. Thanks for your answer. I think my question was a little ambiguous, or I may not have understood your answer. It seems you showed there is a dual that does not contain an isomorphic copy of $C([0,1])$. I'm wondering if $C([0,1])$ embeds into its dual $C([0,1])^*$ $\endgroup$ – Tom Chalmer Aug 6 '18 at 6:18
  • $\begingroup$ $X$ does not embed isomorphically in $X^{*}$ if $X=l^{p}$ with $p>2$ (by the Pitt's Theorem argument). I do not know if $C[0,1]$ embeds in its dual. $\endgroup$ – Kavi Rama Murthy Aug 6 '18 at 7:31

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