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I want to show $\ell_1$ is not reflexive. And I have already shown that $c_0^*$ is not reflexive. And I know there is an isometric isomorphism between $\ell_1$ and $c_0^*$. How to use the isometric isomorphism to show $\ell_1$ is not reflexive?

Can anyone help me out? Thank you in advance!

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  • $\begingroup$ Hint: use contradiction. $\endgroup$ – Paul Sundheim May 9 '18 at 20:22
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You can use the following theorem:

Let $X$ and $Y$ be normed spaces and $T : X \to Y$ an isometrical isomorphism. Then the dual map $T^* : Y^* \to X^*$ is an isometrical isomorphism.

The map $T^*$ acts as $T^*f = f \circ T$ for all $f \in Y^*$.

The proof is outlined here. Basically, you need to show that $(AB)^* = B^*A^*$ and that $I^* = I$. It follows that $(A^*)^{-1} = (A^{-1})^*$ for invertible maps $A$. Now use that if $T$ is an isometrical isomorphism then $T^{-1}$ and $T^*$ are also isometries.

Since $\ell^1 \cong (c_0)^*$ then the theorem implies $(\ell^1)^{**} \cong (c_0)^{***}$. If $\ell^1$ were reflexive, we would have

$$(c_0)^* \cong \ell^1 \cong (\ell^1)^{**} \cong (c_0)^{***}$$

which contradicts the fact that $(c_0)^*$ is not reflexive.

Edit:

As pointed out by @Rhys Steele, we still must check that the canonical embedding $\wedge : (\ell^1)^* \hookrightarrow (\ell^1)^{***}$ is not an isometrical isomorphism.

Take an isometrical isomorphism $T : (c_0)^* \to \ell^1$ and consider $F : (c_0)^* \to (c_0)^{***}$ given by $F : (T^{-1})^{**} \circ \wedge \circ T$.

For $x \in (c_0)^*$ we have

$$F(x) = ((T^{-1})^{**} \circ \wedge \circ T)(x) = (T^{-1})^{**} (\widehat{Tx}) = \widehat{Tx} \circ (T^{-1})^*$$

so for $f \in (c_0)^{**}$ we have

$$F(x)f = (\widehat{Tx} \circ (T^{-1})^*)f = \widehat{Tx}(f \circ T^{-1}) = f(T^{-1}Tx) = f(x)$$

Therefore, $F$ is precisely the canonical embedding $(c_0)^* \hookrightarrow (c_0)^{***}$. If we assume $\wedge : (\ell^1)^* \hookrightarrow (\ell^1)^{***}$ to be an isometrical isomorphism, then so is $F$ which contradicts the fact that $(c_0)^*$ is not reflexive.

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    $\begingroup$ There are non-reflexive Banach spaces that are isomorphic to their bidual via a different map than the canonical embedding. To make this work you also need to check that the isometric isomorphism from $(c_0)^*$ to $(c_0)^{***}$ would be the canonical embedding (which turns out to be not that hard to check). $\endgroup$ – Rhys Steele May 10 '18 at 6:39
  • $\begingroup$ @RhysSteele Ah, you are right. I was supposing that $X$ is reflexive if $X \cong X^{**}$. Btw. do you have a counterexample in mind? $\endgroup$ – mechanodroid May 10 '18 at 8:28
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    $\begingroup$ My understanding is that counterexamples aren't exactly easy to come by. The example I'm aware of was constructed by R.C. James in the 1951 paper "A non-reflexive Banach space isometric with its second conjugate space". $\endgroup$ – Rhys Steele May 10 '18 at 8:42
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The relevant result is:

Theorem: Let $X$ be a normed vector space. Then $X$ is reflexive if and only if $X$ is complete and $X^*$ is reflexive.

Since $c_0$ is complete and $(c_0)^*$ is isometrically isomorphic to $\ell^1$, if $\ell^1$ is reflexive then so is $c_0$ (using the fact that reflexivity is preserved by isomorphism).

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  • $\begingroup$ Thanks for your reply but I want to know why the reflexivity is preserved by isometric isomorphism. $\endgroup$ – Answer Lee May 9 '18 at 20:28
  • $\begingroup$ In that case you should attempt to show that if $J_X$ is the usual embedding of $X$ into its bidual and $T$ is a bounded linear map from $X$ to $Y$ then $T^{**} \circ J_X = J_Y \circ T$. Then apply this with $T$ being the isomorphism. $\endgroup$ – Rhys Steele May 9 '18 at 20:31
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This is no exactly what you ask but I think this is interesting proof that $\ell^{1}$ is not reflexive.

(Kakutani). Let $E$ be a Banach space. Then E is reflexive if and only if $$ B_E = \{x ∈ E; \|x\| ≤ 1\} $$ is compact in the weak topology $\sigma(E, E^{\star})$.

Now put $E=\ell^1$, so $E^{\star}=\ell^{\infty}.$ Consider the sequence $e_n=(0,0,\cdots,0, 1, 0, 0\cdots)$ in $\ell^1$. Suppose that $B_{\ell^1}$ is compact in the $\sigma(\ell^1, \ell^{\infty})$ topology. So there must exists a convergent subsequence $(e_{n_k})$ and $x\in \ell^1$ s.t for all $f\in \ell^\infty$ we have that $\left<f,e_{n_k} \right>\to \left<f,x \right>.$ So consider $f\in \ell^{\infty}$ of the following form: $$ f=(0,\cdots,0,\underbrace{-1}_{n_1}, 0, \cdots,0,,\underbrace{1}_{n_2}, 0, \cdots, 0,\cdots ,\underbrace{-1}_{n_3}, 0, \cdots) $$ then $\left<f,e_{n_k} \right>=(-1)^k$ which does not converges giving a contradition.

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  • $\begingroup$ It might be worth remarking that you are making use of the Eberlein-Smulian theorem to conclude that weak compactness implies weak sequential compactness. $\endgroup$ – Rhys Steele May 9 '18 at 20:58

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