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Let $E,F$ be Banach spaces. The dual spaces (sets of all linear and bounded functionals) are denoted $E',F'$. Now take a linear, bounded operator $T:E \to F$. Then $T':F' \to E', x' \mapsto x'\circ T$ denotes the adjoint operator of $T$.

Now, one can show that if $T$ is an isomorphism (meaning: a linear homeomorphism), then $T'$ is one too, i.e., $E$ and $F$ are isomorphic implies $E'$ and $F'$ are isomorphic too. I want to prove (or a find a counterexample) that the other implication holds too, i.e. $T'$ is an isomorphism than $T$ must be too. I think it is true but I don't know how to prove it and hope someone can help me out.

Note that the statement is wrong in general (i.e. if $E,F$ are not necessarlily complete). Take $E$ to be uncomplete, $F$ its completion and $T:E \to F, x \mapsto x$ the canonical map. Than $T'$ becomes an isomorphism, while $T$ is none.

$\mathbf{EDIT:}$ This question used to be bigger and divided in three parts. The previous $2.)$ part is now the whole question.

Tina
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    This question is similar to: A Banach space with multiple preduals. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. – J. De Ro Aug 29 '24 at 15:09
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    @J.DeRo That only answers OP’s question 3, and then only if isomorphism in the OP’s question is interpreted as isometric isomorphism, which is not standard terminology in Banach space theory. – David Gao Aug 29 '24 at 15:15
  • @J.DeRo Thats a good comment thank you, but thats not a complete solution because by isomorphism i meant only linear homeomorphism. – Tina Aug 29 '24 at 15:22
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    I think David is implicitly saying this, but the examples in the question linked above of $c$ and $c_0$ (convergent sequences and sequences tending to $0$) while not isometric are isomorphic in the sense that there is a continuous linear isomorphism $c\cong c_0$. -- if $(x_n)$ is a sequence and $x_n\to \ell$ then $x_n = y_n+\ell$ where $y_n \to 0$, so setting $T(x_n) = (\ell,y_1,y_2,\ldots)$ gives a linear isomorphism, and $T$ and its inverse have norm $2$. – krm2233 Aug 29 '24 at 15:28

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This is true. Assume $T'$ is an isomorphism with inverse $S: E' \to F'$. Now, to check $T$ is an isomorphism, it suffices to check it is bounded below (and therefore both injective and with closed range) and has dense range. Let $x \in E$. By Hahn-Banach, there is a $\varphi \in E'$ of norm $1$ s.t. $\varphi(x) = \|x\|$. Hence,

$$\|x\| = \varphi(x) = (T'S\varphi)(x) = (S\varphi)(Tx) \leq \|S\|\|Tx\|$$

So $T$ is bounded below. To show its range is dense, assume otherwise. Then by Hahn-Banach, there exists a $\varphi \in F'$ of norm $1$ s.t. $\varphi(\text{range}(T)) = 0$. But then $T'\varphi = \varphi \circ T = 0$, so $T'$ is not injective, a contradiction.

David Gao
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  • Good answer! Do you have any reference for bounded below implies injective and closed range. – Tina Aug 29 '24 at 16:01
  • @TimonKolt Ah, that's great, I was trying and failing to find that reference. Regardless, I have already added a proof for the latter fact in the answer. – David Gao Aug 29 '24 at 16:13
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    @TimonKolt I don't have a reference for bounded below implies injective and closed range right now. But it is quite a standard fact and not hard to prove. I'd suggest you try to prove it yourself. If you get stuck, I can add in a proof later as well, or you can search for a reference - it shouldn't be hard to find. (Hint for proving closed range: it suffices to prove the range is complete, and you should use the fact that the domain is complete as well.) – David Gao Aug 29 '24 at 16:16
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    Yes thank you! I figured it out by myself – Tina Aug 29 '24 at 16:29