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Let $X$ and $Y$ be Banach spaces, and fix a bounded linear operator $A \in \mathcal{B}(X, Y)$. Choose $\mu \in Y^{*}$, and define a functional $A^{*} \mu: X \rightarrow \mathbf{F}$ by $\left(A^{*} \mu\right)(x)=\mu(A x)$, for $x \in X$. I want to show that the mapping $A^*\colon \mu \mapsto A^{*} \mu$ is a bounded linear mapping of $Y^{*}$ into $X^*$. Linear part is easy and my thought about bounded part is \begin{align}\|A^*\|&=\sup_{\|\mu\|=1}\|A^*\mu\|=\sup_{\|\mu\|=1}(\sup_{\|x\|=1}\|(A^*\mu)(x)\|)\\&=\sup_{\|\mu\|=1}(\sup_{\|x\|=1}\|\mu(Ax)\|)=\sup_{\|\mu\|=1}\|\mu A\|=\|A\| \end{align} Is that correct? Also, is there a way to show $\|A\|=\|A^*\|$?Any help is appreciated.

A.Γ.
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Maskoff
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[The last part was added later and answer to that part is in my comments below].

The proof is OK if the last equality is replaced by an inequality :$ \|\mu (A) \|\leq \|A\|$ if $\|\mu \| \leq 1$. I don't think equality holds in general.

Kavi Rama Murthy
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  • Is there a way to show $|A|=|A^*|$ cause this is the other part of the exercise – Maskoff Apr 15 '20 at 08:30
  • You have proved that $|A^{}| \leq |A|$ Change $A$ to $A^{}$. You get $|A^{}| \leq |A^{*}|$. But $|A| \leq |A^{}|$ (Why?)/. Hence equality holds. @maskliesink – Kavi Rama Murthy Apr 15 '20 at 08:33
  • I'm not exactly sure what $A^{**}$ is defined, does it maps $X$ into $Y$? – Maskoff Apr 15 '20 at 09:04
  • It maps $X^{}$ into $Y^{}$ and it is an extension of $A$. @maskliesink – Kavi Rama Murthy Apr 15 '20 at 09:05
  • I'm kinda confused about this, could you please show a little bit more details? – Maskoff Apr 15 '20 at 09:07
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    @maskliesink If you know what $X^$ is then you can define $X^{}=(X^)^$ and $A^{}=(A^)^$. A standard easy exercise (just by definition) is to show that $X$ is a natural part of $X^{}$, and $A^{*}$ "coincides with" (=works as) $A$ on that part. – A.Γ. Apr 15 '20 at 11:57
  • I think I get it now. Thanks for all comments! – Maskoff Apr 17 '20 at 04:51