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Let $X$ be a normed linear space. Then I know that $X$ is imbedded in it's double dual $X^{**}$ via the natural tramsformation $J : X \longrightarrow X^{**}$ (say). I have proved that $J$ is one-to-one, bounded and $\|J\|=1$ i.e. $J$ is continuous. Moreover $J$ is an isometry. Now let us assume that $X$ is reflexive then $J(X)=X^{**}$. Hence $X$ and $X^{**}$ are isomorphic to each other as normed linear spaces. Now $X^{**}$ is a Banach space. From here can we conclude that $X$ is a Banach space?

In general can we say that

If a normed linear space $X$ is isomorphic to a Banach space $Y$ via a linear isomorphism $T:X \longrightarrow Y$. Then $X$ is also a Banach space?

Please help me in this regard. Thank you very much.

EDIT $:$

Since $J$ is an isometry so is $J^{-1}$. Hence $J^{-1}$ is bounded with $\|J^{-1}\|=1$ and hence $J^{-1}$ is continuous. Consider a Cauchy sequence $\{x_n \}$ in $X$. Then by the isometric property of $J$ we can easily see that $\{J(x_n) \}$ is a Cauchy sequence in $X^{**}$. Now since $X^{**}$ is complete. So $\exists$ $\widehat x \in X^{**}$ such that $J(x_n) \rightarrow \widehat x$ as $n \rightarrow \infty$. Then by the sequential criterion for continuity of $J^{-1}$ it then follows that $x_n \rightarrow J^{-1} (\widehat x)$ as $n \rightarrow \infty$, proving that $X$ is complete and hence $X$ is also a Banach space.

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  • $\begingroup$ Yes. If $T$ and $T^{-1}$ are bounded, they preserve Cauchy-ness of a sequence. $\endgroup$ – Berci Oct 2 '18 at 6:34
  • $\begingroup$ That means every reflexive normed linear space is a Banach space. Am I right? $\endgroup$ – Dbchatto67 Oct 2 '18 at 7:59
  • $\begingroup$ Is the converse true? i.e. if X is a Banach space is it necessarily reflexive? $\endgroup$ – Dbchatto67 Oct 2 '18 at 8:00
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    $\begingroup$ It's true that every reflexive normed space is a banach space. However the converse is not true. $L^\infty([0,1])$ is a banach space but not reflexive. $\endgroup$ – eddie Oct 2 '18 at 8:52

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