# Natural tramsformation between a normed linear space and it's double dual.

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.

• Yes. If $T$ and $T^{-1}$ are bounded, they preserve Cauchy-ness of a sequence. – Berci Oct 2 '18 at 6:34
• That means every reflexive normed linear space is a Banach space. Am I right? – Dbchatto67 Oct 2 '18 at 7:59
• Is the converse true? i.e. if X is a Banach space is it necessarily reflexive? – Dbchatto67 Oct 2 '18 at 8:00
• 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. – eddie Oct 2 '18 at 8:52