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.