We say that:
$T:(X,\|\cdot\|_X)\rightarrow (Y,\|\cdot\|_Y)$ is a isometric isomorphism if it is a linear isomorphism, and it is an isometry, that is $\|T(x)\|_Y=\|x\|_X\quad \forall x\in X;$
$T:(X,\|\cdot\|_X)\rightarrow (Y,\|\cdot\|_Y)$ is a topological isomorphism if it is a linear isomorphism and $T$ is continuous with continuous inverse.
My question is: is any isometric isomorphism a topological isomorphism? Since $T$ is an isometry, in particular $\|T(x)\|\leq \|x\|\quad \forall x\in X$, so $T$ is continuous. But what can I say for the inverse? Is it continuous too?
In fact, my problem is to prove that if $E$ is a reflexive n.v.s then $E$ is Banach. So it is sufficient to prove that $J_E:E\rightarrow E''$ is a topological isomorphism, but, by definition of reflexive space, I know that $J_E$ is a isometric isomorphism.
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$\begingroup$It's easy to show that the inverse of an isometry between normed spaces is an isometry. Since an isometry is continuous, you're done.
$\endgroup$ 2 $\begingroup$If you want to show that an isometric isomorphism is continuous with continuous inverse, then the way you start is "Given an isometric isomorphism $T$, a point $x$ and an $\epsilon >0$, ..." and then you use the isometric property of $T$ and the triangle inequality to find your $\delta$, both for $T$ and its inverse.
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