Topological Isomorphism vs Isometric Isomorphism
$\Begingroup$ 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...
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.
2 Answers
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.
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.
Elena Rostova holds a Master's degree in Public Health Journalism. She covers groundbreaking medical research, holistic wellness trends, mental health awareness, and nutritional science.
