Given a real vector space $V$ which is the union of a totally ordered family of vector subspaces $V=\bigcup_{i\in I} V_i$. By that I mean that we assume that $(I,\leq)$ is a totally ordered set and that $V_i\subseteq V_j$ whenever $i \leq j$.
Furthermore, assume that on each $V_i$ there is a norm $\|\cdot\|_{i}$ turning it into a Banach space. And lastly, assume that for each $i\leq j$ the inclusion from $V_i$ to $V_j$ is a topological embedding.
My question is:
Is it always possible to change the norms $\|\cdot\|_{i}$ to equivalent norms $\|| \cdot |\|_i$ such that the inclusion maps become isometric embeddings instead of only topological ones ?
My thoughts so far:
If $I$ consists only of two elements $I=\{1,2\}$, then this is always possible: I found two essentially different approaches here: The hard way: Keep the norm on $V_1$ and adjust the norm on $V_2$ such that the inclusion is an isometric embedding. This is a bit tricky but always posible.
The easy way: Keep the norm on $V_2$ and restrict it to $V_1$ to obtain a new norm on $V_1$. It is equivalent to the given norm on $V_1$ since we assumed that the inclusion is a topological embedding.
Both methods work.
More generally, if $I$ has a maximum, then the "easy way" method above directly generalizes and solves this case.
And if $I=\mathbb N$, one can keep the first norm and then recursively use the "hard way" method to adjust the other norms one at a time.
What about other totally ordered sets, like $I=(\mathbb R,\leq)$ ? I would guess that one could use the "hard way" to go recursively through all the natural numbers and then use the "easy way" method to take care of all the other numbers, but I am not sure about the details here.
For more general totally ordered sets, I have no clue. It would be very interesting to see a counter-example as I have no idea where it should come from...
