The operators $A_n$ have still better smoothness properties.
The new $X$ is no more continuous, although it still has norm 1.
An ingenious alternative proof, shorter but still complicated, can be found in [MR].
This is of course still an implicit characterization of $V$, since ......
It seems that the solution of Problem 1 is still out of reach.
Then we can find a subsequence (still denoted by $a_n$) such that $a_n<1$ for all $n$.
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