[see also: continue, yet
The operators An 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 an) such that an<1 for all n.