Given a vector space V, is it possible to endow it with two nonequivalent norms ‖ ⋅ ‖ 1 and ‖ ⋅ ‖ 2 such that any linear functional on V is continuous in the sense of ‖ ⋅ ‖ 1 if only if it is continuous in the sense of ‖ ⋅ ‖ 2 ?By nonquivalent norms I mean the induced topologies of the norms are different.

Question

Given a vector space   V, is it possible to endow it with two nonequivalent norms   ‖  ⋅      ‖    1   and   ‖  ⋅      ‖    2   such that any linear functional on   V is continuous in the sense of   ‖  ⋅      ‖    1   if only if it is continuous in the sense of   ‖  ⋅      ‖    2  ?By nonquivalent norms I mean the induced topologies of the norms are different.

Liehm1mm · Accepted Answer

For   i  =  1  ,  2, let       V    i   denote   V equipped with   ‖  ⋅      ‖    i  .From our assumption, the map   Id  :      V    1    →      V    2   is sequentially weakly continuous and hence it is continuous. This yields   ‖  ⋅      ‖    2    ≤  C  ‖  ⋅      ‖    1  .Interchanging the role of       V    1   and       V    2   then yields equivalence of norms.To see that a weakly sequentially continuous linear map must be bounded, you can proceed as follows. Suppose   T is weakly sequentially continuous and unbounded. Then there is a sequence       x    n    →  0 such that   ‖  T      x    n    ‖  &amp;gt;  n. Since       x    n    →  0 in norm, it also does so weakly so that   T      x    n   ought converge weakly, but weakly convergent sequences are norm bounded so that we reach a contradiction.

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