With the notations of prop. 6, the relation u(A)⊂A1 implies tu(A1∘)⊂A∘; if further A1 is convex, closed (for σ(F1,G1)) and contains the origin, then these two relations are equivalent.
In fact, the relation u(A)⊂A1 equivalent to A⊂u−1(A1), therefore implies
tu(A1∘)⊂tu(A1∘)⊂(u−1(A1))∘⊂A∘
and conversely the relation tu(A1∘)⊂A∘ implies
A∘∘⊂(tu(A1∘))∘=u−1(A1∘∘)
from (2). When A1=A1∘∘, we deduce that A⊂u−1(A1).
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.