Reversal via Hahn-Banach

Let *X* be a separable Banach space. Consider the following
statements:

**SEP1**- (First Separation) Let
*A*be an open convex set in*X*, let*B*be a separably closed convex set in*X*, and assume . Then*A*and*B*can be separated. **SEP2**- (Second Separation) Let
*A*and*B*be open convex sets in*X*such that . Then*A*and*B*can be strictly separated. **SEP3**- (Third Separation) Let
*A*and*B*be separably closed, convex sets in*X*such that . Assume also that*A*is compact. Then*A*and*B*can be strictly separated. **HB**- (Hahn-Banach) Let
*S*be a subspace of*X*, and let be a bounded linear functional with on*S*. Then there exists a bounded linear functional such that*F*extends*f*and on*X*. **EHB**- (Extended Hahn-Banach) Let
be a continuous
sublinear functional. Let
*S*be a subspace of*X*, and let be a bounded linear functional such that for all . Then there exists a bounded linear functional such that*F*extends*f*and for all .

*Proof. *Reasoning in
,
assume SEP1 and let *A* and *B* be disjoint,
open, convex sets. Let *B*' be the separable closure of *B*.
Clearly *B*' is convex and
.
By SEP1, let *F* and
be such that
on *A* and
on
*B*'. It follows that
on *B*. Thus we have SEP2.
This completes the proof.

*Proof. *Reasoning in
,
assume SEP2 and let *p*, *S*, and *f* be as
in the hypotheses of EHB. Let *A* be the convex hull of

and let

Clearly

We claim that *A* and *B* are disjoint. If not, then for some
,
,
,
,
,
we have *f*(*x*_{1})<1, *p*(*y*_{1})<1, *f*(*x*_{2})>1,
-*p*(-*y*_{2})>1, and

Note that . Hence

yet on the other hand we have

hence . Since at least one of the above inequalities must be strict, we obtain a contradiction. This proves our claim.

By SEP2, there exists a bounded linear functional
such
that *F*(*x*)<1 for all ,
and *F*(*x*)>1 for all .
Clearly *F* extends *f*. It remains to show that
for all .
Suppose not, say *p*(*y*)<*F*(*y*). If *F*(*y*)>0, then
for a suitably chosen *r*>0 we have
*p*(*ry*)<1<*F*(*ry*), a
contradiction. If ,
then for a suitably chosen *r*>0 we
have
*p*(*ry*)<-1<*F*(*ry*). Putting *z*=-*ry* we get
-*p*(-*z*)>1>*F*(*z*),
again a contradiction. This completes the proof.

*Proof. *HB is a special case of EHB with
.

- 1.
- SEP1, the first separation theorem.
- 2.
- SEP2, the second separation theorem.
- 3.
- EHB, the extended Hahn-Banach theorem.
- 4.
- HB, the Hahn-Banach theorem.
- 5.
- .

*Proof. *Lemmas 4.1, 4.2, 4.3 give the
implications SEP1
SEP2 and SEP2
EHB and EHB
HB. The equivalence HB
is the main result of [3]; see also
[14] and [16] and Chapter IV of [20].
Theorem 3.1 gives the implication
SEP1. This completes the proof.