Using the separating hyperplane theorem, we prove the dual norm of the dual norm is the original norm.
1. Extension Theorem
Let be a norm on . The dual norm of a norm is defined by
For any , we show
By definition, we have
Theorem 1. Let be a subspace of and let be a linear functional on satisfying
Then there exists satisfying
One can prove this theorem directly. See  or .However, we prove this theorem using the following lemma:
Lemma 1. Let be an affine set in and let be a non-empty convex subset of , not intersecting . Then there exists a hyperplane in containing and not intersecting .
Proof. By the separating hyperplane theorem, there exist and such that
Then it is a hyperplane in . There are two cases to consider. First, suppose . Then there exists with . Then for , note that since is an affine set. So
On the other hand,
which proves .
Suppose . After a translation, we may assume has so that is a subspace of . Then since . Note that for all . If it were not, then there exists satisfying . If , then for , . Letting , then there exists with , which contradicts . Similarly, leads a contradiction. Hence for all .
Note . So for all and for all . This shows that is a hyperplane in containing and not intersecting .
Thus, both cases gives a desired result. This completes the proof.
Using this lemma, we are ready to prove the theorem.
Proof of Theorem 1 Define
Then is convex, affine subspace of . Consider
Then it is also convex. Then is a hyperplane containing . Note that since
Hence by Lemma 1, there exists a hyperplane containing and . So there exist and such that
with for all . Since , . So . Hence we may assume
Then for all .
Note that if , then for all . Since is linear, is the zero map. So there is nothing to prove. So we assume . Then there exists with .
For nonzero vector , . So
and this shows
If with , then
If for some , then
Since , . So
This completes the proof of theorem.
2. Proof of the duality
Now we are ready to prove the claim (1). If , there is nothing to prove. If , consider . Define by
Clearly it is linear, and
for all . Hence by Theorem 1, there exists with
we conclude that
From (2), if we take , then and , which proves the claim.
3. Some remarks
The motivation of the proof is the philosophy that the Hahn-Banach theorem and the separating hyperplane theorem are closely related. We can prove the separating hyperplane theorem using an extension of Theorem 1. One may ask whether the converse is true. In fact, both statements are equivalent.
The proof of Theorem 1 using Lemma 1 is given in . We modify the lemma and theorem and their proof so that it is suitable to our setting.
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- H. H. Schaefer and M. P. Wolff, Topological vector spaces, Springer, 1966.
- E. Stein and R. Sharkarchi, Functional Analysis: Introduction to Further Topics in Analysis, Princeton University Press, 2011.