# Dual norm in R^n

By | October 20, 2017
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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

So

It remains to prove the reverse inequality. We will show that there exists such that and

(1)

Theorem 1. Let be a subspace of and let be a linear functional on satisfying

Then there exists satisfying

and

One can prove this theorem directly. See [1] or [3].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

Define

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,

Thus,

This implies

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

and so

If for some , then

and so

Since , . So

and

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

(2)

Clearly it is linear, and

Also,

for all . Hence by Theorem 1, there exists with

with

Since

we conclude that

From (2), if we take , then and , which proves the claim.

Therefore,

### 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 [2]. We modify the lemma and theorem and their proof so that it is suitable to our setting.

References

1. H. Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations, Springer, 2011.
2. H. H. Schaefer and M. P. Wolff, Topological vector spaces, Springer, 1966.
3. E. Stein and R. Sharkarchi, Functional Analysis: Introduction to Further Topics in Analysis, Princeton University Press, 2011.

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