I shall start posting some homework solutions that readers may be interested. This post discusses one special property about the Euclidean norm.
Let be a norm on . By identifying with its usual Euclidean inner product, we can study the dual norm of given by .
It is an exercise to show that for all , the dual norm of p-norm in is precisely the q-norm whether . In particular, the dual norm of the Euclidean norm is itself. Next, we are going to show:
Theorem. A norm on equals its dual norm if and only if is the Euclidean norm.
Indeed, this is a direct consequence of the following two lemmas.
Lemma 1. Let be a norm on . Then .
Proof. By Cauchy Schwarz’s inequality, for all , one has
Taking supremum over all of norm 1, we obtain
So one side of the inequality is done. Since the set is compact and is continuous, is attained. We denote the maximizer by . Then and . Hence
which finishes the proof of the lemma.
The proof of the second lemma is a bit nontrivial. We would like to show the readers this part.
Lemma 2. Let be a norm on . Then .
Proof. For all of norm one, one has
so . Hence and we showed one side of inequality.
Similar to the explanation in the proof of Lemma 1, the infimum is attained and we let the minimizer be . It follows that with . By Hahn-Banach theorem, there is with such that . Therefore by Cauchy Schwarz’s inequality. Then as is nonzero, we have
On the other hand, one has so
By (1) and (2) we obtain
In fact, this completes the proof of the lemma.
Proof of the Theorem. If , then by Lemma 1 and Lemma 2, so . If , then by Lemma 1 and Lemma 2, so . We are done.