Dual norm in R^n

I shall start posting some homework solutions that readers may be interested. This post discusses one special property about the Euclidean norm.

Let \|\cdot\| be a norm on \mathbb{R}^n. By identifying \mathbb{R}^n with its usual Euclidean inner product, we can study the dual norm of x\in\mathbb{R}^n given by \displaystyle \|x\|^*=\sup_{\|y\|=1}|\langle x,y \rangle|.

It is an exercise to show that for all 1\leq p\leq \infty, the dual norm of p-norm in \mathbb{R}^n is precisely the q-norm whether \displaystyle \frac{1}{p}+\frac{1}{q}=1. In particular, the dual norm of the Euclidean norm is itself. Next, we are going to show:

Theorem. A norm \|\cdot\| on \mathbb{R}^n equals its dual norm if and only if \|\cdot\| is the Euclidean norm.

Indeed, this is a direct consequence of the following two lemmas.

Lemma 1. Let \|\cdot\| be a norm on \mathbb{R}^n. Then \displaystyle \left(\sup_{\|x\|=1}\|x\|_2\right)^2=\sup_{\|x\|=1}\|x\|^*.

Proof. By Cauchy Schwarz’s inequality, for all x\in \mathbb{R}^n, one has

\displaystyle \|x\|^*=\sup_{\|y\|=1}|\langle x,y\rangle|\leq \sup_{\|y\|=1}\|x\|_2 \|y\|_2 =\|x\|_2 \sup_{\|y\|=1}\|y\|_2.

Taking supremum over all x\in \mathbb{R}^n of norm 1, we obtain

\displaystyle \sup_{\|x\|=1}\|x\|^* \leq \sup_{\|x\|=1}\left(\|x\|_2 \sup_{\|y\|=1}\|y\|_2\right)=\left(\sup_{\|y\|=1}\|y\|_2\right)\left(\sup_{\|x\|=1}\|x\|_2\right)=\left(\sup_{\|x\|=1}\|x\|_2\right)^2.

So one side of the inequality is done. Since the set \{x\in \mathbb{R}^n:\|x\|=1\} is compact and \|\cdot\|_2 is continuous, \displaystyle \sup_{\|x\|=1}\|x\|_2 is attained. We denote the maximizer by \widetilde{x}. Then \|\widetilde{x}\|=1 and \displaystyle \sup_{\|x\|=1}\|x\|_2=\|\widetilde{x}\|_2. Hence

\displaystyle\left(\sup_{\|x\|=1}\|x\|_2\right)^2 = \|\widetilde{x}\|_2^2 =\langle \widetilde{x},\widetilde{x}\rangle=\sup_{\|y\|=1}|\langle \widetilde{x},y \rangle|=\|\widetilde{x}\|^*\leq \sup_{\|x\|=1}\|x\|^*,

which finishes the proof of the lemma. \square

The proof of the second lemma is a bit nontrivial. We would like to show the readers this part.

Lemma 2. Let \|\cdot\| be a norm on \mathbb{R}^n. Then \displaystyle \left(\inf_{\|x\|=1}\|x\|_2\right)^2=\inf_{\|x\|=1}\|x\|^*.

Proof. For all x\in \mathbb{R}^n of norm one, one has

\displaystyle \inf_{\|z\|=1}\|z\|_2 \leq \|x\|_2 = \sqrt{\langle x,x\rangle}\leq \sqrt{\sup_{\|y\|=1} |\langle x,y\rangle|}=\sqrt{\|x\|^*} so \displaystyle \left(\inf_{\|z\|=1}\|z\|_2\right)^2\leq\|x\|^*. Hence \displaystyle \left(\inf_{\|x\|=1}\|x\|_2\right)^2 \leq \inf_{\|x\|=1}\|x\|^* and we showed one side of inequality.

Similar to the explanation in the proof of Lemma 1, the infimum \displaystyle \inf_{\|x\|=1}\|x\|_2 is attained and we let the minimizer be x'. It follows that \displaystyle \inf_{\|x\|=1}\|x\|_2=\|x'\|_2 with \|x'\|=1. By Hahn-Banach theorem, there is w\in \mathbb{R}^n with \|w\|^*=1 such that \langle w,x'\rangle=\|x'\|=1. Therefore 1=\langle w,x'\rangle\leq \|w\|_2 \|x'\|_2 by Cauchy Schwarz’s inequality. Then as w is nonzero, we have

\displaystyle \|x'\|_2^2 \geq\frac{1}{\|w\|_2^2}.       ——-(1) 

On the other hand, one has \|w\|_2^2 =\langle w,w\rangle \leq \|w\|^*\|w\|=\|w\| so

\displaystyle \frac{1}{\|w\|_2^2}\geq \frac{1}{\|w\|}.      ——-(2)

By (1) and (2) we obtain

\displaystyle \|x'\|_2^2 \geq \frac{1}{\|w\|_2^2}\geq \frac{1}{\|w\|}=\frac{\|w\|^*}{\|w\|}=\left\|\frac{w}{\|w\|}\right\|^*\geq \inf_{\|x\|=1}\|x\|^*.

In fact, this completes the proof of the lemma. \square

Proof of the Theorem. If \|\cdot\|=\|\cdot\|_2, then by Lemma 1 and Lemma 2, \displaystyle \sup_{\|x\|=1}\|x\|^*=\inf_{\|x\|=1}\|x\|^*=1 so \|\cdot\|=\|\cdot\|^*. If \|\cdot\|=\|\cdot\|^*, then by Lemma 1 and Lemma 2, \displaystyle \sup_{\|x\|=1}\|x\|_2=\inf_{\|x\|=1}\|x\|_2=1 so \|\cdot\|=\|\cdot\|_2. We are done.

Advertisements
This entry was posted in Functional analysis, Linear Algebra. Bookmark the permalink.

2 Responses to Dual norm in R^n

  1. Ken Leung says:

    Very interesting. I haven’t seen this characterization of the Euclidean norm before~

  2. Sasa Kariz says:

    I think I remember it was mentioned at school.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s