## 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.

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### 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.