Is $\|\mathord\cdot\|_1$ the greatest norm? Is $\|\mathord\cdot\|_\infty$ the littlest?

Question

The question in the title may seem intuitive, because in the set of the $p$-norms on $\mathbb{R}^n$ the inequality: $$\|\mathord\cdot\|_\infty\leq\|\mathord\cdot\|_p\leq\|\mathord\cdot\|_1$$ holds, so I'd like to generalize this property, if possible. So my first question is:

What's the way to make the question in the title rigorous?

What I've thought is the following property: "In the set of the norms on $\mathbb{R}^n$ which have values 1 on the canonical basis, is $\|\mathord\cdot\|_1$ the maximum norm? Is $\|\mathord\cdot\|_{\infty}$ the minimum?"

I don't know if this is the better way to formalize this question, and I don't really have an idea on how to answer to it. Thanks to everybody.

Answer 1

If we impose the requirement that the norm be a symmetric gauge function, then we will get one generalization of your inequality. In particular, Bhatia's Matrix Analysis has the following definition:

$\Phi:\Bbb R^n \to \Bbb R_+$ is a symmetric gauge function if

1. $\Phi$ is a norm
2. $\Phi(Px) = \Phi(x)$ for all permutations $P$
3. $\Phi(\epsilon_1 x_1,\dots,\epsilon_n x_n) = \Phi(x_1,\dots,x_n)$ if $\epsilon_j \in \{\pm 1\}$
4. $\Phi(1,0,\dots,0) = 1$

In some contexts, the 4th "normalization" condition is dropped, but I think that these are the "common sense" properties that you're going for.

Among the gauge functions, we see that the inequality you expected holds. In the following, take $e_j \in \Bbb R^n$ to be the vector with a $1$ as its $j$th entry and $0$s for all other entries. We note that for any $x = (x_1,\dots,x_n)$, we have $$ \Phi(x) = \Phi(x_1e_1 + \cdots + x_n e_n) \leq |x_1| \Phi(e_1) + \cdots + |x_n| \Phi(e_n) = \|x\|_1 $$ and on the other hand, if we suppose without loss of generality that $x_1$ is the largest entry in absolute value, $$ 2 \Phi(x) = \Phi(x_1,x_2,\dots,x_n) + \Phi(x_1,-x_2,\dots,-x_n) \geq \Phi(2x_1,0,\dots,0) = 2|x_1| $$ so that we also have $\Phi(x) \geq \|x\|_\infty$.

Answer 2

You might be able to make a result if you demand the closed unit ball have the full isometry group of the cube, plus demand that the norm of $(1,0,0,0...,0)$ be $1.$

Short of that, try norm $x^2 - \frac{3}{2}xy + y^2.$ Agress along the $x$ and $y$ axes, but the unit vector when $y=x$ is $(\sqrt 2, \sqrt 2)$

Answer 3

For each compact balanced¹⁾ convex neighborhood $K\subseteq\mathbb{R}^n$ of $0$, the Minkowski functional

$$\|v\|_K := \inf\{t>0: v\in tU\}$$

defines a norm on $\mathbb{R}^n$ for which $K$ is the closed unit ball w.r.t. $\|\cdot\|_K$, and any norm on $\mathbb{R}^n$ is realized in this way. That said, your question can be rephrased in terms of such convex sets. Let

$$ \mathcal{C} = \{ \| \cdot \| : \text{norm on $\mathbb{R}^n$ such that $\|\mathrm{e}_i\| = 1$ for $i=1,\cdots,n$}\}. $$

• The requirement $\| \cdot \|_K \in \mathcal{C} $ means that $\mathrm{e}_i \in \partial K$.

• The smallest such $K$ containing all of $\mathrm{e}_1, \cdots, \mathrm{e}_n$ is the closed unit ball for the $1$-norm. So we have $\| \cdot \| \leq \| \cdot \|_1$ for any norm $\| \cdot \| \in \mathcal{C} $.

• On the other hand, we can create incompatible norms in $\mathcal{C}$. The idea is to consider lopsided $K$. So we cannot claim that $\|\cdot\|_{\infty}$ is the smallest norm in the family $\mathcal{C}$ without further restrictions.

• As pointed out by Omnomnomnom, symmetry condition is enough to prevent 'lopsidedness', which is enough to show that $\|\cdot\|_{\infty}$ is smallest among all symmetric norms in $\mathcal{C}$.

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1) A subset $U$ of a vector space $V$ is called balanced if $x \in U$ implies $-x\in U$.