Fix a natural number $n\geq 1.$ Let $\mu$ be a norm on $\mathbb{R}^n$ satisfying $$\mu(0,...,0,\stackrel{i}{1},0,...,0) = 1 \quad\text{for all }1\leq i\leq n.$$ Let $$B_{\mu} = \{(a_1,...,a_n)\in \mathbb{R}^n : \mu(a_1,...,a_n)\leq 1\},$$ that is, ball of $\rho$-norm centered at origin. Let $$B_{\ell^1} = \{(a_1,...,a_n)\in\mathbb{R}^n: |a_1|+\cdots|a_n|\leq 1\}.$$

One can show easily that $B_{\ell^1}\subseteq B_{\mu}$ by using triangle inequality and above equation.

Define dual norm $\rho^*$ on $\mathbb{R}^n$ by $$\mu^*(a_1^*,...,a_n^*) = \sup_{(a_1,...,a_n)\in B_{\mu} }\bigg| \sum_{i=1}^n a_i^*a_i \bigg|.$$

Question: Assuming that $\mu \neq \|\cdot\|_\infty$-norm. Fix $(a_1,...,a_n)\in \mathbb{R}^n$ with $\mu(a_1,...,a_n)=1.$ Is it always true that there exists an extreme point $(a_1^*,...,a_n^*)$ of $B_{\mu^*}$ such that $a_i^*\neq 0$ for all $1\leq i\leq n$ and $$\sum_{i=1}^n a_i^*a_i=1?$$

Recall the following well-known result:

If $X$ is a Banach space, then for any $x\in X,$ there exists an extreme point $x^*$ of $B_{X^*}$ such that $x^*(x) = \|x\|.$

However, if we let $X = \mathbb{R}^n,$ my question requests one more condition on $x^*,$ that is, all its component is nonzero. Can this be done?


1 Answer 1


I think that you assumed that $a_i\ne 0$ for all $i$, otherwise an easy "no" answer is given by the vector $(1,0,\dots,0)$ in the Euclidean space.

Assuming that this correction was made, consider the following space: $\ell_\infty^2\oplus_\infty\ell_2^{n-2}$. This is the space of sequences of length $n$ such that the norm of the sequence is the maximum of the $\ell_\infty$-norm of the first two elements and the Euclidean norm of the remaining elements. It is easy to see that for any of the extreme points in the dual ball one of the first two coordinates is $0$.

This works for dimensions $n\ge 4$. To get a more geometric solution in all dimensions consider the norm $\mu$ whose dual ball is the convex hull of the set of $\{\pm e_1,\dots,\pm e_n\}\cup\{a\}$, where $e_i$ are unit vectors and $a$ is some vector with all coordinates strictly between $-1$ and $1$, which is not in the convex hull of $\{\pm e_1,\dots,\pm e_n\}$. The dual norm for this ball is the desired $\mu$.

  • $\begingroup$ I suppose that $n\geq 4?$ $\endgroup$
    – Idonknow
    Jul 23, 2018 at 6:35
  • $\begingroup$ Just to clarify, in your example, you are letting $\mu^* = \|\cdot\|_\infty \oplus_\infty \|\cdot\|_2$ where the formal norm is on first two components while the latter norm is on remaining $n-2$ compoenents? $\endgroup$
    – Idonknow
    Jul 23, 2018 at 8:24
  • 1
    $\begingroup$ Yes, in the current example $n\ge 4$. One can modify to get $n\ge 3$ by looking at $\ell_\infty^2\oplus_2 \ell_2^{n-2}$. The answer to the second question is "Yes". $\endgroup$ Jul 23, 2018 at 13:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.