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Let $p\in (1,\infty )$. Denote by $\|\cdot \|_p$, the $\ell ^p$ norm on $\mathbb{R}^n$, i.e. for all $x\in \mathbb{R}^n$ one has$$\|x\|_p:=\left (\sum \limits _{i=1}^n|x_n|^p\right )^\frac{1}{p}.$$Assume that $B_p$ is the unit ball in $\mathbb{R}^n$ defined by the $\ell ^p$ norm, i.e.$$B_p=\{x\in \mathbb{R}^n:\|x\|_p\leq 1\}.$$My question is, what is polar of $B_p$? The polar of $B_p$ is denoted by $B_p^\circ$, and it is defined by$$B_p^\circ :=\{v\in \mathbb{R}^n:\langle v,x\rangle \leq 1, ~ \forall x\in B_p\}.$$Clearly $B_{2}^{\circ} = B_{2}$, but I am particularly interested in $B_{3}^{\circ}$ and $B_{4}^{\circ}$.

Edit: Notice that the $\langle v,x\rangle$ represents the usual inner product in $\mathbb{R}^n$, not the dual functional in $\ell ^p$.

commie trivial
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Red shoes
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    It is the ball for the $\ell_q$ norm, where $1/p+1/q=1$. – Ruy Jul 24 '22 at 15:30
  • Yes it is not straightforward. It is basically the content of Holder's inequality. See the section on "counting measure" in https://en.m.wikipedia.org/wiki/H%C3%B6lder%27s_inequality. – Ruy Jul 24 '22 at 15:38
  • Sorry, I was not clear in my question. By $\langle v, x \rangle $; I mean the usual inner product. It has nothing to do with the dual functions of $\ell_p$; that's why the case $p=2$ was clear to me. I will Edit my question. @Ruy – Red shoes Jul 24 '22 at 15:44
  • I guess the answer is still $B_q$, right? @Ruy – Red shoes Jul 24 '22 at 16:14
  • The polar of $B_p$ is indeed the unit ball with respect to the dual function of $\ell_p$, which is $\ell_q$. – Deane Jul 24 '22 at 16:52

1 Answers1

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This is only an idea

As the comment by @Ruy suggests, it is the set $$ B^\circ_p= \{v \in \mathbb R^n \ |\ \|v\|_q = 1, \frac{1}{p} + \frac{1}{q} = 1\} = B_q $$ The norm on $(\mathbb R^n,\|\cdot\|_p)$ establishes an isometry onto the dual space which is $(\mathbb R^n,\|\cdot\|_q)$, where $\|\cdot\|_q$ comes from the operator norm. See this explicit map here.

So that action of $f = (f_1,\dots,f_n)$ on $x = (x_1,\dots,x_n)$ is $$ f(x) = \sum_j f_jx_j = \langle f,x \rangle $$ is exactly as that inner product. So if one asks for the all $f \in \mathbb R^n$ such that on $x \in B_p$ we get $$ \langle f,x \rangle = f(x) \le \sup_{x:\|x\|_p \le 1} f(x) = \|f\|_q = 1 $$ is like asking for all duals with norm 1. The set you want is the unit ball in $\mathbb R^n$ with the (operator) norm which is $\|\cdot\|_q$, with $\frac{1}{p} + \frac{1}{q} = 1$.

I think the absolute value in the definition of the operator norm is only required for complex spaces.

Physor
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  • Yes, I agree on your remark on the absolute value. Indeed, let us fix $f\in\mathbb R^n$. We have that $$\sup_{x} \langle f, x\rangle =\sup_{y}\lvert\langle f,y\rangle\rvert.$$ Of course both $x$ and $y$ are taken in the $\ell^p$ unit ball. Proof: Clearly LHS\le RHS. So take a competitor $y$ for RHS. Replace $y=(y_1, \ldots, y_n)$ by $(y_1\text{signum}(f_1), \ldots, y_n \text{signum}(f_n))$. This does not change $\lVert y\rVert_{\ell^p}$. Moreover $\lvert\langle y, f\rangle\rvert =\langle y, f\rangle$. So each competitor for RHS has a corresponding competitor for LHS, and we conclude RHS\le LHS – Giuseppe Negro Jul 24 '22 at 17:41
  • I'm not an expert in functional analysis yet, but do you agree that the isometry above is also a vector space isomorphism ?, for the case $p=2$ it is clear and famous – Physor Jul 24 '22 at 17:44
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    Which isometry? If you give me the explicit expression of it I can think about that. (I think it will turn out to be a linear map iff $p=2$ - I thought about something very similar several years ago) – Giuseppe Negro Jul 24 '22 at 17:53
  • In the link in my answer. BTW I forgot how I constructed it, I'm trying to remeber. So I have to remove that from my answer – Physor Jul 24 '22 at 17:54
  • OK. As I suspected it is essentially the same as my old question. These maps are nonlinear, unless $p=2$. A nonlinear map cannot be a vector space isomorphism. – Giuseppe Negro Jul 24 '22 at 17:58