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Let $\ell^\infty(2)$ be a subset of $\mathbb{R}^2$ with maximum norm, that is, $\|(x,y)\|_\infty=\max\{|x|,|y| \}.$ Given two Banach spaces $X$ and $Y,$ if we have $X\oplus_2Y,$ we mean that $(x,y)\in X \oplus_2 Y$ with norm $\|(x,y\|_2= \sqrt{\|x\|_X^2 + \|y\|_Y^2}.$

We say that $x^*$ is an extreme point of unit ball $B_{X^*}$ of $X^*$ if whenever $x^*=\lambda y^* +(1-\lambda)z^*$ for some $\lambda\in[0,1]$ and $y^*,z^*\in B_{X^*},$ then $x^* = y^*$ or $x^*=z^*.$ Geometrically speaking, extreme points are 'corners' of unit ball.

Given a Banach space $X$, its set of extreme points $Ext(X)$ is defined to be a subset of closed unit ball of its dual $X^*.$ Also, if $X$ and $Y$ are isometrically isomorphic, we denote it as $X\cong Y.$

Question: Let $E = \ell^\infty(2)\oplus_2 \mathbb{R}$ be given. What is its $Ext(E)?$

First, I guess that $E^* \cong E$ (not very sure) due to $(\ell^2)^* \cong \ell^2$ and $F^* \cong F$ for any finite dimensional space $F$. So $Ext(E) \subseteq E^* \cong E.$

However, I have no idea how $E^*$ look like, thus cannot see its corner points. Any hint would be appreciated.

UPDATED: Based on Daniel Fischer's comment, dual space of $E$ is $\ell^1(2) \oplus_2 \mathbb{R}.$

So my question is reduced to finding extreme points of closed unit ball in $\ell^1(2) \oplus_2 \mathbb{R}.$

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  • $\begingroup$ Should be $\|(x,y)\|_2= \sqrt{\|x\|_X^2 + \|y\|_Y^2}$. You do not need to do anything with $E^*$. Only $E$. You have definition of extreme point of the unit ball of $X^*$, but here you need extreme point of the space $E$ itself not the dual. So I think you need to re-check your definitions. $\endgroup$ – GEdgar Sep 27 '17 at 13:56
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    $\begingroup$ The dual of $\ell^{\infty}(2)$ is $\ell^1(2)$. It happens to be the case that $\ell^1(2)$ and $\ell^{\infty}(2)$ are isometrically isomorphic [that's not the case for dimensions $> 2$], but it's better to use $\ell^1(2)$. Then $E^{\ast} = \ell^1(2) \oplus_2 \mathbb{R}$. [Generally, $(X \oplus_p Y)^{\ast} = X^{\ast} \oplus_q Y^{\ast}$, where $q$ is the conjugate exponent to $p$.] $\endgroup$ – Daniel Fischer Sep 27 '17 at 14:04
  • $\begingroup$ @DanielFischer: Your answer is very helpful and those are what I want. Is it possible that you provide or link proofs to the statements $(\ell^\infty(2))^* \cong \ell^1(2)$ and $(X\oplus_p Y)^* \cong X^* \oplus_q Y^*$ where $q$ is the conjugate exponent to $p?$ $\endgroup$ – Idonknow Sep 27 '17 at 14:11
  • $\begingroup$ @DanielFischer: Do you mean $(\ell^1(2))^*$ and $\ell^\infty(2)$ are isometrically isomorphic? $\endgroup$ – Idonknow Sep 27 '17 at 14:17
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    $\begingroup$ Since we are looking at finite-dimensional spaces, we have $(\ell^{\infty}(n))^{\ast} = \ell^1(n)$ too. Since $\ell^p(2) = \mathbb{R} \oplus_p \mathbb{R}$, $(\ell^{\infty}(2))^{\ast} \cong \ell^1(2)$ is a special case of $X \oplus_p Y)^{\ast} \cong X^{\ast} \oplus_q Y^{\ast}$. Here is a proof (of somewhat more; start at "More generally"). $\endgroup$ – Daniel Fischer Sep 27 '17 at 14:23
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Some points to make first:

  1. Extreme points are not preserved by isomorphisms, so thinking of what the given space is isomorphic to is not helpful. They are preserved by isometric isomorphisms.
  2. A space is strictly convex if $\|a+b\|<\|a\|+\|b\|$ whenever $a,b$ are non-collinear vectors. In a strictly convex space, every boundary point of the unit ball is its extreme point.
  3. If $X$ and $Y$ are strictly convex, then $X\oplus_2Y$ is also strictly convex. Indeed, to have equality $$\|(x_1,y_1) + (x_2,y_2)\| = \|(x_1,y_1)\| + \|(x_2,y_2)\|$$ one needs $x_1,x_2$ to be collinear, $y_1,y_2$ to be collinear, and also the Minkowski inequality $$ \sqrt{(\|x_1\|+\|x_2\|)^2 + (\|y_1\|+\|y_2\|)^2} \le \sqrt{\|x_1\|^2 + \|y_1\|^2} + \sqrt{\|x_2\|^2 + \|y_2\|^2} $$ must turn into equality. All this together implies that the vectors $(x_1,y_1)$ and $(x_2,y_2)$ are collinear.

From 1-3 the solution of the question about $\ell^\infty(2)\oplus_2 \mathbb{R}$ follows.


Given a Banach space $X$, its set of extreme points $\operatorname{Ext}(X)$ is defined to be a subset of closed unit ball of its dual $X^∗$

It's not defined like that, as pointed out in comments.

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  • $\begingroup$ I got the definition of extreme point at Aroujo's paper, page $288.$ theta.ro/jot/archive/2006-055-002/2006-055-002-005.pdf $\endgroup$ – Idonknow Sep 29 '17 at 14:09
  • $\begingroup$ He uses notation $\operatorname{Ext}_B$, probably to avoid typing superscript-in-subscript $\operatorname{Ext}_{B'}$, but does not call it "the set of extreme points of B". It's just author's choice of nonstandard notation for that paper. $\endgroup$ – user357151 Sep 29 '17 at 15:47
  • $\begingroup$ But he call it as subset of closed unit ball of its dual. $\endgroup$ – Idonknow Sep 30 '17 at 1:06

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