I am attempting to find the dual norm of $$\|(x,y)\|_K=\max\{|x|,|y|,|x-y|\}.$$ I have obtained $\|(x',y')\|_K^* = |x'|+|y'|$, but don't think that this is correct. I obtained this as follows : $$K = \{(x,y)\in R^n : \max\{|x|,|y|,|x-y|\} \leq 1\}$$ $$|(x,y).(x',y')| = |xx' + yy'| \leq |x||x'|+|y||y'| \leq \max\{|x-y|,|x|,|y|\}(|x'|+|y'|) \leq |x'|+|y'|,$$ thus concluding that $\|(x',y')\|_K^* = |x'| + |y'|$. I don't think that this is correct because I know that this is the dual norm of $\max\{|x|,|y|\}$. What is the correct dual norm?
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$\begingroup$ You are essentially looking at the norm of a subspace $P$ of $\ell^\infty$ (its a plane inside the 3-dimensional space with the max-norm $\ell^3_\infty$). If you have a look at the section on duals of quotients in Rudin, you will see, that the dual of a subspace is (isomorphic to) the a quotient, here $\ell^3_1 / P^\perp$ (with the quotient norm). It should be a nice exercise for you to figure out exactly, what this gives. $\endgroup$– ARGCommented Nov 14, 2019 at 13:04
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The dual norm of $(a,b)$ is $|a|\vee|b|\vee|a+b|$, where $A\vee B\vee\cdots:=\max(A,B,\dots)$.
Details:
The dual norm of $(a,b)\in\mathbb R^2$ is $$m=m_1\vee m_2, \tag{1} $$ where $$m_1:=\max\{ax+by\colon-1\le x\le 1,0\le y\le 1,-1\le x-y\le 1\} \\ =\max\{ax+by\colon0\le y\le 1,y-1\le x\le 1\}, $$ $$m_2:=\max\{ax+by\colon-1\le x\le 1,-1\le y\le 0,-1\le x-y\le 1\} \\ =\max\{ax+by\colon-1\le y\le 0,-1\le x\le 1+y\}. $$ Further, by the linearity of $ax+by$ in $(x,y)$, $$m_1=\max\{a(y-1)+by\colon0\le y\le 1\}\vee \max\{a+by\colon0\le y\le 1\} \\ =(-a)\vee b\vee a\vee(a+b). \tag{2} $$ Similarly, $$m_2=(-a-b)\vee(-a)\vee a\vee(-b). \tag{3} $$ So, by (1)--(3), indeed $m=|a|\vee|b|\vee|a+b|$.
Another way to compute the dual norm $m$ of $(a,b)$ is to note that $m$, equal the maximum of the linear form $ax+by$ over all points $(x,y)$ in the unit ball $K$, is the maximum of $ax+by$ over all the extreme points of $K$, which are $(-1,-1),(0,-1),(1,0),(1,1),(0,1),(-1,0)$. Here is the picture of the ball $K$:
So, $$m=(-a-b)\vee(-b)\vee a\vee(a+b)\vee b\vee(-a)=|a|\vee|b|\vee|a+b|. $$
answered Nov 12, 2019 at 0:27
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$\begingroup$ What is a good reference for this and norms? $\endgroup$– VS.Commented Nov 12, 2019 at 4:21
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$\begingroup$ @VS. : I am not good at references. Try keywords such "normed spaces", "linear programming", "convex analysis", "extreme points". $\endgroup$ Commented Nov 12, 2019 at 14:13