Dual norm of a subspace of $\ell_\infty^3$

Question

We define a norm on $\mathbb C^2$ as $\|(\alpha,\beta)\|:=\max\left\{|\alpha|,|\beta|,\big|\frac{\alpha+\beta}{\sqrt{2}}\big|\right\}.$ Can the dual norm be calculated explicitly?

Answer 1

$\newcommand{\C}{\mathbb C}\newcommand{\R}{\mathbb R}\newcommand{\si}{\sigma}\newcommand{\al}{\alpha}\newcommand{\be}{\beta}$This is to detail and correct the answer by Onur Oktay, which is based on a nice idea, leading to simplified and speedier calculations of the dual norm. (That Onur Oktay's answer contains at least two mistakes was pointed out in comments by Nathaniel Johnston and me.)

As in Onur Oktay's answer, let $X:=\ell^3_\infty$ and $V:=\{av_1+bv_2:a,b\in\C\}\subset X$, where \begin{equation*} v_1:=(1,0,1/\sqrt2),\quad v_2:=(0,1,1/\sqrt2),\quad v_3:=(1,1,-\sqrt2). \end{equation*} Note that $v_3\cdot v=0$ for all $v\in V$, where $\cdot$ denotes the dot product. Consider
\begin{equation*} V^\perp:=\{x^*\in X^*\colon x^*(v)=0\ \forall v\in V\}, \end{equation*} the so-called annihilator of $V$. Note that, if, as usual, $X^*=(\ell^3_\infty)^*$ is identified with $\ell^3_1$, then $V^\perp$ will be identified with the span $\C v_3=\{tv_3\colon t\in\C\}$ of $\{v_3\}$.

It is then said in Onur Oktay's answer that

$V^*=X^*/V^\perp$, which is isometrically isomorphic to $\C^2$ equipped with the norm \begin{equation*} \|(a,b)\|=\inf\{\|av_1+bv_2-tu\|_{\ell^3_1}: t\in\C\}, \end{equation*}

where $u$ is our $v_3$.

This statement is incorrect. First here is the minor point that $V^*$ is, not equal, but isometrically isomorphic to $X^*/V^\perp$ (Theorem 4.9), with the isometric isomorphism $\si$ given by the formulas \begin{equation*} V^*\ni v^*\mapsto\si(v^*):=x^*+V^\perp \end{equation*} and \begin{equation*} \|x^*+V^\perp\|:=\inf\{\|x^*+y^*\|\colon y^*\in V^\perp\}, \tag{0}\label{0} \end{equation*} where $x^*\in X^*$ is any extension of the continuous linear functional $v^*$. More importantly, the expression for $\|(a,b)\|$ in Onur Oktay's answer is incorrect.

Indeed, consider the isometric isomorphism \begin{equation*} \C^2\ni(\al,\be)\mapsto\iota((\al,\be)):=\al v_1+\be v_2\in V\subset X=\ell^3_\infty \end{equation*} from $\C^2$ onto $V$, with the norm on $\C^2$ induced by $\iota$: \begin{equation*} \|(\al,\be)\|=\|\al v_1+\be v_2\|_\infty=\max(|\al|,|\be|,|\al+\be|/\sqrt2). \tag{1}\label{1} \end{equation*} For any $a$ and $b$ in $\C$, define linear functionals $l_{a,b}\in(\C^2)^*$ and $L_{a,b}\in V^*$ by the conditions \begin{equation*} l_{a,b}((1,0))=a=L_{a,b}(v_1)\quad\text{and}\quad l_{a,b}((0,1))=b=L_{a,b}(v_2); \end{equation*} here, of course, the norm $\|\cdot\|_{(\C^2)^*}$ on $(\C^2)^*$ is dual to the norm on $\C^2$ given by \eqref{1}. Also for $a$ and $b$ in $\C$, define the vector $x_{a,b}\in\R^3$ by the conditions \begin{equation*} x_{a,b}\cdot v_1=a,\quad x_{a,b}\cdot v_2=b,\quad x_{a,b}\cdot v_3=0, \end{equation*} so that \begin{equation*} x_{a,b}=\tfrac14\,(3 a-b,3 b-a,\sqrt{2}(a+b)). \end{equation*} So, the vector $x_{a,b}\in\R^3$ is the canonical representation of the extension -- say $\tilde L_{a,b}$ -- of the linear functional $L_{a,b}\in V^*$ such that $\tilde L_{a,b}\in X^*=(\ell^3_\infty)^*\simeq\ell^3_1$ and $\tilde L_{a,b}(v_3)=0$. The canonical representation of $\tilde L_{a,b}$ by $x_{a,b}$ here is of course the one induced by the canonical isometric isomorphism $(\ell^3_\infty)^*\simeq\ell^3_1$, so that $\tilde L_{a,b}(x)=x\cdot x_{a,b}$ for all $x\in\ell^3_\infty$.

Then $l_{a,b}=L_{a,b}\circ\iota$ and, in view of \eqref{0}, \begin{equation*} \|l_{a,b}\|_{(\C^2)^*}=\|L_{a,b}\|_{V^*} =\inf_{t\in\C}\|x_{a,b}+tv_3/4\|_1. \end{equation*} Thus, \begin{equation*} \|l_{a,b}\|_{(\C^2)^*} =\tfrac14\,\min_{t\in\C}(|3 a - b + t|+|3 b-a + t|+\sqrt2\, |a + b - t|). \tag{2}\label{2} \end{equation*}

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Finding the minimum in \eqref{2} is a problem of real algebraic geometry. So, in principle, the dual norm can be computed purely algorithmically; however, this calculation can take too much time -- as it actually is the case with the minimum in \eqref{2}.

However, finding the minimum in \eqref{2} numerically for any particular $(a,b)\in\C^2$ is not a problem. The numerical results obtained according to \eqref{2} completely agree with the numerical results obtained according to my other answer on this page, but the execution time using \eqref{2} is about 6 times as small.

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As Matthew Daws noted in a comment, one could use the vector $y_{a,b}:=(a,b,0)$ in place of $x_{a,b}=\tfrac14\,(3 a-b,3 b-a,\sqrt{2}(a+b))$ -- in the sense that the linear functional $X\ni x\mapsto x\cdot y_{a,b}$ is, just as the linear functional $\tilde L_{a,b}$, an extension of the linear functional $L_{a,b}$. This results in an expression for $\|l_{a,b}\|_{(\C^2)^*}$ looking slightly simpler than the one in \eqref{2}: \begin{equation*} \|l_{a,b}\|_{(\C^2)^*} =\inf_{s\in\C}\|y_{a,b}+sv_3\|_1 =\min_{s\in\C}(|a + s|+|b + s|+\sqrt2\, |s|), \tag{2a}\label{2a} \end{equation*} which seems to match the final expression in Nathaniel Johnston's answer. In fact, the ultimate expression in \eqref{2a} can be obtained from that in \eqref{2} by substitution $t=a+b+4s$.

Answer 2

Let $X=\ell^{3}_{\infty}$ and let $V=\{av_1+bv_2:a,b\in\mathbb{C}\}$ be the subspace spanned by the vectors $$v_1 = (1, 0, \frac{1}{\sqrt{2}}) \hspace{9mm} v_2 = (0, 1, \frac{1}{\sqrt{2}}).$$ Let $u = (1,1,-\sqrt{2})$ so that $u\perp V$. Then, $V^* = X^*/V^{\perp}$, which is isometrically isomorphic to $\mathbb{C}^2$ equipped with the norm \begin{eqnarray} \|(a,b)\| &=& \inf\{\|av_1+bv_2-tu\|_{\ell^3_1}: t\in\mathbb{C}\} \\ &=& \inf\{|a - t| + |b - t| + |\frac{a + b}{\sqrt{2}} + \sqrt{2}t| : t\in\mathbb{C} \} \\ &=&\min\left\{ |b-a| + \frac{ |3a + b|}{\sqrt{2}} \hspace{1mm},\hspace{1mm} |b-a| + \frac{ |a + 3b|}{\sqrt{2}} \hspace{1mm},\hspace{1mm} \frac{ |a + 3b| + |3a+b|}{2} \right\}\\ \end{eqnarray}

Answer 3

This answer uses a lot of the same ideas as Onur Oktay's answer, but I believe corrects some problems with it. If we define $\mathbf{e}_1 = (1,0)$, $\mathbf{e}_2 = (0,1)$, $\mathbf{u} = (1,1)/\sqrt{2}$, and $S = \{\mathbf{e}_1,\mathbf{e}_2,\mathbf{u}\}$, then your norm is $$ \|\mathbf{v}\| = \max_{\mathbf{w} \in S}\big\{|\langle \mathbf{w}, \mathbf{v}\rangle|\big\}. $$ By Theorem 2 of this paper (sorry for the self-citation; there's probably an earlier reference for this result, I just don't know what it is) it follows that the dual norm is $$ \|\mathbf{v}\|^\circ = \min\big\{ |x|+|y|+|z| : \mathbf{v} = x\mathbf{e}_1 + y\mathbf{e}_2 + z\mathbf{u} \big\}. $$ This can be made more explicit by noting that the linear system $\mathbf{v} = x\mathbf{e}_1 + y\mathbf{e}_2 + z\mathbf{u}$ has a $1$-dimensional solution set given by $$ (x,y,z) = (v_1, v_2, 0) + t(1,1,-\sqrt{2}), $$ where $t \in \mathbb{C}$ is arbitrary. It follows that $$ \|\mathbf{v}\|^\circ = \min\big\{ |v_1+t|+|v_2+t|+\sqrt{2}|t| : t \in \mathbb{C} \big\}. $$ I'm not aware of a closed-form solution to this optimization problem, but it is convex and can be solved numerically very quickly by the CVX package for MATLAB, for example. Here is code that does the job and shows, for example, that this dual norm evaluated on the vector $(1,1+i)$ equals $\sqrt{5}$, which is attained when $t = -(3+i)/5$:

>> v = [1;1+i];

>> cvx_begin sdp quiet
>>     variable t complex;
>>     minimize abs(v(1)+t) + abs(v(2)+t) + sqrt(2)*abs(t)
>> cvx_end

>> cvx_optval

cvx_optval =

    2.2361

>> t

t =

    -0.6000 - 0.2000i

Answer 4

This is a problem of real algebraic geometry. So, in principle, the dual norm can be computed purely algorithmically; however, this calculation can take too much time.

In Mathematica, this algorithm is implemented in the command Maximize. It has now been working for some time, with no result yet (click on the image of a Mathematica notebook below to magnify it):

Mathematica has so far been unable to even complete the calculation of the norm of a particular linear functional.