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?



Dual norm of a subspace of $\ell_\infty^3$
$\newcommand{\C}{\mathbb C}\newcommand{\R}{\mathbb R}\newcommand{\si}{\sigma}\newcommand{\al}{\alpha}\newcommand{\be}{\beta}$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](https://www.powells.com/book/functional-analysis-2nd-edition-9780070542365)), 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\*}




---


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.




---


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 $t=a+b+4s$.


