Let $X(x,y)\in E = \Bbb R^2$ with the norm $||X||_1 = |x | + |y |$
Find $E^*$ and its norm, where $E^*$ is the topological dual, and compute $F((1,0))$ where $F(u) =\{L\in E^*, ||L||_{E^∗} = ||u||_E, \left<L, u\right> = ||u||^2_E \}$.
I found the following results:
$Dim\ E< \infty$ so $E^*$ is the set of linear functions $\Bbb R^2\to \Bbb R$ as in finite dimentional spaces the linear functions are continuous.
The norm on $E^*$ is $||f||_{E^*} = Sup_{||X||\le 1 }|f(X)|$.
Then $F((1,0))= \{f:\Bbb R^2\to \Bbb R: f(x,y)=xf(1,0)+yf(0,1)=x+yf(0,1)$ and $Sup_{x^2+y^2 \le 1}x^2+(yf(0,1))^2= 1\}$
Are there results correct? I am not sure whether we can give more characteristics of $F((1,0))$?
Thank you for your help or hints.
We can write $L(x,y)=ax+by$ for some $a,b \in \mathbb R$. $\|L\|=\sup \{|ax+by|:|x|+|y| \leq 1\}=\max \{|a|,|b|\}$. [See proof below]. Hence $F((1,0)=\{L(x,y)=ax+by: a+b(0)=1,\max \{|a|,|b|\}=1 \}$. Thus $F((1,0)$ consists of all linear functionals $L(x,y)=ax+by$ with $a=1$ and $|b| \leq 1$.
Proof of the fact that $\sup \{|ax+by|: |x|+|y|\leq 1\}=\max \{|a|,|b|\}$: LHS $\leq $RHS is obvious since $|ax+by| \leq \max \{|a|,|b|\}(|x|+|y|)$. For the reverse inequality make the choices $(x,y)=(1,0)$ and $(x,y)=(0,1)$ to see that LHS is $\geq $ both $|a|$ and $|b|$.