Consider $\mathbb{R}^{2}$ with norm $||•||_\infty $ and let $Y= \{ (y_1,y_2) \in \mathbb{R}^{2} : y_1+y_2=0\}. $ If $g:Y \to \mathbb{R}$ is defined by $g(y_1,y_2)=y_2$ for $(y_1, y_2) \in Y$ then
$(a) \ g$ has no unique Hahn Banach extension to $\mathbb{R}^{2}$ .
$ (b) \ g$ has unique Hahn Banach extension to $\mathbb{R}^{2}$.
$(c) \ $ Every linear functional $f:\mathbb{R}^{2} \to \mathbb{R}$ satisfying $f(-1,1)=1$ is a Hahn Banach extension of $g$ to $\mathbb{R}^{2}$.
$(d)$ The functionals $f_1, f_2 : \mathbb{R}^{2} \to \mathbb{R}$ given by $f_1(x_1, x_2)=x_1$ and $f_2(x_1,x_2)=-x_1$ are both Hahn Banach extensions of $g$ to $\mathbb{R}^{2}$.
Could you please help me. This question has been answered here $g$ has Hahn-Banach Extension to $\mathbb{R}^2$ uniquely? but I am not convinced with the answer. Could you please elaborate it more. Could you please tell me why $(a), (b), (c)$ are wrong? My functional analysis is not that good and I am studying it by my self. Please help. Thank you very much.
Best Answer
Partial answer: I will not go into the proof of d) from the previous post but I will show that a),b) and c) are all false. The fact that there are two different Hahn - Banach extensions make it obvious that a) and b) are false. Let us show that c) is false. For any real number $t$ define a linear functional $F_t$ by $F_t (x,y)=-x+t(x+y)$. This is a linear functional such that $F_t(-1,1)=1$. If c) is true then $F_t $ is a Hahn - Banach extension of $g$ for any real number $t$. By definition of Hahn - Banach extension we must have $\|F_t\| =\|g\|$ for all $t$. However, $(1,1)$ is a unit vector so $\|F_t\| \geq |F_t(1,1)|$ by definition of norm of an operator. Thus, $\|g\| =\|F_t\| \geq 2t-1$ for all real numbers $t$ which is obviously a contradiction.