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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.

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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.

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  • $\begingroup$ I am convinced with your answer for $(c)$. But I am not getting as you said there are two distinct Hahn Banach extensions make it obvious that a) and b) are false. But only b) should be fasle. Because there two distinct Hahn Banach extensions it implies not unique. So only b) should be false. $\endgroup$ – Ziya Sep 24 '18 at 14:43
  • $\begingroup$ @Ziya I think a) is not worded properly. I think what they rally meant to write was: a) it is not true that $g$ has a unique Hahn Banach extension to $\mathbb R^{2}$. Can you now see why a) is false? $\endgroup$ – Kavi Rama Murthy Sep 24 '18 at 23:17

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