$g$ has Hahn-Banach Extension to $\mathbb{R}^2$ uniquely? $\mathbb{R}^2$ with sup-norm consider $Y=\{(y_1,y_2):y_1+y_2=0\}$, $g:Y\to\mathbb{R}$ is defined by $g(y_1,y_2)=y_2$, I want to know whether 
$1.$ 
$g$ has Hahn-Banach Extension to $\mathbb{R}^2$ uniquely?
$2$.
every linear functional that satisfies $f(-1,1)=1$ is a H-B extension of $g$ on $\mathbb{R}^2$?
$3.$ 
$f_1(x_1,x_2)=x_2,f_2(x_1,x_2)=-x_1$ are H-B extension of $g$?
Thank you for help, I am facing Functioanl Analysis problem after a long time, thank you for helping me recollecting.
I remember of the statement of H-B Extension Theorem: If I have a linear functional defined on some subspace then I can extend it on the whole vector Space.
 A: The statement of Hahn-Banach is 

Given a normed vector space $X$ and a subspace $Y \subseteq X$, then every linear continuous $y^* \in Y^*$ can be extended to the whole of $X$, i. e. to a $x^* \in X^*$ with $x^*|_Y = y^*$, with the same norm, i. e. $$ \|x^{ * } \| = \|y^{* }\|.$$

Note that not every extension is a Hahn-Banach-extension. 
Let's compute $g$'s norm, for $y \in Y$ we have $|y_1| = |y_2|$, that is $\|y\| = |y_1|$, giving 
$$ \|g\| = \sup_{\|y\| \le 1} |g(y)| = \sup_{\|y\| \le 1} |y_2| = 1 $$
An extension of $f$ of $g$ is uniquely determined by its value on $(1,1)$, as for $x \in \mathbb R^2$
\begin{align*}
  f(x_1, x_2) &= f\left(\frac{x_1 + x_2}2 + \frac{x_1 - x_2}2, \frac{x_1 + x_2}2 - \frac{x_1 - x_2}2\right)\\
       &= f\left(\frac{x_1 + x_2}2, \frac{x_1 + x_2}2\right) + f\left(\frac{x_1 - x_2}2, -\frac{x_1 - x_2}2\right)\\
       &= \frac{x_1 + x_2}2\cdot f(1,1) + g\left(\frac{x_1 - x_2}2, -\frac{x_1 - x_2}2\right)\\
       &= \frac{x_1 + x_2}2 \cdot f(1,1) -\frac{x_1 - x_2}2\\
       &= \left(\frac 12 f(1,1) + \frac 12\right)x_1 + \left(\frac 12 f(1,1)- \frac 12\right)x_2 
\end{align*}
So we have $\def\abs#1{\left|#1\right|}\def\norm#1{\left\|#1\right\|}$
\begin{align*}
  \|f\| &= \frac 12 \abs{f(1,1) + 1} + \frac 12\abs{f(1,1) - 1}\\
        &\ge \frac 12 \abs{f(1,1) + 1 - f(1,1) + 1}\\
        &= 1
\end{align*}
by the triangle inequality. We have equality iff $f(1,1) + 1$ and $1-f(1,1)$ have the same sign, that is iff $|f(1,1)| \le 1$. So any $f_\alpha$, 
$$ f_\alpha(x) = \frac{x_1}2(\alpha + 1) + \frac{x_2}2(\alpha - 1), \qquad \alpha \in [-1,1] $$
is a H-B-extension. So 1., 2. are wrong and 3. is true.
