Consider the normed linear space $\Bbb R^2$ with this norm $\|(x,y)\|=|x|+|y|$.
Define $f$ to be a bounded linear functional on $X=\{(x,y):x=y\}$ by $f(x,y)=3x$.
If $g$ is a Hahn-Banach extension of $f$ on $\Bbb R^2$ given by $g(x,y)=\alpha x+\beta y$ then find $\alpha-\beta$.
My try:
Since $g$ is an extension of $f$ so $g|_X=f\implies g(x,x)=(\alpha+\beta)x=3x\implies \alpha+\beta =3$
Now $\|f\|=\sup \{f(x,x):\|(x,x)\|=1,(x,x)\in \Bbb R\}=3\implies \|g\|=3$
So $\sup \{g(x,y):\|(x,y)\|=1,(x,y)\in \Bbb R\}=3\implies \alpha x+\beta y\le 3$ where $|x|+|y|=1$.
Take $x=0,y=1\implies \alpha x+\beta y\le 3\implies \beta\le 3$.
How to find the value of $\alpha,\beta $ from above.
Please give some hints.