Find the value of $\alpha-\beta$ if $g(x,y)=\alpha x+\beta y$ is a Hahn-Banach extension of $f$

Question

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.

Answer 1

I think you have $\|f\| = \sup \{|3x| : 2|x| = \|(x,x)\| = 1\} = 3/2$, not $\|f\| =3$.

Then, since $\alpha + \beta =3$ and since $\max \{|\alpha |, |\beta|\} = 3/2$, you must have $\alpha = \beta = 3/2$ (why?). Hence, $\alpha - \beta =0$.

Here, I used that the dual norm to the $\ell^1$ norm is the $\ell^\infty $ norm.