Hello! I have got problems with the following exercise: Let $\displaystyle f$ be the continuous linear functional defined on the subspace $\displaystyle U=\{(x,y,z): x+2y=0, z=0\}$ in the Banach space $\displaystyle (\mathbb{R}^3,||.||_1)$ with $\displaystyle f((x,y,z)):=x$. Find different linear extensions on the whole space with same norm as $\displaystyle f$!

I have calculated the norm of $\displaystyle f$ on $\displaystyle U$, it is $\displaystyle \frac{2}{3}$. My first idea was to extend $\displaystyle (2,-1,0)$ to the basis $\displaystyle v_1=(2,-1,0,), v_2=(0,1,0), v_3=(0,0,1)$ and define $\displaystyle f(v_2)=f(v_3)=0$.

But does this extension have the same norm? If yes, how can i show this? How do i find another extension? Hoping for your advice,

Recursion