3

Consider $\mathbb R^2$ with $||.||_1$ norm and $M=\{(x,0):x\in \mathbb R \}$. Define $g:M \to \mathbb R$ by $g(x,0)=x$.

Then which one is the Hahn-Banach extension $f$ of $g$?

  1. $f(x,y)=2x $
  2. $f(x,y)=x-2y$
  3. $f(x,y)=x+2y$
  4. $f(x,y)=x+y$

According to Hahn Banach theorem on normed linear space: If I have a linear functional defined on some subspace then I can extend it on the whole vector Space with same norms.

I think $||(x,y)||_1=|x|+|y|,(x,y)\in \mathbb R^2$

Here $||g||=\sup_{x\neq0} \frac {|g(x,0)|}{||(x,0)||_1}$,$x\in \mathbb R$

$$\implies||g||=1$$

Now $||f||= \sup_{(x,y)\neq0} \frac {|f(x,y)|}{||(x,y)||_1}$,$(x,y)\in \mathbb R^2$. Further i can't proceed. What is the norm of $f$ and which one is the correct answer?

Martin R
  • 113,040
user303866
  • 31

1 Answers1

2

(1) is wrong because $f$ does not extend $g$.

For (2) and (3), consider $f(0, 1)$ to conclude that $||f|| > 1 = ||g||$.

In case (4): $$ |f(x, y)| = | x + y | \le |x| + |y| = ||(x,y)||_1 $$ with equality for $x = y = 1$, therefore $||f|| = 1$.

Remark: You can compute $||f||$ also in the cases (1) to (3) by determining the maximum of $|f(x,y)|$ on the set $|x| + |y| = 1$ which is a rotated square, but that is not necessary to find the correct solution.

Martin R
  • 113,040