Hahn-Banach extension of function on $\mathbb{R}^2$ with $\|\;.\,\|_1$-norm

Question

Consider $\mathbb{R}^2$ with $\|\;.\,\|_1$-norm and $M=\{(x,0) \mid x \in \mathbb{R}\}$. Define $g:M \to \mathbb{R}$ by $g(x,y)=x$. Then a Hahn-Banach extension $f$ of $g$ is given by

a) $f(x,y)=2x$

b) $f(x,y)=x+y $

c) $f(x,y)=x-2y $

d) $f(x,y)=x+2y$

This question looks incoherent as written. Probably there are some typos where "$f(x+y)$" should instead read "$f(x,y)$". In this case, the most obvious HB extension $f$ of $g$ is given by the rule $f(x,y)=x+y$. The other functionals all have norm 2, and thus are not HB extensions. — Ben W, Jan 11 '17 at 14:58
Possible duplicate of "https://math.stackexchange.com/questions/1893823/hahn-banach-extension-of-g" — nmasanta, Jul 06 '22 at 09:29

score 0 · Answer 1 · answered Jan 12 '17 at 10:21

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The subspace is spanned by only one vector viz. $(1,0)$, so any linear functional on $M$ is just a real scalar. Although any linear functional on $M$ has uncountably many HB extensions.

answered Jan 12 '17 at 10:21

Tanmoy Paul

101
1

score 0 · Answer 2 · answered Jan 12 '17 at 10:30

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See Ben Wallis's comment above:

The most obvious HB extension $f$ of $g$ is given by the rule $f(x,y)=x+y$. The other functionals all have norm $2$, and thus are not HB extensions.

answered Jan 12 '17 at 10:30

Ben Grossmann

225,327

@BenWallis feel free to add your own answer with more detail – Ben Grossmann Jan 12 '17 at 10:30
thanx..plz explain it thorouhly.. – Halima.Khatun Jan 13 '17 at 06:40
Then tell me what part you don't understand. What's confusing you here? – Ben Grossmann Jan 13 '17 at 09:12
I can not understand f(x,y)=x+y? what is actual reason – Halima.Khatun Jan 13 '17 at 10:16

Hahn-Banach extension of function on $\mathbb{R}^2$ with $\|\;.\,\|_1$-norm

2 Answers2