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

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    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. – 2017-01-11
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    yes...thanks for correction – 2017-01-11

2 Answers 2

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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.

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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.

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    @BenWallis feel free to add your own answer with more detail – 2017-01-12
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    thanx..plz explain it thorouhly.. – 2017-01-13
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    Then tell me what part you don't understand. What's confusing you here? – 2017-01-13
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    I can not understand f(x,y)=x+y? what is actual reason – 2017-01-13