$G = \{(x_n) \in l_1: x_{2n+1} = 0, \forall n \in \mathbb{N} \}$
Let $f: G \to \mathbb{K}$ be a continuous linear functional, $f \neq 0$.
Show that the Hahn Banach extension of $f$ is not unique.
My attempt
Let $g: l_1 \to \mathbb{K}$ be the Hahn Banach extension of $f$.
We have:
$$g(x_1,x_2,x_3,x_4, \cdots)= g(x_1,0,x_3,0, \cdots) + g(0,x_2,0,x_4, \cdots)
= g(x_1,0,x_3,0, \cdots) + f(0,x_2,0,x_4, \cdots) = \sum_{n=0}^{\infty}
g(e_{2n+1})x_{2n+1} + f(0,x_2,0,x_4, \cdots) $$
I realised the following: if $g(e_{2n+1}) = 0, \forall n \geq 1$ and $g(e_1) = \|f\|$, then
$|g(e_n)| \leq \|f\| , \forall n \geq 0$ and
$$\|g\| = \sup |g(e_n)| = \|f\|$$
Now, if $g(e_{2n+1}) = 0, \forall n \geq 0, n \neq 1,$ and $g(e_3) = \|f\|$, then
$|g(e_n)| \leq \|f\| , \forall n \geq 0$ and
$$\|g\| = \sup |g(e_n)| = \|f\|$$
Thus $g$ has at least two Hahn Banach extensions.
Am I right?
Best Answer
You solution looks good to me. I separately solved it the following way:
We can notice that $l_1$ is a direct sum of the closed subspaces $G = \{(x_n) \in l_1: x_{2n+1} = 0, \forall n \in \mathbb{N} \}$ and $H = \{(x_n) \in l_1: x_{2n} = 0, \forall n \in \mathbb{N} \}$. Also, $||x+y|| = ||x|| + ||y||$ $ \forall x \in G, y \in H$.
Thus, given bounded functionals say $f,g$ on $G,H$ respectively, we get its linear extension (call it $E$) on $l_1$. $E$ is bounded as:
$|E(x+y)| \leq |E(x)| + |E(y)| = |f(x)| + |g(y)| \leq C_1||x|| + C_2||y|| \leq C||x+y||$.
Therefore, in the given problem any bounded linear functional on H(there are infinitely many of them) gives rise to a linear functional on $l_1$. Thus, the extension is not unique in the problem.