$G = \{(x_n) \in l_1: x_{2n+1} = 0, \forall n \in \mathbb{N} \}$

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?


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.

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  • $\begingroup$ Could you explain how you got equality of the norms and how you're calling extension of both f and g to be E? $\endgroup$ – Sahiba Arora Sep 7 '16 at 16:40
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    $\begingroup$ You get equality of norms as $l^1$ norm is just sum of absolute values of terms in the sequence and elements in H have odd terms 0 and elements in G have even terms 0. $\endgroup$ – Aman Sep 7 '16 at 16:49
  • $\begingroup$ Oh, I got it. Silly question. What about my other question? $\endgroup$ – Sahiba Arora Sep 7 '16 at 16:58
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    $\begingroup$ In general if $V= U \oplus W$ (direct sum of vector spaces) and if $f,g$ are linear functionals on $U$ and $W$ respectively, then you can get a unique linear functional $E$ on $V$ such that $E|_{U} =f$ and $E|_{W} = g$. $E$ can be given as $E(x+y) = E(x) + E(y)$ where $x \in U, y\in W$. $\endgroup$ – Aman Sep 7 '16 at 17:08

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