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Let X := $\mathbb{C}^{3}$ equipped with the norm $|(x, y, z)|_{1} := |x| + |y| + |z|$ and $Y := \{(x, y, z) ∈ X|x + y = 0, z = 0\}$. Find at least two extensions of $ℓ(x, y, z) := x$ from $Y$ to $X$ which preserve the norm. What if we take $Y := \{(x, y, z) ∈ X|x + y = 0\}$?

My first approach (I am pretty unfamiliar with extensions of linear functionals, so I try to be as precise as possible): clearly our linear functional $l(x,y,z)$ is bounded since $|l(x,y,z)| = |x| \leq |x| + |y| + |z| = |(x,y,z)|_{1}$. At this point, we can say that Hahn-Banach tells us that the existence of such an extension (which preserves the norm!) is guaranteed.

So, we have $||l|| \leq 1$. I am a little bit unsure about this point but by taking $y = z = 0$, we see that we get equality and I think therefore we can conclude that $||l|| = 1$ (even though I am open to more elaborate suggestions about that).

Now, my guess would be that $l(x,y,z) = y$ and $l(x,y,z) = z$ are two extensions of $l$ which also should preserve the norm. However, I am not really able to show that. Can anybody help me?

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You are indeed correct about the norm of your first extension $l_1 : X \to \mathbb{C}: (x, y, z) \mapsto x.$ To obtain a second viable extension, note that on $Y$ we have that $x = -y,$ so we may define $l_2 : X \to \mathbb{C}: (x, y, z) \mapsto -y.$ Clearly $l_1 \neq l_2,$ $l_2$ is an extension of your original functional on $Y$ and $\lVert l_2 \rVert = 1$ by a similar argument to the one you made in your original post.

In fact, we can completely characterise the functionals that extend your original one as such: let $\lambda : X \to \mathbb{C}$ extend $l$ and let $\alpha = \lambda((1, 0, 0)), \gamma = \lambda((0, 1, 0)), \beta = \lambda((0, 0, 1)).$ Then $\lambda(x, y, z) = a x + b y + c z.$ We want $\lambda \vert_X = l,$ i.e. $\alpha x + \gamma y + \beta z = x$ for all $(x, y, z) \in Y.$ Of course, since $z = 0$ on $Y,$ we can choose any $\beta \in \mathbb{C}.$ Moreover, since $x = -y$ on $Y,$ we deduce that $\alpha - \gamma = 1.$ Thus, the set of all functionals that extend $l$ is $$\{\lambda_{\alpha, \beta}: X \to\mathbb{C}: (x, y, z) \mapsto \alpha x + (\alpha - 1) y + \beta z : \alpha, \beta \in \mathbb{C}\}.$$

In case we are interested in extending the original functional $l$ defined on $Y$ when we don't include $z = 0$ in its definition, note that the above calculation also shows that $\beta = 0,$ so the set of solutions in this case will be modified accordingly (just take $\beta = 0$ in the set above).

I will let it up to you to find those exact functionals out of the ones above that actually preserve the norm, which will be only a matter of simple computation. I hope this helps. :)

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  • $\begingroup$ The second extension $l_2$ makes sense to me but I am struggling with $l_1$. So, as first extension we can choose the linear functional itself? I thought that an extension of $l$ is supposed to be different from $l$? It sounds a little bit like cheating to me. :-D $\endgroup$
    – MathGeek
    Dec 11, 2023 at 2:16
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    $\begingroup$ Alright then, if $l_1$ doesn't sit right with you, you can also take something along the lines of $l_\alpha(x, y, z) = \alpha x - (1 - \alpha) y$ for any $\alpha \in \mathbb{C}.$ In fact, we are able to completely characterise the functionals that extend your initial one. I will edit my original response in view of this fact. $\endgroup$ Dec 11, 2023 at 2:25
  • $\begingroup$ Thank you! I'm completely fine with that one! However, what is now the difference between our two spaces Y? We never used that $z = 0$ in the first $Y$. Does that mean that the extensions for both $Y$s are the same? $\endgroup$
    – MathGeek
    Dec 11, 2023 at 2:27
  • $\begingroup$ Ah, I didn't see your second space $Y$, one sec. $\endgroup$ Dec 11, 2023 at 2:41
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    $\begingroup$ Ok, that was hopefully my last edit. Just one last minute remarks that is worth mentioning on the side: in general if you want to extend a functional defined on a space $Y$ to the entirety of a finite-dimensional space, the resulting space of solutions will have dimension equal to the codimension of $Y$ (can you see why?). If my answer was helpful, please do consider accepting it :) $\endgroup$ Dec 11, 2023 at 2:51

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