# Finding two extensions of this linear functional

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?

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. :)

• 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 Dec 11, 2023 at 2:16
• 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. Dec 11, 2023 at 2:25
• 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? Dec 11, 2023 at 2:27
• Ah, I didn't see your second space $Y$, one sec. Dec 11, 2023 at 2:41
• 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 :) Dec 11, 2023 at 2:51