Given the subspace $Y=\{(x,y,z): z=0 \}$ of $\mathbb R^3$, I would like to determine all norm-preserving linear extensions for the functional

$$f(x):= a \cdot x$$

where $a $ is some fixed vector of the form $(a_1,a_2,0)$

Here norm-preserving linear extension $g $ is any functional defined on all of $\mathbb R^3 $, such that $g(x)=f(x)$ for $x \in Y$, and such that $\|g \| = \|f \| $. And such an extension is possible by the Hahn-Banch Theorem.

I'm not sure how to approach the problem but I believe I can see the following:

The function $f $ would in fact be defined also as a function from $\mathbb R^3 $ and would still be linear as such. Thus this is one possible extension?

Is it possible to extend $f $ so that it is "an other function on $\mathbb R^3 \setminus Y $"? Or will this break down the linearity?

Much grateful for any help provided!

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  • $\begingroup$ Using the Euclidean norm? $\endgroup$ – copper.hat Jan 8 at 20:24
  • $\begingroup$ Yes I'm using the Euclidean norm! $\endgroup$ – MrFranzén Jan 8 at 20:30

The norm of the functional $f(x) = a^T x$ is $\|a\|$. On $Y$, $f$ has norm $\sqrt{a_1^2 + a_2^2}$, hence if extended to the entire space, we must have $\bar{a} = (a_1,a_2,a_3)^T$ for some $a_3$. Hence the only way the norm can be preserved is if $a_3=0$.

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    $\begingroup$ Thank you! But how do we make sure that only extensions of the form $g(x) = \bar{a} x $ are possible? $\endgroup$ – MrFranzén Jan 8 at 20:52
  • $\begingroup$ All linear functionals on $\mathbb{R}^3$ are of the form $b^Tx$ for some $b$. It is immediate that $b_1=a_1$, $b_2=a_2$ by taking $x=(1,0,0)^T $ and $x=(0,1,0)^T$. $\endgroup$ – copper.hat Jan 8 at 20:54
  • $\begingroup$ Ah that fact is due to Riesz theorem, isn't it so? $\endgroup$ – MrFranzén Jan 8 at 20:55
  • $\begingroup$ Or the fact that $e_1,e_2,e_3$ form a basis for $\mathbb{R}^3$. $\endgroup$ – copper.hat Jan 8 at 20:59

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