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Suppose that $F:X\rightarrow\mathbb{C}^n$ is a nonzero, linear, continuous function on a normed vector space $X$. Let $\mathbb{C}^n$ be endowed with some norm so that $\|F\| = 1$. Is it true that for any $x\in X$ we have $\|x + \ker(F)\| = \|F(x)\|$?

This is true for linear functionals $f\in X^*$, as was brought up in this ME post: Distance of $x$ to kernel of bounded linear functional is the norm of the functional at $x$?.

Can someone provide a proof or reference that it works in general? Or perhaps a counterexample if it is false?

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  • $\begingroup$ What do you mean by $\| x+\ker F\|$? $\ker F$ is a subspace of $X$, not an element. Note that the other question mentions the distance between a point and a subset, but that's not the same as the distance between two points. $\endgroup$ – Arnaud D. Sep 30 '17 at 10:25
  • $\begingroup$ $\|x + ker(F)\| = \inf\{\|x + y\| : y\in \ker(F)\}$. This is the distance from the subset $x + \ker(F)$ to the origin. $\endgroup$ – Question Asker Sep 30 '17 at 10:32
  • $\begingroup$ @ArnaudD. Thank you for pointing this out. The $\|x\|$ was a transcription error. The question was intended to be a direct generalization of the one in the posted link. $\endgroup$ – Question Asker Sep 30 '17 at 10:47
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This is false. Take for example $$F:\mathbb{C}^2\to \mathbb{C}^2 : (z_1,z_2)\mapsto \left(z_1,\frac{z_2}{2}\right)$$ with the usual (euclidean) norm on $\Bbb C^2$. You can check easily that $\|F(z_1,z_2)\|\leq \|(z_1,z_2)\|$, with equality if and only if $z_2=0$; thus $\|F\|=1$ for the induced norm. Moreover the kernel of $F$ only contains the origin, so the distance of $(z_1,z_2)$ to the kernel is simply its norm. Then if $x=(0,1)$, you find that $\operatorname{dist}(x,\ker F)=\|x\|=1$ but $\|F(x)\|=\frac{1}{2}$.

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  • $\begingroup$ I am so sorry. I originally stated this as $\|x + ker(F) \| = \frac{\|F(x)\|}{\|F\|}$ but copied something wrong when stating the question. $\endgroup$ – Question Asker Sep 30 '17 at 10:46
  • $\begingroup$ @QuestionAsker Well, that doesn't really make any difference : in my counterexample $x$ has norm $1$, so the result is the same with or without division... $\endgroup$ – Arnaud D. Sep 30 '17 at 10:54
  • $\begingroup$ This was much easier than I thought. Thanks. I think you mean that $\|F\| = 1$ in the second line above. $\endgroup$ – Question Asker Sep 30 '17 at 11:00
  • $\begingroup$ Indeed, thanks for pointing that out ! (that clashes a lot with the next sentence...) $\endgroup$ – Arnaud D. Sep 30 '17 at 11:03

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