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How can one show that a norm-preserving map $T: X \rightarrow X'$ where $X,X'$ are vector spaces and $T(0) = 0$ is linear? Thanks in advance.

Rudy the Reindeer
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pascal
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    Could you clarify what you mean by norm-preserving? Note that if $T$ is not linear, then $|Tx| = |x|$ does not mean that $|Tx - Ty| = |x - y|$, which is what is usually meant (and as noted below, your claim is false if you only require $|Tx| = |x|$ and true in many cases if you actually meant distance-preserving). – Ben Millwood May 17 '12 at 11:30
  • Related threads: http://math.stackexchange.com/q/121046, http://math.stackexchange.com/q/81086, http://math.stackexchange.com/q/16965 – t.b. May 17 '12 at 12:03

2 Answers2

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The claim is false. For example, take $X=X'=\mathbb{R}$, and $T(x)=|x|$ for all $x$.

Chris Eagle
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This claim is true if your map $T$ is surjective and is an isometry. In this case simply apply Mazur-Ulam theorem.

If we do not require that map is surjective, but only norm-preserving, then we can construct a counterexample $$ T:\mathbb{C}\to\mathbb{C}:z\mapsto z e^{i |z|} $$ where $\mathbb{C}$ is considered as a vector space over $\mathbb{R}$.

Norbert
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  • Hold on, the link you gave refers to isometries, but there are functions that are norm-preserving but not isometries (e.g. the norm function itself). Are you sure the theorem applies? – Ben Millwood May 17 '12 at 11:11
  • @benmachine So what is yours definition of norm-preserving? – Norbert May 17 '12 at 11:13
  • $|f(x)| = |x|$ for all $x$. Since $f(x)-f(y)$ need not be $f(x-y)$, this need not mean that $|f(x) - f(y)| = |x - y|$. – Ben Millwood May 17 '12 at 11:16
  • Oh, in this case you are right. Then even for surjective maps we can construct counterexamples. – Norbert May 17 '12 at 11:17
  • Yeah, on second thoughts, norm-preserving as I phrased it is a pretty silly condition. I can't imagine the OP doesn't mean isometry. – Ben Millwood May 17 '12 at 11:25
  • Also, your wikipedia link cites this PDF as a source which in passing mentions that any isometry to a strictly convex ("that is, no sphere contains a line segment") space is also affine, and any inner-product space and $\ell_p$ for $1<p<\infty$ is strictly convex. – Ben Millwood May 17 '12 at 11:36
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    The Mazur-Ulam theorem is a result that only holds for real vector spaces. On complex vector spaces you have many counterexamples, perhaps worth mentioning: $z \mapsto \bar{z}$. – t.b. May 17 '12 at 11:57