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.
Norm-preserving map is linear
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$\begingroup$
functional-analysis
vector-spaces
normed-spaces
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1Could 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). – 2012-05-17
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0Related threads: http://math.stackexchange.com/q/121046, http://math.stackexchange.com/q/81086, http://math.stackexchange.com/q/16965 – 2012-05-17
2 Answers
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The claim is false. For example, take $X=X'=\mathbb{R}$, and $T(x)=|x|$ for all $x$.
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0I think, for nonlinear maps, "norm preserving" has to mean $\|T(x)-T(y)\| = \|x-y\|$. – 2012-05-17
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0@GEdgar, the OP assumes also $T(0)=0$ and so $\|T(x)-T(y)\| = \|x-y\|$ implies $\|T(x)\| = \|x\|$. – 2012-05-17
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1But not conversely. – 2012-05-17
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This claim is true if your map $T$ is surjective and an isometry. In this case simply apply Mazur-Ulam theorem.
If we require that map to be surjective, but only norm-preserving, then we can construct counterexample $$ T:\mathbb{C}\to\mathbb{C}:z\mapsto z e^{i |z|} $$ where $\mathbb{C}$ is consideres as vector spaces over $\mathbb{R}$.
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0Hold 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? – 2012-05-17
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0@benmachine So what is yours definition of norm-preserving? – 2012-05-17
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0$|f(x)| = |x|$ for all $x$. Since $f(x)-f(y)$ need not be $f(x-y)$, this need not mean that $|f(x) - f(y)| = |x - y|$. – 2012-05-17
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0Oh, in this case you are right. Then even for surjective maps we can construct counterexamples. – 2012-05-17
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0Yeah, on second thoughts, norm-preserving as I phrased it is a pretty silly condition. I can't imagine the OP doesn't mean isometry. – 2012-05-17
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0Also, your wikipedia link cites [this PDF](http://www.helsinki.fi/~jvaisala/mazurulam.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
– 2012-05-17
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0The Mazur-Ulam theorem is a result that only holds for *real* vector spaces. On complex vector spaces you have many counterexamples, perhaps worth mentioning: $z \mapsto \bar{z}$. – 2012-05-17