Let $T$ be a linear map defined on a subspace $S$ of some Euclidean vector space $V$, such that $$ d(T(v_1),T(v_2))\leq d(v_1,v_2) \quad \text{for } v_1,v_2\in S $$ where $d(v_1,v_2)$ is the Euclidean distance on $V$. Is there always a linear map $T'$ defined on $V$ such that $$ d(T'(v_1),T'(v_2))\leq d(v_1,v_2) \quad \text{for all } v_1, v_2\in V $$ and that $T'(v)=T(v)$ for all $v\in S$?
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Yes, this is always possible in $\Bbb R^n$ (finite dimensional Euclidean space). In particular, we may note that $V = S \oplus S^\perp$, and so each $v \in V$ can be uniquely expressed as $v = v_S + v^\perp$ with $v_S \in S$ and $v_\perp \in S^\perp$. We may then define $$ T'(v) = T'(v_S + v^\perp) = T'(v_S) $$ Verify that this $T'$ satisfies the requirements given.
Ben Grossmann
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I see a problem: $d(T'(v_1),T'(v_2))=d(T(v_{S1}),T(v_{S2}))\leq d(v_{S1},v_{S2})\neq d(v_{1},v_{2})$. – NessunDorma Mar 17 '17 at 17:29
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@NessunDorma note that $$d(v_{1S},v_{2S}) = |v_{1S} - v_{2S}| = |(v_1 - v_2)_S| \leq |v_1 - v_2| = d(v_1,v_2)$$ – Ben Grossmann Mar 17 '17 at 17:45