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# Linear isometry

Let . If:

• for all

then prove that is linear.

Solution

For convenience, identify with here. Then note that for any such function , also a solution for any point on the unit circle. Also is a solution. Note that and hence we can wlog assume that . So is a point on the unit circle with distance to . Hence , so w.l.o.g. assume that . But then for any , both and have the same distance to and . So supposing , all lie on the perpendicular bisector between these points and in particular and are collinear which clearly is absurd. Hence for all which proves the claim.

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