Isometry Problem: Does Convergence in Norm Imply Convergence Under Map?

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SUMMARY

The discussion centers on the implications of isometric mappings in normed spaces, specifically whether convergence in the norm of space X guarantees convergence in space Y under an isometric map f: X → Y. It is established that if f is a linear isometry, then the property holds true, meaning that \(\|f(x_2) - f(x_1)\|_Y = \|x_2 - x_1\|_X\). However, if f is not linear, this property may not necessarily hold, indicating a critical distinction in the behavior of isometries based on linearity.

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  • Understanding of isometric mappings in metric spaces
  • Familiarity with linear transformations in vector spaces
  • Knowledge of convergence concepts in normed spaces
  • Basic principles of functional analysis
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Mathematicians, students of functional analysis, and anyone studying the properties of isometries and convergence in normed spaces will benefit from this discussion.

wayneckm
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Hello all,


My question is if [tex]f : X \mapsto Y[/tex] is an isometry which preserves norm, i.e. [tex]\left\| f(x) \right\| _{Y} = \left\| x \right\|_{X}[/tex], does this imply [tex]\left\| f(x_2) - f(x_1) \right\| _{Y} = \left\| x_2 - x_1 \right\|_{X}[/tex]?

Or, essentially is it sufficient to gurantee convergence in [tex]X[/tex]-norm implies convergence in [tex]Y[/tex]-norm under the map [tex]f[/tex]?

Thanks a lot.


Wayne
 
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If your f is linear, then this will be true. Otherwise, it may fail...
 

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