- #1

AxiomOfChoice

- 533

- 1

[tex]

\left| (||x|| - ||y||) \right| \leq ||x-y||

[/tex]

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- Thread starter AxiomOfChoice
- Start date

- #1

AxiomOfChoice

- 533

- 1

[tex]

\left| (||x|| - ||y||) \right| \leq ||x-y||

[/tex]

- #2

arildno

Science Advisor

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- 135

[tex]||x+y||\leq{||}x||+||y||[/tex]

Now, set x=z-y

Then, we get:

[tex]||z||\leq{||z-y||}+||y||[/tex],

For completely arbitrary z and y.

utilize this to derive the desired relation, i.e:

[tex]-||u-v||\leq{||u||}-||v||\leq{||}||u-v||[/tex]

- #3

poutsos.A

- 102

- 1

[tex]

\left| (||x|| - ||y||) \right| \leq ||x-y||

[/tex]

||x||=||x-y+y||[tex]\leq[/tex]||x-y||+||y|| ======>

||x||-||y||[tex]\leq[/tex]||x-y||............1

||y||= ||y-x+x||[tex]\leq[/tex]||x-y||+||x||======>

||y||-||x||[tex]\leq||x-y||[/tex] =======>

||x||-||y||[tex]\geq[/tex]-||x-y||............2

from (1) and (2) we get:

-||x-y||[tex]\leq ||x||-||y||\leq ||x-y|| \Longleftrightarrow|(||x||-||y||)|\leq||x-y||[/tex]

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