F(x)=||x|| is Lipschitz function

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Discussion Overview

The discussion centers around the properties of the function f(x) = ||x|| in the context of normed vector spaces, specifically its Lipschitz continuity as derived from the triangle inequality. Participants explore the proof of this property and its implications, as well as analogues in metric spaces.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant states that f(x) = ||x|| is a Lipschitz function from a normed vector space V into [0,∞) based on the triangle inequality.
  • Another participant questions how this follows from the triangle inequality and requests a proof.
  • A participant suggests that the "variant triangle inequality" is relevant and mentions having seen a proof in R, but seeks clarification on proving it for normed vector spaces.
  • Several participants provide expressions of the variant triangle inequality, indicating its form and implications in the context of normed spaces.
  • One participant inquires about the existence of an analogue of the "variant triangle inequality" in general metric spaces and how the proof might differ.
  • Another participant proposes that the same proof applies in metric spaces, leading to a question about whether the distance function d is always a Lipschitz function.
  • There is a clarification that the Lipschitz function in the metric space context is f(x) = d(x,0) rather than d(x,y).

Areas of Agreement / Disagreement

Participants express uncertainty regarding the proof of the Lipschitz property for ||x|| in normed spaces and the applicability of the variant triangle inequality in metric spaces. There is no consensus on the proof details or the implications for the distance function in metric spaces.

Contextual Notes

The discussion involves various assumptions about the properties of normed vector spaces and metric spaces, and the proofs referenced may depend on specific definitions and contexts that are not fully resolved in the thread.

kingwinner
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"Let (V,||.||) be a normed vector space. Then by the triangle inequality, the function f(x)=||x|| is a Lipschitz function from V into [0,∞)."

I don't understand how we this follows from the triangle inequality. How does the proof look like?

Any help is appreciated!
 
Last edited:
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\Big|\|x\| - \|y\| \Big|\le \|x-y\|
 
OK, I think that's the "variant triangle inequality". I've seen the proof in R (by the regular triangle inequality), but how can we prove it for normed vector spaces?
 
Last edited:
g_edgar said:
\Big|\|x\| - \|y\| \Big|\le \|x-y\|


\Big|\|x\| - \|y\| \Big|\le \|x-y\| <=> -||x-y||\leq||x||-||y||\leq ||x-y|| <=> -||x-y|| +||y||\leq ||x||\leq ||x-y|| +||y|| which is correct since

||x|| =||x-y+y||\leq ||y|| +||x-y|| and

||y|| = ||y-x+x||\leq ||x-y||+ ||x||
 
Thanks!

Just wondering...is there a analogue of the "variant triangle inequality" in general METRIC SPACES? If so, how does it look like and how does the proof change?
 
kingwinner said:
Thanks!

Just wondering...is there a analogue of the "variant triangle inequality" in general METRIC SPACES? If so, how does it look like and how does the proof change?
\left|d(x,0)-d(y,0)\right|\leq d(x,y)

The same proof
 
Last edited:
evagelos said:
\left|d(x,0)-d(y,0)\right|\leq d(x,y)

The same proof

So in the metric space context, does it mean that the function "d" is always a Lipchitz function as well?
 
kingwinner said:
So in the metric space context, does it mean that the function "d" is always a Lipschitz function as well?

The Lipschitz function is f(x) =d(x,0) and not d(x,y)
 

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