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

[kingwinner]

"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!

 

[g_edgar]

$$\Big|\|x\| - \|y\| \Big|\le \|x-y\|$$

 

[kingwinner]

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?

 

[evagelos]

$$\Big|\|x\| - \|y\| \Big|\le \|x-y\|$$

$$\Big|\|x\| - \|y\| \Big|\le \|x-y\|$$ <=> -||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||

 

[kingwinner]

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?

 

[evagelos]

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

 

[kingwinner]

$$\left|d(x,0)-d(y,0)\right|\leq d(x,y)$$

The same proof

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

 

[evagelos]

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)