What is the Relationship Between Vectors and Normed Linear Spaces?

[Oster]

I'm trying to do a problem concerning converging sequences in normed linear spaces. Can anyone help me prove that if x=(x1,x2...,xn) is a vector in an n dimensional vector space then |xi| where i=1,2...,n; is always less than or equal to ||x|| (norm of x). Maybe start out by writing x as a sum of n multiples of the basis vectors?

 

[lanedance]

its not quite true, should be less than or equal..

is this the usual norm or just a norm in general?

 

[Oster]

yeah, i forgot to put in the "or equal to". It is any norm in general.

 

[lanedance]

so what do you know about a norm that may help?

in particular, i would look at the triangle inequality

 

[lanedance]

I still can't see it.
Triangle inequality -> ||a+b|| <= ||a|| + ||b||
and ||x-y|| <= ||x-z|| + ||y-z||

how about considering
x1 = (x1,0,0,..) and
x = x1 + u
or maybe even better
x1 = x+(-u)

 

[Oster]

uhhh
So I have ||x1|| = |x1| <= ||x|| + ||u||

 

[Oster]

I still don't see how the triangle inequality implies this =/. I think I proved it using orthogonality and inner product properties in the finite dimensional case.

Assume the negation is true. There exists a non-zero vector x=(x1,x2...xn) such that
|x1| > ||x|| writing out with an orthonormal basis, we get ||x1e1|| > ||x1e1...+xnen||.
On squaring, we'd get x1^2 > x1^2 +... + xn^2 whiich is a contradiction?

 

[lanedance]

so this assumes the standard norm - is that ok?