How do I prove the continuity of the norm in any n.l.s.?

[gtfitzpatrick]

Homework Statement

x[
Prove the continuity of the norm; ie show that in any n.l.s. N if x_n \rightarrow x then \left|\left|x_n\left|\left| \rightarrow \left|\left|x\left|\left|

The Attempt at a Solution

i don't know where to start this
from the definition of convergence x_n \rightarrow x as n\rightarrow \infty if \left|\left|x_n - X\left|\left| \rightarrow 0 as n\rightarrow \infty

do i use this to prove it or am i barking up the wrong tree?

 

[Dick]

Sure you use the definition of convergence. Can you prove the 'reverse triangle inequality'? abs(||a||-||b||)<=||a-b||.

 

[gtfitzpatrick]

let |a| ≥ |b|,
However, we can write a as a - b + b or a = (a - b) + b

Then |a| = |a - b + b|. But, |a - b + b| \leq |a - b| + |b| by the Triangle Inequality, and so we have that |a| \leq |a - b| + |b| . Now, subtract |b| from both sides. This gives us: |a|-|b| \leq |a - b|

right?

 

[Dick]

let |a| ≥ |b|,
However, we can write a as a - b + b or a = (a - b) + b

Then |a| = |a - b + b|. But, |a - b + b| \leq |a - b| + |b| by the Triangle Inequality, and so we have that |a| \leq |a - b| + |b| . Now, subtract |b| from both sides. This gives us: |a|-|b| \leq |a - b|

right?

Sure. That's it. Makes your proof easy, right?

 

[gtfitzpatrick]

thanks a million, i guess I'm just not looking at them right, but now i see