Show the norm ||x|| is less or equal to A|x| for some constant A

  • Thread starter Ryker
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  • #1
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Homework Statement


Show that [tex]\|x\| \leq A|x| \forall x \in \mathbb{R},[/tex] where [tex]A \geq 0.[/tex]

Homework Equations


We know the norm is a function [tex]f: {\mathbb{R}}^{d} \to \mathbb{R},[/tex] such that:
[tex]a) f(x) = 0 \iff x = 0,[/tex]
[tex]b) f(x+y) \leq f(x) + f(y),[/tex] and
[tex]c) f(cx) = |c|f(x) \forall c \in \mathbb{R}[/tex]

The Attempt at a Solution


Ugh, I'm completely stumped here, and don't know where to begin. I know that for the Euclidean norm this is trivial, but I don't know how to even begin showing this in general. In particular, I don't see where I could grasp the absolute value (or the dot product), or how to start comparing it to the norm.

Any help would be greatly appreciated.
 

Answers and Replies

  • #2
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I'll just post a hint : try A=|f(1)|.
 
  • #3
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Hey, thanks for the hint, although I actually figured it out yesterday a couple of hours after posting :smile: I did it via the unit vectors and then expanding upon the properties b) and c). I assume this is what you were going for with the hint, as well, right?
 
  • #4
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I did it via the unit vectors and then expanding upon the properties b) and c). I assume this is what you were going for with the hint, as well, right?

Yes, I thought on similar lines.
 

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