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

[Ryker]

Homework Statement

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

Homework Equations

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

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.

 

[Eynstone]

I'll just post a hint : try A=|f(1)|.

 

[Ryker]

Hey, thanks for the hint, although I actually figured it out yesterday a couple of hours after posting 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?

 

[Eynstone]

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.