Proving Normed Vectorspace: K > 0 Solution

[Hummingbird25]

Hi

I'm tasked with proving the following:

Let S be an open interval S and $$f: S -> \mathbb{R}^n$$ be a continuous function.

Let $$|| \cdot ||$$ be norm on $$\mathbb{R}^n$$. show

1) There exist a K > 0 such that $$||x|| \leq K||x||_1 ; \ x \in \mathbb{R}^n, ||x||_1 = \sum_{i=1} ^n |x_i|$$.

My Solution:

According to the definition the norm of a vector x in R^n is the non-negative scalar $$||x|| = \sqrt{x_1^{2} + x_2^{2} + \cdots x_n^2}$$

The L1-norm can be written as $$||x||_1 = |x_1| + |x_2| + \cdots + |x_n|$$

Expanding the inequality:

$$\sqrt{x_1^{2} + x_2^{2} + \cdots x_n^2} \leq K|x_1| + K|x_2| + \cdots + K|x_n|$$

Is it then concludable from this that since K>0, then the right side of the inequality sign will always be larger than the left side?

Sincerley Yours
Humminbird25

 

[StatusX]

Is it then concludable from this that since K>0, then the right side of the inequality sign will always be larger than the left side?

Can you expand on how you think this argument would work?