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I'm tasked with proving the following:

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

Let [tex]|| \cdot ||[/tex] be norm on [tex]\mathbb{R}^n[/tex]. show

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

My Solution:

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

The L1-norm can be written as [tex]||x||_1 = |x_1| + |x_2| + \cdots + |x_n|[/tex]

Expanding the inequality:

[tex]\sqrt{x_1^{2} + x_2^{2} + \cdots x_n^2} \leq K|x_1| + K|x_2| + \cdots + K|x_n|[/tex]

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

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# Normed vectorspace

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