Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Normed vectorspace

  1. Oct 3, 2006 #1

    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
    Last edited: Oct 3, 2006
  2. jcsd
  3. Oct 3, 2006 #2


    User Avatar
    Homework Helper

    Can you expand on how you think this argument would work?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook