Results 1 to 2 of 2

Thread: Metric of C[0,1]

  1. #1
    Super Member
    Joined
    Mar 2006
    Posts
    705
    Thanks
    2

    Metric of C[0,1]

    This is from my classnote:

    Define $\displaystyle C[0,1]$ to be all continuous functions on [0,1], and define its metric be $\displaystyle p(f,g)= \int ^1 _0 \mid f-g \mid dx $

    Show that if $\displaystyle f \neq g $, then $\displaystyle p(f,g) \neq 0 $

    Proof.

    if $\displaystyle f \neq g $, then $\displaystyle f(x_0) \neq g(x_0) $ for some $\displaystyle x_0 \in [0,1] $

    Consider the function $\displaystyle h(x) = f(x)-g(x) $, we see that [tex]h(x) [tex] is continuous on $\displaystyle x_0$

    Now, let $\displaystyle \epsilon = \mid f(x_0)-g(x_0) \mid $, then $\displaystyle \exists \delta > 0 \ s.t. \ \mid x - x_0 \mid < \delta $

    with $\displaystyle \mid [ f(x) - g(x)]-[f(x_0)-g(x_0)] \mid < \frac { \epsilon }{2} $

    Then we have $\displaystyle \mid f(x) - g(x) \mid \geq \frac {\epsilon }{2} $ on $\displaystyle [x_0 - \delta , x_0 + \delta ] $ (Why is that?)

    Then we have $\displaystyle p(f,g)= \int ^1_0 \mid f-g \mid dx \geq \int ^{x_0+ \delta} _{x_0 - \delta } \mid f-g \mid dx $$\displaystyle \geq 2 \delta \frac { \mid f(x_0) - g(x_0) \mid }{2} > 0 $
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Apr 2009
    From
    México
    Posts
    721
    Quote Originally Posted by tttcomrader View Post

    Then we have $\displaystyle \mid f(x) - g(x) \mid \geq \frac {\epsilon }{2} $ on $\displaystyle [x_0 - \delta , x_0 + \delta ] $ (Why is that?)

    [/tex]
    By the triangle inequality, and properties of the absolute value we have:

    $\displaystyle \vert f(x_0)-g(x_0) \vert - \vert f(x)-g(x) \vert \leq \vert f(x)-g(x) - [f(x_0)-g(x_0)] \vert < \frac{ \epsilon }{2}$ now, using the first and last parts of the inequality we have: $\displaystyle \vert f(x_0)-g(x_0) \vert -\frac{ \epsilon }{2} < \vert f(x)-g(x) \vert$ and using the fact that $\displaystyle \vert f(x_0)-g(x_0) \vert = \epsilon$ you're finished
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: Sep 17th 2011, 03:44 PM
  2. [SOLVED] Euclidean metric is a metric
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Jan 14th 2011, 12:13 AM
  3. Limit of function from one metric space to another metric space
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Sep 17th 2010, 02:04 PM
  4. Metric of C[0,1]
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Sep 10th 2009, 12:39 AM
  5. standard metric and discrete metric
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: Mar 24th 2009, 07:25 AM

Search Tags


/mathhelpforum @mathhelpforum