1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Metrics on continuous functions question

  1. Oct 2, 2012 #1
    Hey guys,

    I have been working on the following question:

    http://imageshack.us/a/img407/4890/81345604.jpg [Broken]

    For part a

    f and g are continuous on I

    => there exists e > 0 and t_0 s.t.

    0<|{f(t) - g(t)} - {f(t_0) - g(t_0)}| < e

    using |a-b| >= |a| - |b|,

    |{f(t) - g(t)} - {f(t_0) - g(t_0)}| >= |{f(t) - g(t)}| - |{f(t_0) - g(t_0)}|

    and

    |{f(t) - g(t)}| - |{f(t_0) - g(t_0)}| >= |{f(t) - g(t)}| - |f(t_0)| + |g(t_0)|

    but i'm not sure where to go from here, I have tried a lot of different steps but I end up going in circles or getting back to the triangle inequality haha.

    Does anyone have any ideas on how to do this? Is there some simple trick that I am just missing?

    For part b,

    Proving σ>=τ is pretty straight forward, just applying the Schwarz inequality but I'm having trouble showing that ρ>σ

    How can you relate that sup to the integral expression σ?

    Thanks in advance
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Oct 2, 2012 #2

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    For part (a): can you show that if [itex]F:[0,1]\rightarrow \mathbb{R}[/itex] is continuous and if [itex]F(t_0)>0[/itex] then there [itex]F(t)\geq \frac{1}{2} F(t_0)[/itex] for all t in some neighborhood of [itex]t_0[/itex].

    Use continuity.

    For (b): apply that [itex]|f(t)-g(t)|\leq \rho(f,g)[/itex] and use that inequality in the integral.
     
  4. Oct 3, 2012 #3
    Hey micromass, thanks for replying!

    So the continuity of F(t) gives you that (t_0 in [0,1])

    For every ε>0 there exists a δ>0 such that |F(t) - F(t_0)|<ε for 0<|t-t_0|<δ

    |F(t) - F(t_0)|≥|F(t)|-|F(t_0)|≥0

    |F(t)|≥|F(t_0)|

    I havn't been able to figure out how the 1/2 comes into it,

    Is |F(t)|-|F(t_0)|≥0 not correct?

    Is it just a matter of saying, then for t in [t_0 - d, t_0 + d] for some d it is possible that
    |F(t)|-(1/2)|F(t_0)|≥0
     
  5. Oct 3, 2012 #4

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    It's not correct. Why should [itex]|F(t)|-|F(t_0)|\geq 0[/itex] hold? I agree that [itex]|F(t)-F(t_0)|\geq |F(t)|-|F(t_0)|[/itex] and that [itex]|F(t)-F(t_0)|\geq 0[/itex]. But I don't see how that follows.
     
  6. Oct 3, 2012 #5
    Ah right

    I'm really have trouble getting past this step

    |F(t) - F(t_0)|≥|F(t)|-|F(t_0)|

    Today I have been trying things like this,

    |F(t)|-|F(t_0)|≥|F(t)|-2|F(t_0)|

    |F(t)|-2|F(t_0)|≥(1/2)|F(t)|-2|F(t_0)|

    |F(t)|-|F(t_0)|≥(1/2)|F(t)|-2|F(t_0)|

    (1/2)|F(t)|-|F(t_0)|≥-2|F(t_0)|

    |F(t)|≥-2|F(t_0)|

    But keep getting unhelpful things like ^ haha

    Is the next step something like what I did above?
     
  7. Oct 3, 2012 #6

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    You need to use continuity of F. Because F is continuous you know that: for each [itex]\varepsilon>0[/itex], there exists an interval I such that [itex]t_0\in I[/itex] and such that if [itex]t\in I[/itex], then [itex]|F(t)-F(t_0)|<\varepsilon[/itex].

    Now, choose a special value for [itex]\varepsilon[/itex].
     
  8. Oct 3, 2012 #7
    I tried picking different ε before but I couldn't see how having

    |F(t)| - |F(t_0)|≤|F(t) - F(t_0)|<ε

    Could get to F(t)≥(1/2)F(t_0)

    By saying F(t_0)>0 were you hinting that ε=F(t_0)?
     
  9. Oct 3, 2012 #8

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    It will be easier if you start with

    [tex]|F(t_0)|-|F(t)|\leq |F(t_0)-F(t)|<\varepsilon[/tex]
    [/QUOTE]

    Could get to F(t)≥(1/2)F(t_0)

    By saying F(t_0)>0 were you hinting that ε=F(t_0)?[/QUOTE]

    Something like that. What if you try [itex]\varepsilon=|F(t_0)|/2[/itex] ?
     
  10. Oct 3, 2012 #9
    Ohhhhhhhh

    |F(t_0)| - |F(t)|<F(t_0)/2

    Which is only true if

    |F(t)|≥(1/2)|F(t_0)| !

    Thanks heaps!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook