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Homework Help: 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)}|


    |{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
    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


    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
  5. Oct 3, 2012 #4
    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,






    But keep getting unhelpful things like ^ haha

    Is the next step something like what I did above?
  7. Oct 3, 2012 #6
    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
    It will be easier if you start with

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

    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

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

    Which is only true if

    |F(t)|≥(1/2)|F(t_0)| !

    Thanks heaps!
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