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Continuity implies boundedness in an interval proof

  1. Sep 1, 2011 #1
    1. The problem statement, all variables and given/known data

    If a function [itex]f[/itex] is continuous at a point [itex]x[/itex], then [itex]f[/itex] is bounded on some interval centered at [itex]x[/itex]. That is, [itex]\exists M \geq 0[/itex] s.t. [itex]\forall y[/itex], if [itex]|x - y| < \delta[/itex], then [itex]|f(y)| \leq M[/itex]

    2. Relevant equations



    3. The attempt at a solution

    Let [itex]\varepsilon > 0[/itex]. Since [itex]f[/itex] is continuous at [itex]x[/itex], [itex]\exists \delta > 0[/itex] s.t. [itex]\forall y[/itex], if [itex]|x - y| < \delta[/itex], then [itex]|f(x) - f(y)| < \varepsilon[/itex]. Now,

    [itex]|f(x) - f(y)| < \varepsilon \iff[/itex]

    [itex]- \varepsilon < f(x) - f(y) < \varepsilon \iff[/itex]

    [itex]f(x) - \varepsilon < f(y) < f(x) + \varepsilon[/itex].

    Stated differently,

    [itex]f(y) \in (f(x) - \varepsilon, f(x) + \varepsilon)[/itex].

    We are trying to find an [itex]M \geq 0[/itex] s.t. [itex]|f(y)| < M[/itex], or [itex]f(y) \in (-M, M)[/itex]. This is where I get a little stuck. I realize that we may choose any [itex]\mu > 0[/itex] and say that [itex]M = f(x) + \varepsilon + \mu[/itex]. But I run into trouble when doing this. I want something like [itex]M = f(x) \pm (\varepsilon + \mu)[/itex]. This would cover the interval that I am trying to get [itex]M[/itex] into, but I don't exactly know how to say it (except by the way just mentioned of course). Is there a way to write this as a SINGLE value (as opposed to the [itex]\pm[/itex] showing up)?

    EDIT: What about [itex]M = f(x) + |\varepsilon + \mu|[/itex]?

    Thank you for your help!
     
  2. jcsd
  3. Sep 1, 2011 #2

    micromass

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    Can't you just do

    [tex]M=\max\{f(x)+\varepsilon,-f(x)+\varepsilon\}[/tex]

    I see no use for the [itex]\mu[/itex]...
     
  4. Sep 1, 2011 #3
    If you have to prove that a function is continuous, you have to show that certain conditions are satisfied for every value of [itex]\epsilon > 0[/itex]. However, if you're given that a function is continuous and you have to prove something about it, you can select a specific, convenient value for [itex]\epsilon > 0[/itex]. So let [itex]\epsilon = 1[/itex]. Since f is continuous at x, [itex]\exists \delta > 0[/itex] such that if [itex]|x - y| < \delta[/itex], then [itex]|f(x) - f(y)| < 1[/itex]. Can you take it from there?
     
  5. Sep 1, 2011 #4
    Since [itex]M \geq 0[/itex], where is the negative [itex]f(x)[/itex] coming from and how would it fit into choosing an [itex]M[/itex]?

    Right, but I could just plug in [itex]\varepsilon = 1[/itex] into what I had written so far and I would still be in the same boat (unless I am just completely misunderstanding). I am getting stuck on picking an [itex]M[/itex] that will work. I can see how to do it if everything is positive ([itex]M = f(x) + \varepsilon[/itex]), but I don't see how to handle it if it starts to dip into the negatives.
     
  6. Sep 1, 2011 #5
    An additional hint:

    [tex]|f(y)| = |f(y) - f(x) + f(x)| \leq |f(y) - f(x)| + |f(x)|[/tex]
     
  7. Sep 1, 2011 #6

    micromass

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    The negative f(x) is to be able to deal with negative function values.

    If [itex]|f(y)|\leq -f(x)+\varepsilon[/itex], then [itex]f(x)-\varepsilon\leq f(y)[/itex]. So the f(y) is bounded below.
     
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