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Continuity proof check please

  1. Sep 5, 2011 #1
    1. The problem statement, all variables and given/known data

    Given two functions [itex] f [/itex] and [itex] g [/itex], if [itex] f [/itex] and [itex] g [/itex] are continuous at a point [itex] x [/itex], then the function [itex] h = fg [/itex] is continuous at [itex] x [/itex].

    2. Relevant equations

    Lemma 1
    If a function [itex] f [/itex] is continuous at a point [itex] x [/itex], then f is bounded on some interval centered at [itex] x [/itex]. That is, there exists an [itex] M \geq 0 [/itex] and a [itex] \delta > 0 [/itex] such that for all [itex] y [/itex],
    [itex] |x - y| < \delta \implies |f(y)| \leq M. [/itex]

    3. The attempt at a solution


    Let [itex] f [/itex] and [itex] g [/itex] be functions that are continuous at a point [itex] x [/itex].
    Define a new function [itex] h [/itex] as [itex] h = fg [/itex].
    Let [itex] \varepsilon > 0 [/itex] and [itex] \delta_1 > 0 [/itex].
    If [itex] f(x) = 0 [/itex], then it is trivially true that for all [itex] y [/itex],
    [itex] \displaystyle |x - y| < \delta_1 \implies |f(x)||g(x) - g(y)| < \varepsilon, [/itex]
    so assume [itex] f(x) \neq 0 [/itex].
    Now, suppose [itex] \displaystyle \mu = \frac{\varepsilon}{|f(x)|} [/itex].
    Clearly [itex] \mu [/itex] is a positive real number.
    Since [itex] g [/itex] is continuous at [itex] x [/itex], for all [itex] \varepsilon [/itex], there exists a [itex] \delta_1 [/itex] such that
    [itex] |x - y| < \delta_1 \implies |g(x) - g(y)| < \mu. \quad \quad (1) [/itex]
    Let [itex] M = 0 [/itex]. Lemma 1 states that, since [itex] g [/itex] is continuous, there exists a [itex] \delta_2 > 0 [/itex] such that [itex] |x - y| < \delta_2 \implies |g(y)| \leq M = 0. [/itex] So,
    [itex] |x - y| < \delta_2 \implies |g(y)| = 0. \quad \quad (2) [/itex]
    Let [itex] \delta = \text{min}( \delta_1, \delta_2 ) [/itex]. Then [itex] |x - y| < \delta [/itex] implies
    [itex] |h(x) - h(y)| = |f(x)g(x) - f(y)g(y)| [/itex]
    [itex] = |f(x)g(x) - f(x)g(y) + f(x)g(y) - f(y)g(y)| [/itex]
    [itex] \leq |f(x)g(x) - f(x)g(y)| + |f(x)g(y) - f(y)g(y)| [/itex]
    [itex] = |f(x)||g(x) - g(y)| + |g(y)||f(x) - f(y)| [/itex]
    [itex] < |f(x)| \mu + 0 = \varepsilon. [/itex] QED

    Pointing out any mistakes / giving constructive criticism is welcome. Thank you very much!
     
  2. jcsd
  3. Sep 5, 2011 #2

    micromass

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    You can't just set M=0. All lemma 1 says is that there exists an M, but that M is potentially very large. You have no control over what value M has. So saying that M=0 is not ok.
     
  4. Sep 5, 2011 #3
    Could I say that [itex]M = \frac{\varepsilon}{2|f(x) - f(y)|}[/itex]?
    Then [itex]M \geq 0[/itex]

    Then I would have to reset [itex]\mu = \frac{\varepsilon}{2|f(x)|}[/itex] in order for everything to cancel and add up to [itex]\varepsilon[/itex]
     
  5. Sep 5, 2011 #4

    micromass

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    No, you cannot choose M. You can choose epsilon though.
     
  6. Sep 5, 2011 #5
    So the lemma is pretty pointless in this case? Can I try something like:

    I know that since [itex]f[/itex] is continuous, I can choose a [itex]\delta > 0[/itex] such that

    [itex]|x - y| < \delta \implies |f(x) - f(y)| < \frac{\varepsilon}{2|g(y)|}[/itex]?
     
  7. Sep 5, 2011 #6

    micromass

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    No since your "epsilon" is [itex]\frac{\varepsilon}{2|g(y)|}[/itex] and is dependent on y. It shouldn't be dependent on y. It can depend on x, though.
     
  8. Sep 5, 2011 #7
    OK, since [itex]g[/itex] is continuous, we may say that there exists a [itex]\delta_1[/itex]

    [itex]|x - y| < \delta_1 \implies |g(y)| < |g(x)| + 1[/itex].

    Also, since [itex]f[/itex] is continuous, there exists a [itex]\delta_2[/itex] such that

    [itex]|x - y| < \delta_2 \implies |f(x) - f(y)| < \frac{\varepsilon}{2(|g(x)| + 1)}[/itex].

    This would cancel fine and would add to [itex]\varepsilon[/itex] in the end, but this way seems so roundabout. Is there no nicer way?
     
  9. Sep 5, 2011 #8

    micromass

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    Thay seems ok! :smile:
     
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