1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: 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
    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
    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
    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
    Thay seems ok! :smile:
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook