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!

Epsilon-Delta proof. Have I done it right?

  1. Jan 23, 2008 #1
    I'm going to prove that if f and g are continuous functions, then so is fg.

    We define [itex]fg \; :\; E \rightarrow \bb{F}[/itex] by
    [itex](fg)(t) \;=\; f(t)g(t)[/itex] for all [itex]t \in E[/itex]
    (F is an ordered field and E is a subset of F).

    Proof
    [tex]|(fg)(x) - (fg)(a)| = |f(x)g(x) - f(a)g(a)| =[/tex]

    Add and subtract f(x)g(a)
    [tex]|f(x)g(x) - f(x)g(a) \;+\; f(x)g(a) - f(a)g(a)| \leq [/tex]

    We use the triangle inequality
    [tex]\leq |f(x)g(x) - f(x)g(a)| \;+\; |f(x)g(a) - f(a)g(a)| = [/tex]

    We factor out f(x) and g(a)
    [tex]|f(x)||g(x) - g(a)| \;+\; |g(a)||f(x) - f(a)|[/tex]

    From the definition pf continuity we have
    [itex]\delta_1[/itex] such that [itex]|f(x) - f(a)| < \frac{\epsilon}{2}[/itex] for [itex]|x-a|[/itex]

    [itex]\delta_2[/itex] such that [itex]|g(x) - g(a)| < \frac{\epsilon}{2}[/itex] for [itex]|x-a|[/itex]

    But we still have |f(x)| and |g(a)|.

    From the definition, we can deduce [*] - This part I'm not really sure about.
    [tex]|f(x) - f(a)| < \frac{\epsilon}{2} \;\Rightarrow\; |f(x)| < \frac{\epsilon}{2} + |f(a)| [/tex]

    [tex]|g(x) - g(a)| < \frac{\epsilon}{2} \;\Rightarrow\; |g(a)| < \frac{\epsilon}{2} + |g(x)| [/tex]

    And we can choose new delta-values:
    [tex]\delta_3[/tex] such that [tex]|f(x) - f(a)| \;<\; \frac{\epsilon}{2(\frac{\epsilon}{2}+ |f(a)|)}[/tex] for [tex]|x-a|[/tex]

    [tex]\delta_4[/tex] such that [tex]|g(x) - g(a)| \;<\; \frac{\epsilon}{2(\frac{\epsilon}{2}+ |g(x)|)}[/tex] for [tex]|x-a|[/tex]

    In conclusion, we take [itex]\delta_5 = \min(\delta_3, \delta_4)[/itex]
    And get:
    [tex]|f(x)||g(x) - g(a)| \;+\; |g(a)||f(x) - f(a)| <[/tex]

    [tex](\frac{\epsilon}{2} + |f(a)|\cdot \frac{\epsilon}{2(\frac{\epsilon}{2}\cdot |f(a)|)}) + (\frac{\epsilon}{2} + |g(x)|\cdot \frac{\epsilon}{2(\frac{\epsilon}{2}\cdot |g(x)|)}) = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon[/tex]

    And the proof is concluded. Quad Erad Demonstrandum... if it is right, of course.
    Hope it wasn't too long!
     
    Last edited: Jan 23, 2008
  2. jcsd
  3. Jan 23, 2008 #2
    You factorized properly. Take a delta such that |f(x)-f(a)| < 1 as x -> a. Then you can get a bound on |f(x)|, say M. Then you choose another delta so that |f(x)-f(a)| < epsilon/(2|g(a)|+1). Then choose another delta such that |g(x)-g(a)| < epsilon/(2M). Then take the smaller of all three deltas, and you get your result.
     
  4. Jan 23, 2008 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You do need the fact that, since f is continuous, |f(x)| has an upper bound in order to handle the fact that |f(x)| is variable, not a constant like |g(a)|.
     
  5. Jan 23, 2008 #4
    We now choose three [itex]\delta[/itex]'s, so
    [tex]\delta_1[/tex] such that [tex]|f(x) - f(a)| < 1 \;\Rightarrow\; |f(x)| < M[/tex] (since f is continous, |f(x)| has an upper bound).

    [tex]\delta_2[/tex] such that [tex]|g(x) - g(a)| < \frac{\epsilon}{2M}[/tex]

    Obviously, [itex]|g(a)| < |g(a)| + 1[/itex]
    [tex]\delta_3[/tex] such that [tex]|f(x) - f(a)| < \frac{\epsilon}{2(|g(a)| + 1)}[/tex]

    In conclusion, we take [itex]\delta = \min(\delta_1, \delta_2, \delta_3)[/itex]
    And get:
    [tex]|f(x)||g(x) - g(a)| \;+\; |g(a)||f(x) - f(a)| <[/tex]

    [tex](M\cdot \frac{\epsilon}{2M}) + (|g(a)| + 1\cdot \frac{\epsilon}{2(|g(a)| + 1)}) = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon[/tex]

    Quad Erad Demonstrandum.
    Thanks for the help!
     
  6. Jan 23, 2008 #5
    To make the proof complete, you should explain that the +1 in epsilon/(2|g(a)|+1) is used because |g(a)| might be zero.

    Also, you said "since f is continous, |f(x)| has an upper bound". Continuity alone is not enough, |f(x)| has an upper bound on what set? It certainly does not necessarily have an upper bound on R.
    You can remove any doubt by getting an explicit expression for M in terms of f(a) by using the fact that |f(x)-f(a)| < 1 (which is easy). You see, |f(x)| is bounded not only because f is continuous but also because (a - delta1, a + delta1) is a subset of [a - delta1, a + delta1] and [a - delta1, a + delta1] is compact, but I don't think you should talk about this for this proof. Getting an explicit expression for M will remove any questions.
     
    Last edited: Jan 23, 2008
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Epsilon-Delta proof. Have I done it right?
  1. Delta epsilon proof (Replies: 5)

Loading...