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Homework Help: 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).

    [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


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    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
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