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Product of two continous functions

  1. Feb 10, 2008 #1
    There is a proof in my book that asks us to prove that the product of two continuous functions is continuous. If anyone could help please reply back, thanks!
     
  2. jcsd
  3. Feb 10, 2008 #2
    If h(x) is undefined at some point c (and thus not continuous), will it ever have a product that is defined at that point?
     
  4. Feb 11, 2008 #3

    HallsofIvy

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    It is not clear to me what SticksandStones response has to do with the question!

    im2fastfouru, you know, of course, that the product, fg, will be continuous at a as long as the lim, as x goes to a, of f(x)g(x) is equal to f(a)g(a). Therefore, you must prove that, given any [itex]\epsilon> 0[/itex], there exist [itex]\delta> 0[/itex] such that if |x-a|< [itex]\delta[/itex], then |f(x)g(x)- f(a)g(a)|< [itex]\epsilon[/itex]. Using, of course, the fact that f and g are both continuous at a, that is, that given any [itex]\epsilon> 0[/itex], there exist [itex]\delta> 0[/itex] such that if |x-a|< [itex]\delta[/itex] such that |f(x)- f(a)|< [itex]\epsilon[/itex] and similarly for g.

    I would recommend you look at |f(x)g(x)- f(x)g(a)+ f(x)g(a)- f(a)g(a)|.

    After all, your book asked you to prove this, not us!
     
  5. Feb 11, 2008 #4
    I thought he was asking something entirely different; I apologize.:redface:
     
  6. Feb 11, 2008 #5

    lurflurf

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    hint
    write
    f(x+h)=f(x)+[f(x+h)-f(x)]
    g(x+h)=g(x)+[g(x+h)-g(x)]
    note
    |f(x+h)-f(x)|<eps1
    |g(x+h)-g(x)|<eps2
    |f(x+h)-f(x)|,|g(x+h)-g(x)|<eps=max(eps1,eps2)
    also recall
    |a+b+c|<|a|+|b|+|c|
     
  7. Feb 12, 2008 #6
    Why on earth would he do so? Halls hints are quite straightforward, he actually almost did all of what was needed to do.>< All he needs to do is show that If(x)I does not get too large as x-->a(or sth), in other words the op just needs to show that f(x) is bounded, which is easy to show since lim(x-->a)f(x)=f(a), so any function who has a limit is bounded at that limit point. all he needs to do is let e-eplylon= 1, and it is easy to show that f(x) is bounded. If(x)I=If(x)-f(a)+f(a)I<=If(x)-f(a)I+If(a)I< 1 +If(a)I

    For the OP: do you see how to go now? Like Halls said, you should give some effort on your own next time, if you want to recieve more help. People here won't just give us answers(including me since i often ask for help).
     
    Last edited: Feb 12, 2008
  8. Feb 12, 2008 #7

    HallsofIvy

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    lurflurf may has misread "continuous" as "differentiable".
     
  9. Feb 12, 2008 #8

    mathwonk

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    the point (of continuity) is simply that if two numbers are respectively near two other numbers, then the products are also near each other.

    to see this, let the numbers be a+h and b+k and compare the product of ab to that of (a+h)(b+k), when h and k are small.
     
  10. Feb 18, 2008 #9

    lurflurf

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    To solve the problem and because it is fun. I agree, I provided a different (though very slightly) view.
    theorem f is continuous if and only if
    lim_h->0 (f(x+h)-f(x))=0
    where (for picking nits) we accept as implicit f(x) exist and limit_h->0 f(x+h) exist
    A useful framework

    we desire to show that

    if
    |f(x+h)-f(x)|<eps1
    |g(x+h)-g(x)|<eps2
    then
    |f(x+h)g(x+h)-f(x)g(x)|<eps
    write
    f(x+h)=f(x)+[f(x+h)-f(x)]
    g(x+h)=g(x)+[g(x+h)-g(x)]
    thus
    |f(x+h)g(x+h)-f(x)g(x)|=|[f(x+h)-f(x)]g(x)+f(x)[g(x+h)-g(x)]+[f(x+h)-f(x)][g(x+h)-g(x)]|
    <|[f(x+h)-f(x)||g(x)|+|f(x)||g(x+h)-g(x)|+|f(x+h)-f(x)||g(x+h)-g(x)]|
    <eps1|g(x)|+|f(x)|eps2+eps1*eps2
    letting
    eps1=(eps/3)/|g(x)| if |g(x)|>0
    eps2=(eps/3)/max(eps1,|f(x)|) if |f(x)|>0
    we easily see the result is true
     
    Last edited: Feb 19, 2008
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