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Analysis, silly question

  1. Mar 31, 2009 #1
    1. The problem statement, all variables and given/known data
    Given a convergent sequence:
    [tex]a_n \rightarrow a[/tex]
    and a continuous function:
    [tex]f:\mathbb{R}\rightarrow\mathbb{R}[/tex]
    show that there exists an [tex]N\in\mathbb{N}[/tex] such that [tex]\forall n>N[/tex]:
    [tex]f(a_n)\geq\frac{f(a)}{2}[/tex]



    2. Relevant equations
    Usual definitions for limit of a sequence and continuous function.


    3. The attempt at a solution
    I've tried playing around with it but I really don't understand what to do, I know its an easy question and I feel pretty stupid but I'm stuck. Can anyone just get me started on this one?
    So far I've rearranged it so it looks like:
    [tex]f(a)-f(a_n)\leq f(a_n)[/tex]
    Which sort of looks like a limit without the abs values...
    I figure that the function of a sequence is a sequence again, btu what am I trying to say here? That it is a divergent sequence?
     
  2. jcsd
  3. Mar 31, 2009 #2
    False. Let f(x) = -1 (constant function). Let a_n =0 for all n. Here, f(a_n)=f(0)=-1 <-1/2 for all n, yet a_n:n=1,2,.... converges in the domain if f.
    [Is there another condition?]
     
  4. Mar 31, 2009 #3
    Sorry, sorry, yes there is another condition:
    [tex]f(a)>0[/tex]
     
  5. Mar 31, 2009 #4
    To not give the answer away, I would just suggest that the problem is now a matter of
    fitting into the \epsilon - \delta definition of continuous functions. Note that f(a)/2 < f(a).
     
  6. Apr 1, 2009 #5

    HallsofIvy

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    Take [itex]\epsilon> a/2[/itex] in the definition of "continuous function".
     
  7. Apr 2, 2009 #6
    whaaaa.? So much for the 'for all epsilon >0' part. Whaaa? Well, not my problem anyway....
     
  8. Apr 2, 2009 #7

    HallsofIvy

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    Yes, "For all epsilon> 0", something is true so it is true for any specific value of epsilon. Here we are given that f is continuous so we are free to choose any value of epsilon we like.
     
  9. Apr 2, 2009 #8
    I have an answer now, basically because [tex]a_n\rightarrow a[/tex] and f is continuous then [tex]f(a_n)\rightarrow f(a)[/tex], so there exists n bigger than N such that [tex]|f(a_n)-f(a)|<\epsilon[/tex] so then just choose [tex]\epsilon = f(a)/2[/tex] do some rearranging and the inequality pops out.

    In other words, [tex]f(a_n)[/tex] is a convergent sequence.
     
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