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Uniform continuity and derivatives

  1. Jul 25, 2007 #1

    daniel_i_l

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    Gold Member

    1. The problem statement, all variables and given/known data
    1) f is some function who has a bounded derivative in (a,b). In other words, there's some M>0 so that |f'(x)|<M for all x in (a,b).
    Prove that f is bounded in (a,b).
    2) f has a bounded second derivative in (a,b), prove that f in uniformly continues in (a,b).

    2. Relevant equations



    3. The attempt at a solution

    1) For all x in (a,b), (f(x)-f(a))/(x-a) = f'(c) =>
    |f(x)-f(a)| = |f'(c)||x-a| =< M|b-a| (c is in (a,x))
    and so
    f(a) - M|b-a| =< f(x) >= f(a) + M|b-a| and so f(x) is bounded in (a,b).

    2)Using (1) we see that f'(x) is bounded in (a,b). So there's some M>0 so that |f'(x)|<=M. And so for all x,y in (a,b)
    (f(x)-f(y))/(x-y) = f'(c) =< |M| and so
    |f(x)-f(y)| =< |x-y|M and it's easy to see that f is UC in (a,b).

    Did I do that right? It some what bothers me that in (1) I also could have proved that f is UC which is a stronger result than the fact that it's bounded the same way that I proved it in (2). Is that right? If so then why did they only ask to prove that it was bounded?

    Thanks.
     
  2. jcsd
  3. Jul 25, 2007 #2
    Theorem: Let [tex]f[/tex] be continous on a (possibly infinite) interval [tex]I[/tex]. Let [tex]J[/tex] be the open interval without the endpoints of [tex]I[/tex]. If [tex]f[/tex] is differentiable on [tex]J[/tex] and [tex]|f'|\leq M[/tex] then [tex]f[/tex] is uniformly continous on [tex]I[/tex].

    Proof:Chose [tex]M>0[/tex] so that [tex]|f'(x)|\leq M[/tex] for all [tex]x\in J[/tex]. For any [tex]\epsilon > 0 [/tex] choose [tex]\delta = \frac{\epsilon }{M}[/tex].
    Choose [tex]a<b[/tex] with [tex]b-a<\delta[/tex]. By Mean Value Theorem: [tex]\exists x \in (a,b)[/tex] so that [tex]f'(x) = \frac{f(b)-f(a)}{b-a}[/tex].
    Thus,
    [tex]|f(b)-f(a)| = |f'(x)||b-a|\leq M|b-a| < M\delta = M\cdot \frac{\epsilon}{M} = \epsilon[/tex].

    -Wolfgang.
     
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