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Continuous and smooth on a compact set implies differentiability at a point

  1. Dec 30, 2011 #1
    I'm trying to prove that if a function is continuous on [a,b] and smooth on (a,b) then there's a point x in (a,b) where f'(x) exists. The definition of smoothness is: f is smooth at x iff [itex] \lim_{h \rightarrow 0} \frac{ f(x+h) + f(x-h) - 2f(x) }{ h } = 0 [/itex].

    I'm starting with the simpler case where f(a) = f(b). I know that f has a max x and a min y on [a,b], and that if both occur at a or b, then f is a constant function. So I'm assuming wlog that the max x occurs somewhere in (a,b).

    So at x,
    [itex] - \frac{ (f(x) - f(x+h)) } {h} \leq 0 [/itex] and
    [itex] \frac{f(x-h) - f(x)}{h} \leq 0 [/itex].

    This is where I'm getting stuck. I'd like to show that the limits of the one sided difference quotients as [itex] h \rightarrow 0- [/itex] exist and then apply smoothness to show that these must be equal, hence f is differentiable at x. However, I haven't been able to show that they exist - taking the infima of both sides doesn't seem to yield anything, so I'm guessing I might be missign something else. Thanks in advance.
  2. jcsd
  3. Dec 31, 2011 #2


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    What, exactly, is your definition of "smooth"?

  4. Dec 31, 2011 #3
    The definition I'm using for smooth is:

    A function [itex]f : \mathbb{R} \rightarrow \mathbb{R} [/itex] is said to be smooth at a point [itex]x \in \mathbb{R} [/itex] iff it is defined in a neighborhood of x and

    [itex] \lim_{h \rightarrow 0} \frac{ f(x+h) + f(x-h) - 2f(x) }{ h } = 0 [/itex].
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