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Uniform differentiability

  1. Sep 30, 2008 #1
    Let [tex]f:]a,b[\to\mathbb{R}[/tex] be a differentiable function. For each fixed [tex]x\in ]a,b[[/tex], we can define a function

    [tex]
    \epsilon_x: D_x\to\mathbb{R},\quad\quad \epsilon_x(u) = \frac{f(x+u) - f(x)}{u} \;-\; f'(x)
    [/tex]

    where

    [tex]
    D_x = \{u\in\mathbb{R}\backslash\{0\}\;|\; a < x+u < b\}.
    [/tex]

    Now we have [tex]\epsilon_x(u)\to 0[/tex] when [tex]u\to 0[/tex] for all [tex]x[/tex], but let us then define a following collection of functions for all [tex]|u|<b-a[/tex].

    [tex]
    \epsilon_u:E_u\to\mathbb{R},\quad\quad \epsilon_u(x) = \epsilon_x(u)
    [/tex]

    where

    [tex]
    E_u = \{x\in ]a,b[\;|\; a < x + u < b\}.
    [/tex]

    For all [tex]\delta > 0 [/tex] there exists [tex]U>0[/tex] so that [tex]]a+\delta, b-\delta[\subset E_u[/tex] when [tex]|u| < U[/tex]. So now it makes sense to ask, that under which conditions does the collection [tex]\epsilon_u|_{]a+\delta, b-\delta[}[/tex] approach zero uniformly when [tex]u\to 0[/tex], for all [tex]\delta > 0 [/tex]?

    For example, could f being continuously differentiable be enough?
     
  2. jcsd
  3. Dec 28, 2008 #2
    This thread is reasonably recent, so I thought I might reiterate your question. I was looking for conditions for which uniform differentiability holds and I found your thread as a Google search result. Have you found what you're looking for?
     
  4. Dec 28, 2008 #3
    Actually it happened, that one guy did try to explain this to me, and the result seemed to be that uniform differentiability is very common. It could be that continuous differentiability is sufficient (or then I remembered incorrectly. In the next post I needed continuity of the second derivative). Unfortunately I was too tired and not so interested at that moment, so I didn't listen to that guy back then... :biggrin: Perhaps I should to try to return to this now....
     
    Last edited: Dec 28, 2008
  5. Dec 28, 2008 #4
    It could be I proved now that if [itex]f''[/itex] exists and is continuous, then [itex]f[/itex] is uniformly differentiable. From the mean value theorem it follows that we have some mapping

    [tex]
    \{(x,u)\in\mathbb{R}^2\;|\; u\neq 0,\; a<x<b,\; a<x+u<b\},\quad (x,u)\mapsto \xi_{x,u}
    [/tex]

    such that

    [tex]
    |\xi_{x,u} - x| \leq u
    [/tex]

    and

    [tex]
    \frac{f(x+u)-f(x)}{u} = f'(\xi_{x,u}).
    [/tex]

    So

    [tex]
    |\epsilon_u(x)| = |f'(\xi_{x,u}) - f'(x)| \leq \Big|\frac{f'(\xi_{x,u}) - f'(x)}{\xi_{x,u} - x}\Big| \;|u|
    [/tex]

    If [itex]f''[/itex] is continuous, then by using mean value theorem again, we obtain some upper bound [itex]M[/itex] such that

    [tex]
    \Big|\frac{f'(\xi_{x,u}) - f'(x)}{\xi_{x,u} - x}\Big| < M
    [/tex]

    for all [itex]x,\xi_{x,u}\in[a+\frac{\delta}{2},b-\frac{\delta}{2}][/itex]. This condition follows when [itex]x\in [a+\delta, b-\delta][/itex] and [itex]u\in ]-\frac{\delta}{2},0[\;\cup\; ]0,\frac{\delta}{2}[[/itex]. Then

    [tex]
    |\epsilon_u(x)| < M\;|u|
    [/tex]

    for all [itex]x\in [a+\delta, b-\delta][/itex] and relevant [itex]u[/itex].

    JinM, if you were interested in this, perhaps you can check the proof for mistakes? :wink:
     
    Last edited: Dec 29, 2008
  6. Dec 28, 2008 #5
    Nice. I like your result -- your notation in the original post is foreign to me though, so I have to dig into that to check your proof -- although I'm sure its fine. :)

    I also made a little bit of research, and another interesting result dictates loosely that f is uniformly differentiable iff f' is uniformly continuous; f is a differentiable function by assumption, of course. The proof follows easily -- thought I'd share this.
     
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