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Uniform continuity and Bounded Derivative

  1. Jul 16, 2011 #1
    Hi, All:

    Let f R-->R be differentiable. If |f'(x)|<M< oo, then f is uniformly continuous, e.g.,

    by the MVTheorem. Is this conditions necessary too, i.e., if f:R-->R is differentiable

    and uniformly continuous, does it follow that |f'(x)|<M<oo ?

    Thanks.
     
  2. jcsd
  3. Jul 16, 2011 #2

    micromass

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    Hi Bacle! :smile:

    This is not true. Consider the function

    [tex]f:]0,1[\rightarrow\mathbb{R}: x\rightarrow \sqrt{x}[/tex]

    then this is uniform continuous (continuous on compact interval). But the derivative [itex]f^\prime(x)[/itex] grows very large if x gets closer to 0.

    In fact, the condition you mention is equivalent to the Lipschitz-condition (for differentiable functions). That is, if [itex]f:[a,b]\rightarrow \mathbb{R}[/itex] is continuous and differentiable on ]a,b[, then the following are equivalent
    • [itex]|f~{}^\prime(x)|\leq k~[/itex] for all x in ]a,b[
    • f is k-Lipschitz, that is [itex]|f(x)-f(y)|\leq k|x-y|~~[/itex] for all x,y in [a,b]

    The proof uses the mid-value theorem.

    It is easy to see that being Lipschitz is stronger than uniform continuity.
     
  4. Jul 16, 2011 #3

    gb7nash

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    Unless I'm reading something wrong, the statement in the original post requires differentiability on all R. This example isn't applicable.
     
  5. Jul 16, 2011 #4

    micromass

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    Fine, consider

    [tex]f:[0,1]\rightarrow \mathbb{R}:x\rightarrow x^{3/2}\sin(1/x)[/tex]

    this extends to an uniform continuous, differentiable function on [itex]\mathbb{R}[/itex], but it's derivative is not bounded.
     
  6. Jul 16, 2011 #5
    Thanks, micromass; I agree, because continuous+compact implies uniformly-continuous;

    I was thinking more of what if the extension outside of

    [a,b] is non-trivial (e.g., a constant function, or without exhausting , by covering

    the remainder by compact sets and pasting/smoothing at the endpoints), if the condition on

    the derivative must be satisfied in (-00,a)\/(b,oo), i.e., if the function is not defined

    piecewise in countably-many compact intervals.
     
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