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Math Help - Uniform Continuity

  1. #1
    Super Member redsoxfan325's Avatar
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    Uniform Continuity

    A bounded derivative implies uniform continuity. However, there are certain instances of functions (most notably \sqrt{x}) that are uniformly continuous and have unbounded derivatives. I attempt to strengthen the theorem below:

    Theorem: An everywhere continuous function f is uniformly continuous on \mathbb{R} if its derivative is bounded on (-\infty,-m] and [n,\infty) ( m,n\in\mathbb{R}).

    Proof: On [-m,n] we have a continuous function defined on a compact set, so it is uniformly continuous and \exists~\delta_1... On (-\infty,-m] and [n,\infty), f has a bounded derivative so there exist \delta_2 and \delta_3 such that... Therefore taking \delta=\min\{\delta_1,\delta_2,\delta_3\} proves that the entire function is uniformly continuous on \mathbb{R}.

    So what's my question? My question is whether the above theorem is an if-and-only-if statement or whether the implication only goes in one direction. If not, what hypotheses can we add so that it does go both ways?

    Thoughts? Counterexamples?
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  2. #2
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    Quote Originally Posted by redsoxfan325 View Post
    A bounded derivative implies uniform continuity. However, there are certain instances of functions (most notably \sqrt{x}) that are uniformly continuous and have unbounded derivatives. I attempt to strengthen the theorem below:

    Theorem: An everywhere continuous function f is uniformly continuous on \mathbb{R} if its derivative is bounded on (-\infty,-m] and [n,\infty) ( m,n\in\mathbb{R}).

    Proof: On [-m,n] we have a continuous function defined on a compact set, so it is uniformly continuous and \exists~\delta_1... On (-\infty,-m] and [n,\infty), f has a bounded derivative so there exist \delta_2 and \delta_3 such that... Therefore taking \delta=\min\{\delta_1,\delta_2,\delta_3\} proves that the entire function is uniformly continuous on \mathbb{R}.
    About this proof: perhaps you know that, but for a "full" proof you should take care of the case when the two points x,y such that |x-y|<\delta lie in different intervals (like x<n<y). Either you say |f(x)-f(y)|\leq |f(x)-f(n)|+|f(n)-f(y)|\leq 2\varepsilon. Or you use the following trick: you define \delta_1 with uniform continuity on [-m-1,n+1] and choose \delta=\min(\delta_1,\delta_2,\delta_3,1) so that |x-y|<\delta implies that x,y are in the same interval (either (-\infty,-m],, [-m-1,n+1] or [n,+\infty)).


    So what's my question? My question is whether the above theorem is an if-and-only-if statement or whether the implication only goes in one direction. If not, what hypotheses can we add so that it does go both ways?

    Thoughts? Counterexamples?
    Counterexample: you will be delighted to hear that the function f:\mathbb{R}\to\mathbb{R} defined by f(x)=x^2\sin\frac{1}{|x|^{4/3}} ( x\neq 0), f(0)=0, is derivable on \mathbb{R}, yet its derivative is unbounded on any neighbourhood of 0.

    In particular, f is uniformly continuous and derivable on [-1,1] while its derivative is unbounded (but finite, that's the difference with \sqrt{x}).

    You can then define a derivable periodic function g that looks like f repeatedly. So that the derivative of g is unbounded on any [m,+\infty), while g is uniformly continuous (it is continuous and periodic, so...).
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  3. #3
    Super Member redsoxfan325's Avatar
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    So what can we add to the theorem I proposed so that it is an iff statement?
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