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Hmmm Arzela-Ascoli, maybe?

  1. Oct 10, 2011 #1
    Suppose you know the following about [itex]g: \mathbb R \to \mathbb R[/itex]:
    1. [itex]g\in C^1[/itex];
    2. [itex]g\in C^2[/itex], expect at finitely many points [itex]\{x_1,\ldots,x_n\}[/itex], and [itex]|g''(x)| \leq M[/itex] (except at those points).

    How can you show that there exists a sequence [itex]f_k[/itex] with the following properties?
    1. [itex]f_k \to g[/itex] uniformly;
    2. [itex]f'_k \to g'[/itex] uniformly;
    3. [itex]f_k \in C^2[/itex], [itex]|f''_k(x)| \leq M[/itex], and [itex]f''_k \to g''[/itex] outside [itex]\{x_1,\ldots,x_n\}[/itex].

    Arzela-Ascoli is pretty much the only theorem I know of that talks about sequences of functions with uniformly bounded derivatives, but I can't for the life of me see how you could use it to construct the sequence [itex]f_k[/itex].
     
  2. jcsd
  3. Oct 10, 2011 #2
    Hmmm...well, what if instead of looking at [itex]\mathbb R[/itex], we restrict our attention to some compact subset of [itex]\mathbb R[/itex]? Does that help at all?
     
  4. Oct 10, 2011 #3

    micromass

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    Take [itex]f_k=g[/itex] for each k. Or do you want [itex]f_k[/itex] to be [itex]C^2[/itex] at each point?? In that case, first approximate [itex]g^{\prime\prime}[/itex] by continuous functions. Then take integrals.
     
  5. Oct 10, 2011 #4
    micromass, thanks for your response.

    Since [itex]g''[/itex] is continuous (except on a set of measure zero) on the compact interval [itex][0,t][/itex], it is integrable there and can therefore be written as the pointwise limit of a sequence of continuous functions (with compact support) - call these functions [itex]f''_k[/itex]. We know these functions are integrable, too, and that [itex]f'_k(s) = \int_0^s f''_k(q)dq[/itex] for [itex]s\in [0,t][/itex]. We also have
    [tex]
    |f'_k(s)| = \left| \int_0^s f''_k(q)dq \right| \leq \int_0^s |f''_k(q)|dq \leq Ms \leq Mt,
    [/tex]
    such that the sequence [itex]\{f'_k\}[/itex] is uniformly bounded on [itex][0,t][/itex]. But how are we supposed to show that [itex]f'_k \to g'[/itex] uniformly? I mean, do we even know [itex]\int_0^s g''(q)dq = g'(s)[/itex]?
     
  6. Oct 10, 2011 #5

    micromass

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    So you must prove that [itex]\int_0^s f^{\prime\prime}_k\rightarrow \int_0^s g^{\prime\prime}[/itex] uniformly. This is easily proved from the definition of uniform continuity. Use that

    [tex]|\int_0^s{f^{\prime\prime}_k} - \int_0^{s-h}{g^{\prime\prime}}| \leq \int_{s-h}^s{|f^{\prime\prime}_k - g^{\prime\prime}|} \leq h \sup_{x\in [s-h,h]}{ |f^{\prime\prime}_k(x)-g^{\prime\prime}(x)|}[/tex]



    Fundamental theorem of calculus??? (if you add a -g'(0) there)
     
  7. Oct 10, 2011 #6
    But [itex]g''[/itex] is not continuous at finitely many points in the interval, so we can't apply the (traditional) FTC. Maybe we can apply Lebesgue's FTC, but how do we know [itex]g''[/itex] is absolutely continuous?
     
  8. Oct 10, 2011 #7

    micromass

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    Continuity if [itex]g^{\prime\prime}[/itex] isn't necessary.

    It can be proven that if [itex]f:[a,b]\rightarrow \mathbb{R}[/itex] has a primitive F on [a,b], then

    [itex]\int_a^b{f}=F(a)-F(b)[/itex]

    Continuousness of f isn't important. This is a slight generalization of the usual fundamental theorem of calculus. A proof can be found in Bartle "a modern theory of integration" Theorem 4.5.
     
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