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Are these proofs correct(bounded and finite variation).

  1. Jun 19, 2006 #1
    First of all if you read this and the latex is all messed upp I am probably working on getting it right so plz be patient till I get it right. No need to post a comment that it doesnt work. Thanks :wink:

    I havent taken a pure maths class in over 2,5 years so I can hardly remember how to write proofs:yuck:

    Problem 1.

    Let [tex] C \in \mathbb{R} [/tex] be a arbitrary number. Show that the function
    [tex] f:[a,b]\rightarrow \mathbb{R} [/tex]

    given by [tex] f(t)=cos(ct) [/tex]

    is of bounded variation. i.e it satisifies the condition

    [tex]Sup V_f (t) < \infty [/tex]


    [tex] V_f (t) = \lim_{n\rightarrow \infty} Sup_{ t_k^n,t_{k-1}^n \in \pi_n }
    \sum_{k=1}^n |{(f(t_k^n)-f(t_{k-1}^n)}| [/tex]

    with [tex] \pi_n : t_0^n < ...... < t_n^n [/tex]

    Since [tex] Cos(ct) [/tex]

    is differentiable we can rewrite [tex]V_f (t)[/tex] with the mean value theorem

    [tex] V_f (t) = \lim_{n\rightarrow \infty} Sup_{ t_k^n,t_{k-1}^n \in \pi_n }
    \sum_{k=1}^n \|{(f(t_k^n)-f(t_{k-1}^n)}| = \lim_{n\rightarrow \infty} Sup_{ t_k^n,t_{k-1}^n \in \pi_n } \sum_{k=1}^n |f^{'} (G)| (t_k^n - t_{k-1}^n) [/tex]

    wich is equal to(according to the definition of the riemann integral)

    [tex] \int_{a}^{b} |f^{'} (x)| dx [/tex]

    with [tex]f(x)=f(t)=cos(ct) [/tex] we get

    [tex] \int_{a}^{b} |-csin(ct)| dx \leq \int_{a}^{b} |c| dx = |c|(b-a)[/tex]


    [tex] Sup V_f (t) = Sup |c|(b-a) <\infty [/tex]
    Last edited: Jun 20, 2006
  2. jcsd
  3. Jun 20, 2006 #2
    problem 2
    show that the function [tex]g:[0,\infty) \rightarrow \mathbb{R}[/tex]
    given by [tex]g(t)=sin(ct)[/tex] is of finite variation.

    In the same manner as in problem one we get

    [tex]V_f ((0,t)) = \int_{0}^{t} |f'(t)|dt = \int_{0}^{t} |ccos(ct)|dt \leq \int_{0}^{t} |c|dt=|c|t<\infty [/tex] for all t equal to or larger than zero. (how do I get the equal to or larger than symbol in latex? and the "for all" symbol?)

    wich shows that the function is of finite variation
    Last edited: Jun 20, 2006
  4. Jun 20, 2006 #3
    problem 3
    If the functions [tex]f,g:[0,\infty) \rightarrow \mathbb{R}[/tex] are of finite variation, show that any linear combination of them,
    [tex]\alpha f + \beta g:[0,\infty) \rightarrow \mathbb{R}[/tex] [tex] \alpha,\beta \in \mathbb{R}[/tex] are of finite variation.

    [tex]V_{f,g} ((0,t))=\lim_{n\rightarrow \infty} \Sup_{ t_k^n,t_{k-1}^n \in \pi_n }\sum_{k=1}^n |{\alpha f(t_k^n)+\beta g(t_k^n)-\alpha f(t_{k-1}^n)-\beta g(t_{k-1}^n)|=[/tex]
    [tex]\lim_{n\rightarrow \infty}Sup_{ t_k^n,t_{k-1}^n \in \pi_n }\sum_{k=1}^n| \alpha (f(t_k^n)-f(t_{k-1})+\beta (g(t_k^n)-g(t_{k-1})| \leq[/tex]

    (using triangle inequality)

    [tex] \lim_{n\rightarrow \infty}Sup_{ t_k^n,t_{k-1}^n \in \pi_n }\sum_{k=1}^n |\alpha (f(t_k^n)-f(t_{k-1})|+|\beta (g(t_k^n)-g(t_{k-1})| = [/tex]
    (can I just move the alpha and beta outside of the lim,sup and sum like this?)
    [tex]|\alpha| \lim_{n\rightarrow \infty}Sup_{ t_k^n,t_{k-1}^n \in \pi_n }\sum_{k=1}^n |(f(t_k^n)-f(t_{k-1})|+ |\beta|\lim_{n\rightarrow \infty}Sup_{ t_k^n,t_{k-1}^n \in \pi_n }\sum_{k=1}^n |(g(t_k^n)-g(t_{k-1})| =[/tex]

    [tex]|\alpha| V_f((0,t))+|\beta| V_f((0,t)) <\infty[/tex] for all t equal to or larger than zero since [tex]V_f((0,t))<\infty[/tex] and [tex]V_g((0,t)) <\infty[/tex] for all t equal to or larger than zero.
    Last edited: Jun 21, 2006
  5. Jun 21, 2006 #4
    I hope I havent made any error that makes the whole thing incomprehensible :(
  6. Jun 21, 2006 #5


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    Homework Helper

    For the first question, the basic idea of the proof seems correct, although I'm not sure how Vf is a function of t, it seems more a function from intervals into the real numbers. And I'm not sure what you're taking the sup of when you write sup Vf(t) or sup |c|(b-a). Besides these details it looks correct. For the second, do they mean "of bounded variation on the set [0,t], for all t in R"? If so, this is just a special case of the last problem. The third proof seems correct as well.
  7. Jun 21, 2006 #6
    yes you are quite correct I just made some errors :)

    Thanks alot for checking, appriciate it.
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