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Math Help - bounded variation

  1. #1
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    bounded variation

    Let f(x) be defined by f(0)=0 and f(x)=x^2 sin(1/x) for x \neq 0. Prove that f is of bounded variation on [-1,1].
    please help me on this one.
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  2. #2
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    Quote Originally Posted by Kat-M View Post
    Let f(x) be defined by f(0)=0 and f(x)=x^2 sin(1/x) for x \neq 0. Prove that f is of bounded variation on [-1,1].
    please help me on this one.

    f'(x)=2x\sin 1\slash x-\cos 1\slash x \Longrightarrow |f'(x)|\le 2+1=3 on [-1,1], so using the mean value theorem for differentiable functions, |f(x)-f(y)|\le 3|x-y| . Now take any partition of the given interval and from the above you get at once that f has bounded variation there.

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    f'(x)=2x\sin 1\slash x-\cos 1\slash x \Longrightarrow |f'(x)|\le 2+1=3 on [-1,1], so using the mean value theorem for differentiable functions, |f(x)-f(y)|\le 3|x-y| . Now take any partition of the given interval and from the above you get at once that f has bounded variation there.

    Tonio
    \Sigma_1^n |f(x_i)-f(x_{i-1})| \leq 3\Sigma_1^n |x_i -x_{i-1}|=3*2=6<\infty.

    Is is correct?
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  4. #4
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    Quote Originally Posted by Kat-M View Post
    \Sigma_1^n |f(x_i)-f(x_{i-1})| \leq 3\Sigma_1^n |x_i -x_{i-1}|=3*2=6<\infty.

    Is is correct?

    Yup...it'd be a nice exercise for you to prove that in fact x^p\sin 1\slash x has bounded variation iff p>1 ...

    Tonio
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