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Continuously differentiable

  1. Jan 3, 2009 #1
    I'm having trouble with this inequality:

    let f be (real valued) continuously differentiable on [0,1] with f(0)=0, prove that

    sup[tex]_{x\in[0,1]}[/tex] [tex]\left|f(x)\right|[/tex] [tex]\leq[/tex] [tex]\int^{1}_{0}\left|f\acute{}(x)\right| dx [/tex]

    Thanks for any help.
  2. jcsd
  3. Jan 3, 2009 #2


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    Gold Member

    Clarifying questions: Do you intend to find the supremum of the function or the interval? Furthermore, is the second f(x) term an f(x) or f'(x)? (It looks like there is a little tick next to it but I cannot tell for sure)
  4. Jan 3, 2009 #3
    Sorry for the discrepancy, the problem is to show that the supremum of |f(x)| over [0,1] is less than or equal to the integral from 0 to 1 of |f ' (x)|, where f ' (x) is the derivative of f.
  5. Jan 4, 2009 #4
    Look at this:

    |f(t)| = \left|\int_0^t f'(x) \mbox{ d}x \right| \leq \int_0^t |f'(x)|\ \mbox{ d}x \leq \int_0^1 |f'(x)|\ \mbox{ d}x

    Can you finish this?
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