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Calculus Problem - Properties of a Function

  1. Jan 30, 2013 #1
    Hello

    I have a mathematical problem that I could not solve. Could you please give me some hints how to solve it?

    Let [itex]f: [0,1] \rightarrow \mathbb{R}[/itex] be a continous and on [itex](0,1)[/itex] a differentiable function with following properties:

    a) [itex]f(0) = 0[/itex]
    b) there exists a [itex]M>0[/itex] with [itex]|f'(x)| \leq M |f(x)| [/itex] for all [itex]x \in (0,1)[/itex]

    Now the problem is: Show that [itex]f(x) = 0[/itex] is true for all [itex]x \in [0,1][/itex]

    There is a hint given but it doesn't help me The hint is: Consider the set [itex]D = \{ x \in [0,1]: ~ f(t) =0 [/itex] for [itex]t \in [0,x] \}[/itex] and show that the the supremum of this set is [itex]1[/itex].

    Thanks for help
    Greetings
     
  2. jcsd
  3. Jan 30, 2013 #2

    jbunniii

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    What if we look at the interval ##I = [0, \frac{1}{2M}]## or ##I = [0, 1]##, whichever is smaller. This ensures that any point in ##I## will be no larger than ##\frac{1}{2M}##.

    As ##f## is continuous on ##I##, so is ##|f|##, so ##|f|## achieves a maximum at some point ##x \in I##. If ##x = 0## then we're done. Otherwise, we can apply the mean value theorem to ##f## on ##[0,x]## and, with a bit of work, derive a contradiction.
     
  4. Jan 30, 2013 #3

    jbunniii

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    P.S. The above will show that under the given hypotheses, we will have ##f(x) = 0## for all ##x \in I##. If ##I = [0,1]## then we're done. Otherwise, this shows that ##f(x) = 0## for all ##0 \leq x \leq \frac{1}{2M}##. In that case, you can repeat the argument for ##[\frac{1}{2M}, \frac{2}{2M}]## and so forth until you have covered all of ##[0,1]##.
     
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