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Two real analysis problems: proving constancy and a uniform convergence problem

  1. Oct 13, 2011 #1
    The problem statement

    Let [itex]f:[a,b]→\mathbb{R}[/itex] be differentiable and assume that [itex]f(a)=0[/itex] and [tex]\left|f'(x)\right|\leq A\left|f(x)\right|, x\in [a,b].[/tex]
    Show that [itex]f(x)=0,x\in [a,b][/itex].

    The attempt at a solution

    It was hinted at that the solution was partly as follows. Let [itex]a \leq x_0 \leq b[/itex]. For all [itex]x\in [a,x_0][/itex] [tex]\left|f(x)\right|\leq (x_0-a)\underset{[a,x_0]}{\sup}\left|f'(x)\right|\leq A(x_0-a)\underset{[a,x_0]}{\sup}\left|f(x)\right|.[/tex] Then we should ask what happens for [itex]A(x_0-a)<1[/itex].

    The first inequality follows from the mean value theorem and the second from the given equation. When [itex]A(x_0-a)<1[/itex], then [tex]\left|f(x)\right|< \underset{[a,x_0]}{\sup}\left|f(x)\right|.[/tex] I do not see how this would imply the proposed relation.

    EDIT: I have already solved the next problem.

    The problem statement

    Is the series [tex]\sum^{\infty}_{n=1} \frac{x^n(1-x)}{n}[/tex] uniformly convergent on [itex][0,1][/itex]?

    The attempt at a solution

    I suspect it is. I want to use the Weierstrass M-test, but obviously the sequence [itex]M_n=1/n[/itex] doesn't work as a bound since [tex]\sum^{\infty}_{n=1} \frac{1}{n}[/tex] does not converge. I can't figure how to produce a tighter bound than [itex]M_n=1/n[/itex].
    1. The problem statement, all variables and given/known data
     
    Last edited: Oct 13, 2011
  2. jcsd
  3. Oct 13, 2011 #2
    I just wanted to say that I have solved the second problem. By finding the maximum of [itex]x^n(1-x)[/itex] by differentiating it and setting it to zero, we can get the inequality [itex]x^n(1-x) < 1/n[/itex]. I feel a little silly for not considering that before.

    I am still breaking my brain over the first problem, though. If anyone has any suggestion on how to solve it, I'd love to hear from you.
     
  4. Oct 13, 2011 #3

    Bacle2

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    What's A, Sensei? Is it a positive constant?
     
  5. Oct 13, 2011 #4

    Bacle2

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    Maybe this will help: Since f(0)=f(1)=0, check to see what happens with values like
    1/4, 1/2 and 3/4; if these values do not go towards 0, the series cannot converge uniformly. Similarly, try to see where the maximum is for the genersl product x(1-x)n to get an idea of how convergence happens.
     
  6. Oct 13, 2011 #5
    A is indeed a positive constant. Sorry for the ambiguity.

    Also, I have solved the second problem, so I don't need help with that anymore.
     
  7. Oct 14, 2011 #6
    First prove it for the special case where [a,b] = [0,1]. Then let [itex] g: [0,1] \rightarrow \mathbb{R} [/itex] be defined by [itex] g(x) = f(a + (b-a)x) [/itex]. Using your special case proof, conclude that g = 0, from which your theorem follows.
     
  8. Oct 14, 2011 #7
    Alright, that is a nice strategy. But I still have to prove the special case, which I still can't seem to do.

    I have made the following attempt at proving the general case in one go.

    Choose [itex]c[/itex] such that [itex]a<c<b[/itex] and [itex]A/(c-a)>1[/itex]. Let us assume for contradiction that there exists an [itex]x_0\in [a,c][/itex] satisfying [itex]f(x_0)\neq 0[/itex]. By the mean value theorem there then exists an [itex]x_1\in (a,x_0)[/itex] satisfying [itex]|f'(x_1)|=|f(x_0)|/(x_0-a)[/itex]. But [itex]|f(x_1)|\geq A|f'(x_1)|=A|f(x_0)|/(x_0-a)> |f(x_0)|>0[/itex] We can repeat this procedure to construct a sequence [itex](x_n)[/itex]. In general there exists an [itex]x_n\in (a,x_{n-1})[/itex] satisfying [itex]|f(x_n)|>0[/itex].

    Now if I can prove that the sequence [itex](x_n)[/itex] converges to [itex]a[/itex], then I am done. This is because then [itex]\lim x_n=a[/itex] but [itex]\lim f(x_n)\neq f(a)[/itex], which implies that [itex]f[/itex] is not continuous, which contradicts the hypothesis of [itex]f[/itex] being differentiable. This would allow me to say that for all [itex]x\in [a,c][/itex] [itex], f(x)=0[/itex]. But then I can apply the same argument to the interval [itex][c, a+2(c-a)][/itex] (the interval of length [itex](c-a)[/itex] which lies next to [itex][a,c][/itex]) and so on.

    The problem is, I can't prove that [itex]\lim x_n=a[/itex].

    [itex][/itex]
    [tex][/tex]
     
    Last edited: Oct 14, 2011
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