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Thread: Derivative Inequality

  1. #1
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    Derivative Inequality

    I'm trying to think this out, but I just can't get it to make any sense. Let f be continuous on [a,b] (a<b) and differentiable on (a,b). If f(a) = 0 and abs(f'(x)) <= c*abs(f(x)) on (a,b), show that abs(f'(x)) <= (c^2)*(x-a)*abs(f(d)), a < d < x. Does anyone know what is needed to be done here? Thanks in advance.
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  2. #2
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    I think I can use the Mean Value Theorem in the form f(x) = f'(d)(x-a) where a < d < x. OK, so I get:
    abs(f'(x)) <= c*abs(f'(d)(x-a)) = c*(x-a)*abs(f'(d)). What do I do now?
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  3. #3
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    Quote Originally Posted by agentZERO View Post
    I'm trying to think this out, but I just can't get it to make any sense. Let f be continuous on [a,b] (a<b) and differentiable on (a,b). If f(a) = 0 and abs(f'(x)) <= c*abs(f(x)) on (a,b), show that abs(f'(x)) <= (c^2)*(x-a)*abs(f(d)), a < d < x. Does anyone know what is needed to be done here? Thanks in advance.
    I am going to restate a problem a little bit because I usually use the same notation over and over again.

    1)Let $\displaystyle f(x)$ be a continous function on $\displaystyle [a,b]$ differenciable on $\displaystyle (a,b)$.
    2)Let $\displaystyle f(a)=0$.
    3)There exists $\displaystyle M\in \mathbb{R}^+$ so that $\displaystyle |f'(x)|\leq M|f(x)|$ for all $\displaystyle x\in (a,b)$.

    To show: For any $\displaystyle x\in (a,b)$ we can choose $\displaystyle c\in (a,x)$ so that $\displaystyle |f'(x)|\leq M^2(x-a)|f(c)|$.

    Proof: By #1 use the mean value theorem on $\displaystyle f$ on the interval $\displaystyle [a,x]$ that means $\displaystyle f(x) - f(a) = f'(c)(x-a)$ for some $\displaystyle a<c<x$. Using #2 we can write $\displaystyle f(x) = f'(c)(x-a)$ thus $\displaystyle |f(x)|\leq |f'(c)|(x-a) \leq M(x-a)$. By #3 we know $\displaystyle |f'(x)|\leq M|f(x)|$ by what we just got it means $\displaystyle |f'(x)|\leq M|f(x)|\leq M|f(c)|(x-a)$. Q.E.D.
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  4. #4
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    Thank you very much.
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  5. #5
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    I also have to show that
    abs(f(x)) <= (c^n)*[(x-a)^n]*M where M = max(abs(f(x))) on [a,b].

    I'm going to use my notation:
    I use the MVT f(x) = f'(d)(x-a).
    Then, abs(f(x)) <= abs(f'(d))*(x-a) <= c*abs(f(d))*(x-a) <= c(x-a). I understand that M will be greater than or equal to all abs(f(x)), but how do I get the c and the (x-a) to the nth power?
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  6. #6
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    Alright, I figured the last inequality out. I just need one last thing,
    Prove that f = 0 on [a,b] given the conditions at the beginning, namely:
    -f continuous on [a,b] (a<b)
    -f differentiable on (a,b)
    -f(a) = 0
    -abs(f'(x)) <= c*abs(f(x)) on (a,b)
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