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Math Help - Prove that f=0

  1. #1
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    Prove that f=0

    Let be function differentiable on and satisfy conditions

    Prove that .
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  2. #2
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    Let g(x)=e^{-Kx}f(x), thus the only thing we need to prove is that g(x)=0.

    We have that g'(x)=e^{-Kx}\big(f'(x)-Kf(x)\big), but given |f'|\le K|f| we get that g(x)g'(x)\le0 which is the same as \left( \frac{g^{2}(x)}{2} \right)'\le 0, thus \big(g(x)\big)^2\le0 then g(x)=0 for x\ge0.

    We proved a stronger result, given f:[0,\infty)\to\mathbb R differentiable.
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  3. #3
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    Yes, this is it. Congratulations for nice solution. And little correction:
    from
    <br />
\left( \frac{g^{2}(x)}{2} \right)'\le 0,<br />
    you can't conclude
    <br />
\big(g(x)\big)^2\le0<br />

    but just, by integration, that g^2(x)=C=const,
    and than from g(0)=0 that C=0.
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  4. #4
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    how can we simply assume?

    Quote Originally Posted by Krizalid View Post
    Let g(x)=e^{-Kx}f(x), thus the only thing we need to prove is that g(x)=0.

    We have that g'(x)=e^{-Kx}\big(f'(x)-Kf(x)\big), but given |f'|\le K|f| we get that g(x)g'(x)\le0 which is the same as \left( \frac{g^{2}(x)}{2} \right)'\le 0, thus \big(g(x)\big)^2\le0 then g(x)=0 for x\ge0.

    We proved a stronger result, given f:[0,\infty)\to\mathbb R differentiable.
    assuming for just g(x)? Doesnot it mean that the proof is only for g(x)? please clarify and comment.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Pulock2009 View Post
    assuming for just g(x)? Doesnot it mean that the proof is only for g(x)? please clarify and comment.
    e^{f(x)}\ne0
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  6. #6
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    Krizalid's Avatar
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    Quote Originally Posted by ns1954 View Post
    Yes, this is it. Congratulations for nice solution. And little correction:
    from
    <br />
\left( \frac{g^{2}(x)}{2} \right)'\le 0,<br />
    you can't conclude
    <br />
\big(g(x)\big)^2\le0<br />

    but just, by integration, that g^2(x)=C=const,
    and than from g(0)=0 that C=0.
    Actually by definite integration I conclude that, so there's no problem.
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