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Math Help - convexity problem

  1. #1
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    convexity problem

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  2. #2
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    Quote Originally Posted by szpengchao View Post
    For the first part assuming 0 < t,

    f''(t) \geq 0 \Rightarrow \int_0^t f''(t)\, dt \geq 0 \Rightarrow f'(t) \geq f'(0) > 0 \Rightarrow f'(t) > 0

    Rinse and repeat:

    f'(t) > 0 \Rightarrow \int_0^t f'(t)\, dt > 0 \Rightarrow f(t) > f(0)= 0 \Rightarrow f(t) > 0

    Now for t_2 > t_1, f'(t) > 0 \Rightarrow \int_{t_1}^{t_2} f'(t)\, dt > 0 \Rightarrow f(t_2) > f(t_1)

    So f is monotonically increasing and yet f(0) = f(1) = 0. Thus f is identically 0 throughout [0,1]. Similarly for the next one, except

    \int_{t}^1 f''(t) \geq 0 \Rightarrow f'(t) \leq f'(1) \leq 0 \Rightarrow f'(t) < 0

    Then you can repeat what you did before
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  3. #3
    Moo
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    Hello,

    An insignificant report...

    <br />
f''(t) \geq 0 \Rightarrow \int_0^t f''(t)\, dt \dots<br />
    I'm not sure it's correct to call the variable t in the integrand, just like the variable in the boundaries... I know it's understandable, but well, it's quite disturbing
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  4. #4
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    Quote Originally Posted by Moo View Post
    Hello,

    An insignificant report...

    I'm not sure it's correct to call the variable t in the integrand, just like the variable in the boundaries... I know it's understandable, but well, it's quite disturbing
    I dont understand whats the problem..... Do you mean you dont want the same variable appearing in both places?

    Here its not troublesome, so why bother inventing a new one?
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  5. #5
    Moo
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    Quote Originally Posted by Isomorphism View Post
    I dont understand whats the problem..... Do you mean you dont want the same variable appearing in both places?

    Here its not troublesome, so why bother inventing a new one?
    Nevermind, forget it
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  6. #6
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    any other way

    any other way of doing this by convex function's property?

    and i hav a question, is there any property of convexity in R^n ?

    f''(x)>=0 gives global definition in R
    but this fails in R^n, why???
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