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Math Help - f(x)+f(y)<f(x+y)

  1. #1
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    f(x)+f(y)<f(x+y)

    Let's assume that f:[0,∞[ -> R is differentiable twice and f(0)=0 and f''(x)>0, when x>0.

    Show that f(x)+f(y)<f(x+y) when x,y>0.


    It's really important that I get this done, so any help is highly appreciated!!
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  2. #2
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    Quote Originally Posted by Sasbe View Post
    Let's assume that f:[0,∞[ -> R is differentiable twice and f(0)=0 and f''(x)>0, when x>0.

    Show that f(x)+f(y)<f(x+y) when x,y>0.
    For fixed y, let g(x) = f(x+y)-f(x)-f(y). Then g(0) = 0. Show that g'(x)>0 for all x>0, and deduce that g(x)>0 for all x>0.
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  3. #3
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    Thanks Opalg for the hint. Is f a (strictly) convex function because f''(x)>0? Or is it not because f''(x)=0 when x=0 (because f(0)=0)?

    I'm still pretty confused with this.
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  4. #4
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    Quote Originally Posted by Sasbe View Post
    Thanks Opalg for the hint. Is f a (strictly) convex function because f''(x)>0? Or is it not because f''(x)=0 when x=0 (because f(0)=0)?
    The only reason that you need the condition f''(x)>0 (for x>0) is that it implies that f'(x) is a strictly increasing function, so that f'(x+y) > f'(x) when y>0.

    The value of f''(x) when x=0 is not relevant. The fact that f''(x) > 0 when x>0 does in fact imply that f is strictly convex.
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