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Math Help - Proving that f^2 is convex if f is convex and f>=0

  1. #1
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    Lightbulb Proving that f^2 is convex if f is convex and f>=0

    How do I go about this proof? Should I use the definition of convexity:

    f(ax+(1-a)y)<=af(x)+(1-a)f(y)?
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  2. #2
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    You know that for 0<a<1, f(ax+(1-a)y)<af(x)+(1-a)f(y).

    Since f>0, you can square this inequality to obtain
    f(ax+(1-a)y)<af(x)+(1-a)f(y)+2a(1-a)f(x)f(y).

    Now remember that 2f(x)f(y)<f(x)+f(y) (because (f(x)-f(y))>0 )

    The inequality becomes
    f(ax+(1-a)y)<af(x)+(1-a)f(y)+a(1-a)(f(x)+f(y))=f(x)(a+a-a)+f(y)((1-a)+a(1-a))=
    af(x)+(1-a)f(y)
    and f is convex by definition.

    Note that if f is C, f convex <=> f''>0.
    In this case you have a simpler proof
    (f)'=2ff': (f)''=2f'2+2ff''>0.

    (Note that I used < for
    <br />
\leq<br />
    all over the place)
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  3. #3
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    Ahhh, thank you for your help. I didn't think about 2f(x)f(y)<=f^2(x)+f^2(y).

    Thank you.
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