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Convex Function Inequality

  1. Jun 11, 2012 #1
    1. The problem statement, all variables and given/known data
    Givens: [tex]\forall x\ge 0:\quad f^{ \prime \prime }\left( x \right) \ge 0;\quad f\left( 0 \right) =0[/tex]
    Prove: [tex]\forall a,b\ge 0:\quad f\left( a+b \right) \ge f\left( a \right) +f\left( b \right)[/tex]

    2. Relevant equations
    By definition, f is convex iff [tex]\forall x,y\in \Re \quad \wedge \quad \forall \lambda :\quad 0\le \lambda \le 1\quad \Rightarrow \quad f\left( \lambda x+(1-\lambda )y \right) \le \lambda f\left( x \right) +(1-\lambda )f\left( y \right)[/tex]

    3. The attempt at a solution
    Intuition-wise I see that a convex function's values increase at an increasing rate, but that's equivalent to [tex]f^{ \prime \prime }\left( x \right) \ge 0[/tex]
    I also see that [tex]f\left( 0 \right) =0[/tex] is necessary for the inequality to hold, but I can't find any tools with which I can work on proving the inequality.
    Also I figure [tex]\forall x\ge 0:\quad f^{ \prime }\left( x \right) \ge 0[/tex] and also monotonously increasing.
     
    Last edited: Jun 11, 2012
  2. jcsd
  3. Jun 11, 2012 #2

    micromass

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    OK, so we wish to prove that a convex function is "superadditive".

    First, can you prove that if [itex]t\in [0,1][/itex], that then

    [tex]f(tx)\leq tf(x)[/tex]

    Just apply convexity and use that f(0)=0.
     
  4. Jun 11, 2012 #3
    alright, that's immediate from taking y=0.
    So I now know that [tex]f(\lambda x)\leq \lambda f(x)[/tex]
     
  5. Jun 11, 2012 #4

    micromass

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    Now write

    [tex]f(a)+f(b)=f\left(\frac{a}{a+b}(a+b)\right)+f\left(\frac{b}{a+b}(a+b) \right)[/tex]

    and apply that inequality you just obtained.
     
    Last edited: Aug 2, 2013
  6. Jun 11, 2012 #5
    just figured that bit out
    thanks alot!
     
    Last edited: Jun 11, 2012
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