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Challenge I: Concavity, solved by Millenial

  1. Aug 8, 2013 #1
    Written by micromass:

    I have recently posted a challenge in my signature. The challenge read as follows:

    The first answer I got was from Millenial. He gave the following correct solution:

    This solution is very elegant. But there are other solutions. For example, we can prove the following chain of implications:

    • ##f^{\prime\prime}(x)\leq 0## for all ##x\in \mathbb{R}##.
    • ##f^\prime## is decreasing.
    • ##\frac{f(x)}{x}## is decreasing.
    • For each ##t\in [0,1]## and for all ##x## holds that ##tf(x)\leq f(tx)##.
    • For each ##x## and ##y## holds that ##f(x+y)\leq f(x) + f(y)##.

    The proof is as follows:
    Yet another proof consists of proving that ##f## is a concave function. This means that for each ##x## and ##y## and ##t\in [0,1]## that
    [tex]f(tx + (1-t)y) \geq tf(x) + (1-t)f(y)[/tex]
    Visually, it means that

    [Broken]

    So, the straight line between two points is always below the actual function.

    Here is the proof that any function such that ##f^{\prime\prime}<0## is concave:

    Now, we can immediately prove again that ##tf(x)\leq f(tx)## for each ##x## and ##t\in [0,1]##. Indeed, apply concavity with ##y=0## (and remember that ##f(0) = 0##, then
    [tex]tf(x) = tf(x) + (1-t)f(0) \leq f(tx + (1-t)y)\leq f(tx)[/tex]
    We can then, as before, prove ##f(x+y)\leq f(x) + f(y)##.

    Many congratulations to Millenial for solving the problem!!
     
    Last edited: May 6, 2017
  2. jcsd
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