# Challenge I: Concavity, solved by Millenial

1. Aug 8, 2013

### Greg Bernhardt

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
$$f(tx + (1-t)y) \geq tf(x) + (1-t)f(y)$$
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
$$tf(x) = tf(x) + (1-t)f(0) \leq f(tx + (1-t)y)\leq f(tx)$$
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