A Suspicious Function

Algebra Level 3

Let $f(x)$ denote a function satisfying the 2 properties below:

$\begin{cases} \begin{aligned} f(x+y) &\geq f(x) + f(y) \\ f(x) &\geq x, \end{aligned} \end{cases}$ where $x$ and $y$ are real numbers.

It's obvious that a possible solution for $f(x)$ is $f(x) = x.$ Are there any other possible solutions for $f(x)?$

Yes No

1 solution

Pepper Mint
Oct 7, 2017

First, put x=y=0 in the first inequation: f(0)≥f(0)+f(0),f(0)≤0 And in the second one too: f(0)≥0 Eventually, f(0)=0.

Put y=-x in the first one: f(0)≥f(x)+f(-x),f(x)≤-f(-x), and put -x instead of x in the second one: f(-x)≥-x,x≥-f(-x)

So, x≥-f(-x)≥f(x). Since f(x)≥x, f(x)≥x≥-f(-x)≥f(x)

Thus, f(x)=x. The answer is No .