# Functional inequality $f(x_1+x_2)\ge f(x_1)+f(x_2)$

Given a function $f$ on the interval $0\le x \le 1$. We know that this function is non-negative and $f(1)=1$. Moreover, for any two numbers $x_1$ and $x_2$ such that $x_1\ge 0, x_2 \ge 0$ and $x_1+x_2\le 1$ the inequality $$f(x_1+x_2)\ge f(x_1)+f(x_2).$$ a) Prove that $f(x)\le 2x$

b) Is it true that for all $x:$ $f(x)\le 1.9x$?

### My work so far:

a) The function increases monotonically. Really, if $x_2\ge x_1$ and $x_2\le 1$, that $f(x_2)\ge f(x_1)+f(x_2-x_1)$ and $f(x_2-x_1)\ge 0$. Hence, $f(x_2)\ge f(x_1)$.

I need help here.

b) I need help here too.

• Your proof on monotonictiy seems to be not correct. At least I don't understand it. You'll need $x_1+x_2\le1$, and even then it follows only that $f(x_2)\le f(x_1+x_2) - f(x_1)$. Or do I miss something here? – Dr_Be Jun 16 '16 at 16:24
• $f(x_2)=f(\color{red}{x_1}+\color{blue}{x_2-x_1})\ge \color{red}{f(x_1)}+\color{blue}{f(x_2-x_1)}\ge f(x_1)$ $\Rightarrow f(x_2)\ge f(x_1)$ – Roman83 Jun 16 '16 at 18:00
• OK, thanks. Didn't notice you used the inequality with $x_1$ and $x_2-x_1$. – Dr_Be Jun 17 '16 at 6:39

You showed $f$ is monotone, so in particular $f(x) \le 1$ for all $x \in [0,1]$. In particular, if $x\ge 1/2$ then $f(x) \le 1 \le 2x$ automatically.
Now suppose $x \in [1/4,1/2)$ and $f(x) > 2x$. Then $2x < 1$ so $f(2x)$ is defined, but on the other hand $f(2x) \ge 2f(x) > 2\cdot 1/2 = 1$, contradicting that $f(x) \le 1$.
Next, if $x \in [1/8,1/4)$ and $f(x) > 2x$. Then $4x < 1$ so $f(4x)$ is defined. But then $f(4x) \ge 4f(x) > 4 \cdot 1/4 = 1$, contradicting that $f(x) \le 1$.
For part b, we the answer is no. Let $f(x) = 1$ if $x \ge 1/2$ and $f(x)=0$ otherwise.