If $f(x_{1}+x_{2})+2\ge f(x_{1})+f(x_{2})$, then $f(2^{-x})>2^{-x}+2$?

Question

Let $f(x)$ be monotonic increasing on $[0,1]$ and such that $f(0)=2,f(1)=3$, and for any $x_{1},x_{2},x_{1}+x_{2}\in[0,1]$: $$f(x_{1}+x_{2})+2\ge f(x_{1})+f(x_{2})$$

Question: between $\displaystyle f \left (\frac{1}{2^x} \right)$ and $\displaystyle\frac{1}{2^x}+2,\forall x>0,x\in \mathbb{R}$ which is bigger?

My idea: let $$F(x)=f(x)-2$$ then $$F(0)=0,F(1)=1,F(x_{1}+x_{2})\ge F(x_{1})+F(x_{2}),\forall x_{1},x_{2},x_{1}+x_{2}\in[0,1]$$

The problem is that $x$ is a real number, not an integer.

Answer 1

You were in the right direction. Note that $F(2t)\ge 2F(t)$ for all $t$ such that $2t\le1$. It follows that $$F(2^mt)\ge 2^mF(t)\quad\text{ for }m\in \Bbb N$$ or that $$F(\frac1{2^m})\le \frac{F(1)}{2^m}=\frac1{2^m}.$$ Now for any $t\in(0,1)$, use the decreasing dyadic expansion of $t$ and the monotone of $F$ to conclude.

Edit: Note that we still need to apply the superadditivity of $F$.

Edit 2: There's a flaw in my argument (and of course it's the last step). I will just leave it here and try to fix it then.

Edit 3: I don't know why the previous answer was accepted. The correct answer turns out to be non decisive.

Naturally, we (or at least I) tend to think that $F(x)\le x$ for all $x\in[0,1]$. And that is true for functions like $F(x)=x^2$. However, without the continuity of $F$, there are counter examples. The simplest would be $$F(x)=\begin{cases}0&\text{ if } x\le \frac12\\ 1&\text{ if } x>\frac12\end{cases}$$ or, if we want $F$ to be strictly increasing $$F(x)=\begin{cases}\frac{x}{2} &\text{ if }x\le \frac12\\ \frac{1+x}{2}&\text{ if }x>\frac12.\end{cases}$$

Answer 2

You're almost finished. If $F(x_1+x_2) \geq F(x_1)+F(x_2)$ then take $x_1=x_2$ and get $F(2x) \geq 2F(x)$. Therefore $1=F(1) \geq 2F(1/2)$, which implies $1/2 \geq F(1/2)$. Inductively you can show that $1/2^n \geq F(1/2^n)$.

Answer 3

Set $F=f-2$.

Then $F(0)=0$ and $F(1)=1$, and $F(x+y)\ge F(x)+F(y)$, and $F$ increasing.

So, in particular, $F(2x)\ge 2 F(x)$, if $x\in[0,1/2]$, and thus $$ 2^nF(2^{-n})\le F(1)=1, \,\,\,n\in\mathbb N. $$ Thus $$ F(2^{-n})\le 2^{-n}, $$ and finally $$ f(2^{-n})\le 2^{-n}+2. $$