Let $f : \mathbb R \rightarrow \mathbb R$ satisfy $f(x) \le x$ and $f(x+y) \le f(x)+f(y)$ for all $x,y \in \mathbb R$. Show that $f(x)=x$. 
Let $f : \mathbb R \rightarrow \mathbb R$ satisfy $$f(x) \le x$$ and $$f(x+y) \le f(x)+f(y)$$ for all $x,y \in \mathbb R$. Show that $f(x)=x$.

I already know that $f(0) = 0$ but I don’t know how to do next. Thank you very much for helping!
 A: Let $f:\mathbb{R}$ satisfy
(a) $f(x)\leq x$ for each $x\in\mathbb{R}$, and
(b) $f(x+y)\leq f(x)+f(y)$ for all $x,y\in\mathbb{R}$.
From (b), with $x,y:=0$, we get $f(0)\geq 0$.  However, by (a), we conclude that $f(0)=0$.  
Now, using (a), we have $f(-x)\leq -x$ for each $x\in\mathbb{R}$.  Now, plug in $y:=-x$ in (b) to get
$$0=f(0)=f\big(x+(-x)\big)\leq f(x)+f(-x)\leq x+(-x)=0\,.$$
This implies $f(x)+f(-x)=0$ (and in fact, at this point, it already follows that $f(x)=x$ for all $x\in\mathbb{R}$).  Hence, $f(-x)=-f(x)$ for all $x\in\mathbb{R}$.  
Using (a), we have
$$-f(x)=f(-x)\leq -x\,,$$
or
$$f(x)\geq x$$
for every $x\in\mathbb{R}$.  By (a), we see that $f(x)=x$ for every $x\in\mathbb{R}$.
A: By pluging in $a$ and $-a$:
$\forall a\in \mathbb{R}: f(a)+f(-a)\ge f(0) = 0$
from what was given, $f(a) \le a,f(-a) \le-a$
So: $f(a)-a \ge f(a)+f(-a) \ge 0 \to f(a) \ge a$ (True because $-a \ge f(-a)$     )
We have that $f(a) \le a $
and also that $f(a) \ge a$
So $f(a)=a$
A: From the second inequality, $f(x+h)\leq f(x)+f(h)$ so that,
$$\lim_{h\to 0^+} \frac{f(x+h)-f(x)}{h}\leq \lim_{h\to 0^+} \frac{f(x)+f(h)-f(x)}{h}=\lim_{h\to 0^+} \frac{f(h)}{h}$$
But from the first inequality in the question, $f(h)/h\leq 1$ so that, $$\lim_{h\to 0^+} \frac{f(x+h)-f(x)}{h}\leq \lim_{h\to 0^+} \frac{f(h)}{h}\leq 1$$
Similarly, $$\lim_{h\to 0^+} \frac{f(x)-f(x-h)}{h}\geq \lim_{h\to 0^+} \frac{f(x)-(f(x)+f(-h))}{h}=\lim_{h\to 0^+} \frac{f(-h)}{-h}$$ 
But $f(-h)/(-h)\geq 1$ (inequality reverses when dividing by a negative number and $-h<0$ since $h>0$). Thus we get, 
$$\lim_{h\to 0^+} \frac{f(x)-f(x-h)}{h}\geq \lim_{h\to 0^+} \frac{f(-h)}{-h}\geq 1$$
On the other hand, for $h>0$,
$$\frac{f(x+h)-f(x)}{h}=\frac{f(x+h)-f((x+h)+(-h))}{h}\geq \frac{f(-h)}{-h}\geq 1$$
and
$$\frac{f(x)-f(x-h)}{h}=\frac{f((x-h)+h)-f(x-h)}{h}\leq \frac{f(h)}{h}\leq 1$$
Therefore $f'(x)$ exists and $f'(x)=1$ (particularly, the fact that limit of a function strictly dominated by another function is less than or equal to limit of the other function is used).
This means that $f(x)=x+c$, where the integration constant $c=0$ because $f(0)=0$.
Thanks to Batominovski for supplying the proof for differentiability of this function.
