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!

  • 1
    $\begingroup$ Which contest is it from please ? $\endgroup$ – Ewan Delanoy Apr 26 '20 at 9:53
  • $\begingroup$ It’s just a problem in the book written about contest math. Would you like me to remove the tag? $\endgroup$ – Peerakorn Trechakachorn Apr 26 '20 at 9:55
  • 1
    $\begingroup$ No, no need to remove the tag. Which book ? $\endgroup$ – Ewan Delanoy Apr 26 '20 at 9:55
  • $\begingroup$ It’s in Thai. It is about the problems that could be TSTST in Thailand. $\endgroup$ – Peerakorn Trechakachorn Apr 26 '20 at 9:57
  • 1
    $\begingroup$ After Let, there is no need to inflect the verb. For example, "Let it go" as oppose to "Let it goes." Here is another example, "He lets her take his car" as oppose to "He lets her takes his car." $\endgroup$ – Batominovski Apr 26 '20 at 10:10

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 $x\in\mathbb{R}$.


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 $f(a)=a$


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.

  • $\begingroup$ This would be a nice solution if $f$ is assumed to be differentiable, but there isn't such an assumption. (It is up to you, but I think you should keep this answer, rather than deleting it. It might teach the OP a new trick.) $\endgroup$ – Batominovski Apr 26 '20 at 10:14
  • $\begingroup$ @Batominovski Ah right! I will edit my answer to include the assumption that the function is differentiable, but ya now my answer won't matter much. Thank you for pointing this out. $\endgroup$ – ModCon Apr 26 '20 at 10:16
  • 1
    $\begingroup$ After a bit of thinking, I think your solution can be modified, so that you don't have to assume differentiability. You have proven that $$\lim_{h\to 0^+}\frac{f(x+h)-f(x)}{h}\leq 1$$ and $$\lim_{h\to 0^-}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0^+}\frac{f(x)-f(x-h)}{h}\geq 1\,.$$ On the other hand, for $h>0$, $$\frac{f(x+h)-f(x)}{h}=\frac{f(x+h)-f\big((x+h)+(-h)\big)}{h}\geq \frac{-f(-h)}{h}=\frac{f(-h)}{-h}\geq 1$$ and $$\frac{f(x)-f(x-h)}{h}=\frac{f\big((x-h)+h\big)-f(x-h)}{h}\leq \frac{f(h)}{h}\leq 1\,.$$ Therefore, $f'(x)$ exists, and it is equal to $1$. $\endgroup$ – Batominovski Apr 26 '20 at 10:38
  • $\begingroup$ @Batominovski That's really nice! I should have thought in this direction. $\endgroup$ – ModCon Apr 26 '20 at 10:43
  • $\begingroup$ I wouldn't mind if you use my comment to improve your answer, so the differentiability assumption can be removed from your answer. $\endgroup$ – Batominovski Apr 26 '20 at 10:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.