Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f : \mathbb R \to \mathbb R$ be a real-valued function defined on the set of real numbers that satisfies $$f(x + y) \leq yf(x) + f(f(x))$$ for all real numbers $x$ and $y$.

How can I prove that $f(0) = 0$?

share|improve this question
3  
FWIW: the full IMO problem seems to be (in addition to above): prove that $f(x) = 0$ for all $x \le 0$. –  ShreevatsaR Jul 19 '11 at 18:43
13  
Answering the question mark in the title: this cannot be the easy part, because if you assume $f(0)=0$ then it's easy to solve the rest of the problem! Indeed, put $x=0$ to get $f(y) \le 0$ (for all $y \in \mathbb{R}$), then for any given negative $x$, put $y = -x$ to get $f(x-x) \le -xf(x) + f(f(x))$ which means that $f(f(x)) \ge xf(x)$. Here the LHS is $\le 0$ and the RHS is $\ge 0$ (because $x< 0$ and $f(x)\le 0$), which can only happen if $f(x)=0$. As this is true for any negative $x$, this completes the problem. So proving that $f(0)=0$ is not much easier than solving the entire problem. –  ShreevatsaR Jul 19 '11 at 18:54
    
It seems worth checking the answer given by mlequi here: olimpiade.org/Forum/?qa=1717/imo-2011-problem-3 (currently the second answer in that link). –  Shai Covo Jul 19 '11 at 19:19
    
Assuming that I'm not the only one here who doesn't speak Malai ;-) here's the Google translation to English (unfortunately MathJaX doesn't seem to work there, and the translation is barely intelligible). –  joriki Jul 19 '11 at 19:57
1  
@Chandru: My solution goes along the very same lines. –  Yuval Filmus Jul 19 '11 at 20:17

3 Answers 3

up vote 12 down vote accepted

First, we show that $f(x) \leq 0$ for all $x$.

Suppose that $f(z) > 0$ for some $z$. The functional inequality implies that $$\lim_{x \to -\infty} f(x) = -\infty,$$ since $f$ is bounded by a strictly increasing linear function. We also have $$ f(y) = f(0+y) \leq f(0)y + f(f(0)). $$ Using this for $y = f(x)$, $$ f(0) = f(x-x) \leq -xf(x) + f(f(x)) \leq (f(0) - x)f(x) + f(f(0)).$$ As $x \to -\infty$, the right-hand side tends to $-\infty$, leading to a contradiction.

Second, let $x > 0$. Note that $$ f(0) \leq x f(-x) + f(f(-x)) \leq xf(-x). $$ Thus $f(-x) \geq f(0)/x$. As $x\to\infty$, the righthand side tends to zero. Since $$ f(-x) = f(-x+0) \leq f(f(-x)), $$ we get that there is a sequence of points $x_n = -f(-n)$ tending to zero such that $f(-x_n) \to 0$.

Suppose that $f(f(0)) < 0$. Then $$f(-x_n) = f(0-x_n) \leq -f(0)x_n + f(f(0)).$$ Since $x_n\to 0$, the righthand side tends to $f(f(0)) < 0$, contradiction.

We conclude that $f(f(0)) = 0$. This implies that $$0 = f(f(0)) \leq f(f(f(0))) = f(0), $$ and so $f(0) = 0$.

Third, as noted above, for $x > 0$ we have $$xf(-x) \geq f(0) = 0.$$ Thus $f(-x) = 0$ for all $x \geq 0$.

Addendum: Putting $g(x) = -f(x)$, the functional inequality reduces to $$ g(x+y) \geq yg(x), $$ where now $g\colon \mathbb{R}_+ \to \mathbb{R}_+$. An example is the function $g(x) = \exp(x)$.

share|improve this answer
    
i don't understand how lim(x→−∞)f(x)=−∞. –  Victor Jul 19 '11 at 19:28
    
The function is bounded from above by a strictly increasing linear function. –  Yuval Filmus Jul 19 '11 at 19:31
1  
I don't understand how "we get that there is a sequence of points $x_n$ tending to zero such that $f(-x_n) \to 0$". In the line before, $x$ tended to $\infty$, and I don't see how that combined with the displayed equation in between makes $x$ or $f(-x_n)$ tend to zero. –  joriki Jul 19 '11 at 19:40
3  
@joriki: As $t \to -\infty$, the values $f(t)$ approach zero. Now consider $f(f(t))$. Since $0 \geq f(f(t)) \geq f(t)$, if we take $t$ large and negative, then both $f(t)$ and $f(f(t))$ will be very close to zero. So we can take, say, $x_n = -f(-n)$. –  Yuval Filmus Jul 19 '11 at 20:12
    
Ah, now I see, thanks! –  joriki Jul 19 '11 at 20:54

This is my solution: $$f(f(x)) = f(y+f(x)-y) \le (f(x)-y).f(y) + f(f(y)),\forall x, y (1)$$ swap $x, y$, we have: $$f(f(y))\le (f(y)-x)f(x)+f(f(x)), \forall x, y (2)$$ (1), (2) $\Rightarrow 0 \le 2f(x)f(y)-xf(x)-yf(y), \forall x, y.$

$\Rightarrow -xf(x) \ge (y-2f(x))f(y), \forall x, y \Rightarrow -xf(x)\ge 0, \forall x (*)$ (using $y=2f(x)$)

In the other hand: $f(y)=f(x+y-x)\le (y-x)f(x)+f(f(x)), \forall x, y$. Suppose that there exist $x: f(x)>0$, then $\lim\limits_{y\to-\infty}(y-x)f(x)=-\infty \Rightarrow \lim\limits_{y\to-\infty}f(y)=-\infty \Rightarrow \lim\limits_{x\to-\infty}(-xf(x))=-\infty$! (absurd from $(*)$).

Therefore $f(x) \le 0, \forall x (**)$

From $(*), (**) \Rightarrow\forall x<0: -xf(x)\ge 0 \Rightarrow f(x)\ge 0 \Rightarrow f(x)=0.$

The last one: $0=f(-1)\le f(f(-1))=f(0)\Rightarrow f(0)=0.$ [End of proof]

koreagerman.

share|improve this answer
    
Please post a full solution in one answer (and then delete the other one) –  Mariano Suárez-Alvarez Jul 21 '11 at 15:55
    
Yep, above is a full ten-lines-solution. –  koreagerman Jul 21 '11 at 17:25

$f(0)=0$ is the last part of my solution :) Step1: prove that $-xf(x)\ge 0$, for all $x$

Step2: prove that $f(x)\le 0$, for all $x$, then $f(x)=0$ for all $x<0$

Step3: $0=f(-1)=f(-1+0)\le 0$.
$f(-1)+f(f(-1))=f(0)$, then $f(0)=0$.

share|improve this answer
    
Can you please post a full solution? Thanks. –  Amir Hossein Jul 21 '11 at 9:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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