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 $f(0) = 0$?

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    $\begingroup$ FWIW: the full IMO problem seems to be (in addition to above): prove that $f(x) = 0$ for all $x \le 0$. $\endgroup$ – ShreevatsaR Jul 19 '11 at 18:43
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    $\begingroup$ 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. $\endgroup$ – ShreevatsaR Jul 19 '11 at 18:54
  • $\begingroup$ 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). $\endgroup$ – Shai Covo Jul 19 '11 at 19:19
  • $\begingroup$ 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). $\endgroup$ – joriki Jul 19 '11 at 19:57
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    $\begingroup$ @Chandru: My solution goes along the very same lines. $\endgroup$ – Yuval Filmus Jul 19 '11 at 20:17

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 $g(x) = \exp(x)$.

  • $\begingroup$ i don't understand how lim(x→−∞)f(x)=−∞. $\endgroup$ – Victor Jul 19 '11 at 19:28
  • $\begingroup$ The function is bounded from above by a strictly increasing linear function. $\endgroup$ – Yuval Filmus Jul 19 '11 at 19:31
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    $\begingroup$ 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. $\endgroup$ – joriki Jul 19 '11 at 19:40
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    $\begingroup$ @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, $x_n = -f(-n)$. $\endgroup$ – Yuval Filmus Jul 19 '11 at 20:12
  • $\begingroup$ Ah, now I see, thanks! $\endgroup$ – 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]


  • $\begingroup$ Please post a full solution in one answer (and then delete the other one) $\endgroup$ – Mariano Suárez-Álvarez Jul 21 '11 at 15:55
  • $\begingroup$ Yep, above is a full ten-lines-solution. $\endgroup$ – 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 $f(0)=0$.

  • $\begingroup$ Can you please post a full solution? Thanks. $\endgroup$ – Amir Hossein Jul 21 '11 at 9:47

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