# UPDATED: If $f(x + y) \leq yf(x) + f(f(x))$ for all real numbers $x$ and $y$, prove that $f(0) = 0.$

$$\large\text{UPDATED:}$$ (with completely correct arguments)

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 $$y$$. Prove that $$f(x) = 0$$ for all $$x ≤ 0$$. (IMO $$2011$$)

The purpose of my question is only proof verification. (not knowing the correct solution)

Here, I focus only on the case of $$f(0) = 0.$$ Because this is the main part of the problem and this is very easy to show that, $$f(0) = 0$$ follows $$f(x) = 0$$ for all $$x ≤ 0.$$ I want to prove only the $$f(0)=0$$.

Here are my steps:

Case $$1.$$ $$f(0)\in \mathbb R^+$$

We have,

$$f(0)\leq-xf(x)+f(f(x))$$

$$f(x)\leq xf(0)+f(f(0))$$

Applying $$x \longrightarrow -\infty$$ we get from $$f(x)\leq xf(0)+f(f(0))$$, $$\lim_{x\to -\infty}f(x) = -\infty$$.

Then applying again $$x \longrightarrow -\infty$$, from $$f(0)\leq-xf(x)+f(f(x))$$ we get $$f(0) \longrightarrow-\infty \not \in\mathbb R^+$$, which gives a contradiction.

Case $$2.$$ $$f(0)<0$$ (with the wrong argument, e.g. $$\lambda=0$$)

$$\require{enclose} \enclose{horizontalstrike}{ \text{We have, from}}$$ $$\require{enclose} \enclose{horizontalstrike}{f(x)\leq xf(0)+f(f(0))}$$ $$\require{enclose} \enclose{horizontalstrike}{\text{we deduce}}$$ $$\require{enclose} \enclose{horizontalstrike}{\lim_{x\to +\infty}f(x)=-\infty.}$$ $$\require{enclose} \enclose{horizontalstrike}{\text{Suppose that,}}$$ $$\require{enclose} \enclose{horizontalstrike}{ \lim_{x\to -\infty}f(x)=+\infty.}$$ $$\require{enclose} \enclose{horizontalstrike}{\text{Applying}}$$ $$\require{enclose} \enclose{horizontalstrike}{x\to-\infty}$$ $$\require{enclose} \enclose{horizontalstrike}{\text{from}}$$ $$\require{enclose} \enclose{horizontalstrike}{f(x-1) ≤ -f(x) + f(f(x))}$$ $$\require{enclose} \enclose{horizontalstrike}{\text{we have}}$$ $$\require{enclose} \enclose{horizontalstrike}{\lim_{x\to -\infty}f(x+(-1)) \longrightarrow +\infty}$$. $$\require{enclose} \enclose{horizontalstrike}{\text{But,}}$$ $$\require{enclose} \enclose{horizontalstrike}{\lim_{x\to -\infty} (-f(x) + f(f(x)))=-\infty}$$. $$\require{enclose} \enclose{horizontalstrike}{\text{According our assumption, we applied}}$$ $$\require{enclose} \enclose{horizontalstrike}{\lim_{x\to +\infty}f(x)=-\infty.}$$ $$\require{enclose} \enclose{horizontalstrike}{\text{So, this is a contradiction.}}$$ $$\require{enclose} \enclose{horizontalstrike}{ \text {Suppose that}}$$ , $$\require{enclose} \enclose{horizontalstrike}{ \lim \inf_{x\to -\infty}f(x)=a}$$ $$\require{enclose} \enclose{horizontalstrike}{ \text{and}}$$ $$\require{enclose} \enclose{horizontalstrike}{ \lim \sup_{x\to -\infty}f(x)=b}$$, $$\require{enclose} \enclose{horizontalstrike}{ \text{where}}$$ $$\require{enclose} \enclose{horizontalstrike}{ a,b\in\mathbb{R}}$$ $$\require{enclose} \enclose{horizontalstrike}{ \text{and for any}}$$ $$\require{enclose} \enclose{horizontalstrike}{ \lambda \in [a,b]}$$ $$\require{enclose} \enclose{horizontalstrike}{ \text{we have}}$$ $$\require{enclose} \enclose{horizontalstrike}{ \lambda\leq y\lambda+f(\lambda)}$$. $$\require{enclose} \enclose{horizontalstrike}{ \text{For any}}$$ $$\require{enclose} \enclose{horizontalstrike}{ \lambda}$$ $$\require{enclose} \enclose{horizontalstrike}{ \text{we can always choose a finite}}$$ $$\require{enclose} \enclose{horizontalstrike}{y}$$ $$\require{enclose} \enclose{horizontalstrike}{ \text{such that, where we get}}$$ $$\require{enclose} \enclose{horizontalstrike}{ \lambda\ > y\lambda+f(\lambda)}$$ $$\require{enclose} \enclose{horizontalstrike}{ \text{which gives a contradiction.}}$$ $$\require{enclose} \enclose{horizontalstrike}{ \text{So, we deduce that}}$$ $$\require{enclose} \enclose{horizontalstrike}{ \lim_{x\to -\infty}f(x)=-\infty}$$. $$\require{enclose} \enclose{horizontalstrike}{\text{Then, applying}}$$ $$\require{enclose} \enclose{horizontalstrike}{x\to-\infty}$$ $$\require{enclose} \enclose{horizontalstrike}{\text{from}}$$ $$\require{enclose} \enclose{horizontalstrike}{f(0)\leq-xf(x)+f(f(x))}$$ $$\require{enclose} \enclose{horizontalstrike}{\text{we get}}$$ $$\require{enclose} \enclose{horizontalstrike}{f(0)\longrightarrow -\infty}$$.

Case $$2.$$ $$f(0) \in \mathbb {R^-}$$ (with the correct argument)

We have, from $$f(x)\leq xf(0)+f(f(0))$$ we deduce $$\lim_{x\to +\infty}f(x)=-\infty.$$ From $$f(x + y) \leq yf(x) + f(f(x))$$ we have:

$$\begin{cases} f(x)\leq f(f(x)) \\ f(x) \leq xf(0)+ f(f(0)) \end{cases} \Longrightarrow f(x)\leq f(x)f(0)+f(f(0)) \Longrightarrow f(x)(1-f(0))\leq f(f(0))$$.

Then applying $$x=f(0)$$, we get $$f(f(0))\leq 0$$, which imply $$f(x)\leq 0$$, which gives $$f(f(x))\leq 0$$. In this case, we have $$f(x)<0.$$ Because, if $$f(x)=0$$, from $$f(x)\leq f(f(x))$$, we get $$f(0)\geq 0$$, which gives a contradiction. Then, from $$f(x + y) \leq yf(x) + f(f(x))$$ we have:

$$f(z)\leq(z-x)f(x)+f(f(x)) \Longrightarrow f(x) \leq (x-y)f(y)+f(f(y))\Longrightarrow 0\leq(f(y)-y)f(y) \Longrightarrow f(x)(f(x)-x)\geq 0 \Longrightarrow f(x) \leq x$$

Applying $$x\to-\infty$$ from $$f(0)\leq-xf(x)+f(f(x))$$, we get $$f(0)\longrightarrow -\infty \not \in \mathbb{R^-}$$, which gives again a contradiction.

So, we can deduce that $$f(0)=0$$.

Q.E.D.

Can you verify the new solution?

I just want to make sure that I got $$f (0) = 0$$ correctly.

Thank you!

• math.stackexchange.com/questions/3545978/… Feb 14, 2020 at 17:05
• @Servaes I decided to delete it to avoid involuntary discussions.You said "Your solution is incorrect". But you didn't show any arguments. And you claimed that you found the correct solution with $1$ line. You even presented it as proof. (But, later you deleted the your comment) Feb 14, 2020 at 17:14
• @mathworker21 I cannot understand your question well. Feb 16, 2020 at 22:59
• @lonestudent ah sorry. I did not read the whole solution Feb 16, 2020 at 23:01
• @mathworker21 okay, No Problem, My English is not good. I use a little bit of my own knowledge and a bit of google translate. Sorry for wrong words . Feb 16, 2020 at 23:04

I will do some comments on your redaction.

Case 1. $$f(0)>0$$

We have,

$$f(0)\leq-xf(x)+f(f(x))$$

$$f(x)\leq xf(0)+f(f(0))$$

Let $$x\to -\infty$$ we get from $$f(x)\leq xf(0)+f(f(0)) \Longrightarrow \lim_{x\to -\infty}f(x)=-\infty$$.

For a good redaction, don't mix $$\Rightarrow$$ with a french sentence.

Then applying $$\lim_{x\to -\infty}f(x)=-\infty$$, from $$f(0)\leq-xf(x)+f(f(x))$$ we get $$f(0) \longrightarrow-\infty.$$ So, this is a contradiction.

Edit : OK. Precise clearly that the two terms tends to $$- \infty$$ to avoid fastidious verifications to the reader.

Case 2. $$f(0)<0$$

We have, from $$f(x)\leq xf(0)+f(f(0))$$ we deduce $$\lim_{x\to +\infty}f(x)=-\infty.$$ Suppose that, $$\lim_{x\to -\infty}f(x)=+\infty.$$ Applying $$x\to-\infty$$ from $$f(x-1) ≤ -f(x) + f(f(x))$$ we have $$\lim_{x\to -\infty}f(x+(-1)) \longrightarrow +\infty$$.

You meant : $$\lim_{x\to -\infty}f(x+(-1)) \rightarrow +\infty$$ according to your assumption.

But, $$\lim_{x\to -\infty} (-f(x) + f(f(x)))=-\infty$$. According our assumption, we applied $$\lim_{x\to +\infty}f(x)=-\infty.$$ So, this is a contradiction.

Ok since $$\lim_{x\to -\infty} f(f(x)) = - \infty$$.

Suppose that, $$\lim \inf_{x\to -\infty}f(x)=a$$ and $$\lim \sup_{x\to -\infty}f(x)=b$$, where $$a,b\in\mathbb{R}$$

Ok. (A priori, $$a, b \in \mathbb{R} \cup \{ - \infty \}$$ but you deal with this after ) EDIT : to be more precise, $$a \in \mathbb{R} \cup \{ - \infty \}$$ and $$b\in \mathbb{R} \cup \{ - \infty, +\infty \}$$ ; you forgot the case $$b = +\infty$$ in your reasoning.

and for any $$\lambda \in [a,b]$$ we have $$\lambda\leq y\lambda+f(\lambda)$$.

This argument is interesting but problematic. I believe you have taken a sequence $$x_n \rightarrow - \infty$$ such that $$f(x_n)$$ tends to $$\lambda$$. This kind of argument is possible only if $$f$$ is supposed continuous (intuitively its graphe oscillates continuously between a and b). Furthermore, you cannot have a control on $$f(x_n +y)$$ while doing this (it might be improved by replacing $$y$$ by $$y_n$$). Finally, since again f is not supposed continuous, the behaviour of $$f(f(x_n))$$ might be chaotic and not converge at all to $$f(\lambda)$$.

If $$f$$ is supposed continuous, it is possible to make a (rigorous) proof. (X)

[[ EDIT : I said you needed the continuity for the first step because you have taken "any $$\lambda \in [a, b]$$". I think things will be clearer if I present the argument.

If you have continuity. You have two sequences $$(a_n)$$ and $$(b_n)$$ tending to $$-\infty$$ such that $$\lim_{n\rightarrow \infty} f(a_n) = a, \lim_{n \rightarrow \infty} f(b_n) = b$$ By the intermediate value theorem, you can find a sequence $$(x_n)$$ tending to $$-\infty$$, such that $$f(x_n) \rightarrow \lambda$$, and also (a little more technical) a sequence $$(y_n)$$ with $$sup (y_n) = +\infty$$, $$inf (y_n) = -\infty$$ such that $$f(x_n + y_n) \rightarrow \lambda$$.

Let us suppose $$\lambda \neq 0$$.

Looking at the inequality : $$f(x_n + y_n) \leq y_n f(x_n) + f(f(x_n))$$

You have a limit for the left term, but the right term cannot be minorated : contradiction.

Remark : If $$\lambda = 0$$ the argument not applies. So you have a problem if $$a = b = 0$$.

If you don't have continuity.

I recall some properties of the lim inf : $$\lim \inf_{x \rightarrow - \infty} f(x+y) = \lim \inf_{x \rightarrow - \infty} f(x)$$ $$\lim \inf_{x \rightarrow - \infty} a f(x) = a \lim \inf_{x \rightarrow - \infty} f(x) \text{ if } a \geq 0$$ $$\lim \inf_{x \rightarrow - \infty} a f(x) = a \lim \sup_{x \rightarrow - \infty} f(x) \text{ if } a \leq 0$$ $$\lim \inf_{x \rightarrow - \infty} f(x) + \lim \inf_{x \rightarrow - \infty} g(x) \leq \lim \inf_{x \rightarrow - \infty} f(x) + g(x) \leq \lim \inf_{x \rightarrow - \infty} f(x) + \lim \sup_{x \rightarrow - \infty} g(x)$$ Each inequality here might be strict.

Take the lim inf $$x\rightarrow - \infty$$ in the inequality $$f(x+y) \leq y f(x) + f(f(x))$$ to get : $$a \leq ay + \lim \sup_{x \rightarrow - \infty} f(f(x)) \text{ for } y \geq 0$$ $$a \leq ay + \lim \inf_{x \rightarrow - \infty} f(f(x)) \text{ for } y \leq 0$$

So if you suppose $$\lim \sup_{x \rightarrow - \infty} f(f(x)) < +\infty$$ (which implies $$\lim \inf_{x \rightarrow - \infty} f(f(x)) < +\infty$$) you get a contradiction as soon as $$a \neq 0$$.

Take again the lim inf $$x\rightarrow - \infty$$ in the inequality $$f(x+y) \leq y f(x) + f(f(x))$$, but use this time $$\lim \inf u(x) + v(x) \leq \lim \sup u(x) + \lim \inf v(x)$$ to get :

$$a \leq by + \lim \inf_{x \rightarrow - \infty} f(f(x)) \text{ for } y \geq 0$$ $$a \leq by + \lim \sup_{x \rightarrow - \infty} f(f(x)) \text{ for } y \leq 0$$

With the same hypothesis $$\lim \sup_{x \rightarrow - \infty} f(f(x)) < + \infty$$, you get a contradiction as soon as $$b \neq 0$$.

It seems you need the assumption $$\lim \sup_{x \rightarrow - \infty} f(f(x)) < + \infty$$ to get something with your argument. ]]

For any $$\lambda$$ we can always choose a finite $$y$$ such that, where we get $$\lambda\ > y\lambda+f(\lambda)$$ which gives a contradiction. So, we deduce that $$\lim_{x\to -\infty}f(x)=-\infty$$.

Ok, since the case $$a = - \infty$$, $$b \neq - \infty$$ can be covered by the preceding argument (you should have mentionned it).

Then, applying $$x\to-\infty$$ from $$f(0)\leq-xf(x)+f(f(x))$$ we get $$f(0)\longrightarrow -\infty$$. But, this contradicts with $$f : \mathbb R → \mathbb R$$.

So, we can deduce that $$f(0)=0$$.

For (X), you need to suppose $$f$$ continuous to make a rigorous proof (do it ! ). I must say your redaction looked messy because you did'nt skip lines. There is really little effort to do to improve this.

UPDATE :

Case 2. $$f(0)<0$$ (with the correct argument)

We have, from $$f(x)\leq xf(0)+f(f(0))$$ we deduce $$\lim_{x\to +\infty}f(x)=-\infty.$$ From $$f(x + y) \leq yf(x) + f(f(x))$$ we have:

$$\begin{cases} f(x)\leq f(f(x)) \\ f(x) \leq xf(0)+ f(f(0)) \end{cases} \Longrightarrow f(x)\leq f(x)f(0)+f(f(0)) \Longrightarrow f(x)(1-f(0))\leq f(f(0))$$.

Correct.

Then applying $$x=f(0)$$, we get $$f(f(0))\leq 0$$, which imply $$f(x)\leq 0$$, which gives $$f(f(x))\leq 0$$.

Nice.

In this case, we have $$f(x)<0.$$

It you be nice to add quantifiers. I think you mean : for all $$x \in \mathbb{R}$$.

Because, if $$f(x)=0$$, from $$f(x)\leq f(f(x))$$, we get $$f(0)\geq 0$$,

Be more precise : "if $$f(x) = 0$$ for some $$x \in \mathbb{R}$$". Ok for the argument.

which gives a contradiction. Applying $$x\to-\infty$$ from $$f(0)\leq-xf(x)+f(f(x))$$ we get $$f(0)\longrightarrow -\infty$$. Again a contradiction.

Are you supposing $$-xf(x) \rightarrow - \infty$$ ? It seems not to be necessarily the case (e.g. $$f(x) = - \exp(-x)$$) (XX)

So, we can deduce that $$f(0)=0$$.

Q.E.D.

You have to check (XX).

UPDATE 2 : (knowing $$f < 0$$) :

Then, from $$f(x + y) \leq yf(x) + f(f(x))$$ we have:

$$f(z)\leq(z-x)f(x)+f(f(x)) \Longrightarrow f(x) \leq (x-y)f(y)+f(f(y))\Longrightarrow 0\leq(f(y)-y)f(y) \Longrightarrow f(x)(f(x)-x)\geq 0 \Longrightarrow f(x) \leq x$$

Great. This enables to conclude indeed. Good job.

• in case 1, you say $-xf(x) > 0$; that's wrong Feb 17, 2020 at 4:25
• Corrected. I think you have downvoted my answer. Does it deserve a negative vote ?
– Dlem
Feb 17, 2020 at 4:37
• @lonestudent Without continuity, if $\lim_{x\to -\infty} f(x) = \lambda\in \mathbb{R}$, how to obtain $\lim_{x\to -\infty} f(f(x)) = f(\lambda)$? Feb 17, 2020 at 10:16
• You say in your answer $\liminf_{x\to-\infty}f(x+y) \neq \liminf_{x\to-\infty}f(x)$, where $\liminf_{x\to-\infty}f(x)=a.$ Do I understand correct? Are you sure this is related to the continuity of the function? Feb 20, 2020 at 8:10
• What do you think about math.stackexchange.com/questions/52447/… As I understnad The Solver prove that $f$ is continuous. Feb 20, 2020 at 10:03