The Problem: 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 $2011$ , Pr: $3$)

Here is the solution:

Before we prove that $f (x) = 0$ for all $x ≤ 0$, we will prove that $f (0) = 0.$

$\color{black}{\large\text{Case} \thinspace 1:}$ $f(0)>0$

Let, $\varepsilon >\dfrac{f(f(0))}{f(0)}$ and $x<- \dfrac{\varepsilon+f(f(0))}{f(0)}$ we have, from $f(x)\leq xf(0)+f(f(0))$, $f(x)<-\varepsilon.$ Then from, $f(f(x))\leq f(x)f(0)+f(f(0))$ we have $f(f(x))<0$. Finally, since $0<f(0)\leq f(f(0))$ ,we get $f(0)\leq-xf(x)+f(f(x))<0$ which imply $f(0)<0$, which gives a contradiction.

$\color{black}{\large\text{Case} \thinspace 2:}$ $-\alpha \leq f(0)<0$, where $\alpha\in\mathbb{R^+}.$

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 $f(x)=0$ at least for some $x$, from $f(x)\leq f(f(x))$, we get $f(0)\geq 0$, which gives a contradiction.

Then let, $x<-\sqrt {\alpha}$, from $f(x + y) \leq yf(x) + f(f(x))$ we have:

$$\begin{align} f(z)\leq(z-x)f(x)+f(f(x)) \Longrightarrow f(f(x)) \leq (f(x)-x)f(x)+f(f(x))\Longrightarrow f(x)(f(x)-x)\geq 0 \Longrightarrow f(x) \leq x <-\sqrt {\alpha} \Longrightarrow f(x)< -\sqrt{\alpha}.\end{align}$$

Finally, from $f(0)\leq-xf(x)+f(f(x))$ we get $f(0)<-\alpha$, which gives a contradiction.

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

Then, applying $f(0)=0$, from $f(0)\leq-xf(x)+f(f(x))$ and $f(x) \leq xf(0)+ f(f(0))$ we have,

$$\begin{cases} xf(x)-f(f(x)) \leq 0 \\ f(x) \leq 0 \\ f(f(x)) \leq 0 \end{cases} \Longrightarrow \begin{cases} xf(x)\leq0 \\ f(x)\leq 0 \end{cases} \Longrightarrow \begin{cases} xf(x)\leq0, x\in\mathbb {R} \\ xf(x)\geq 0, x\leq 0 \end{cases} \Longrightarrow xf(x)=0, x\leq 0 \Longrightarrow \color{blue} {\boxed{ f(x)=0, \text{for all} \thinspace x\leq 0.}}$$

End of the Proof.

Can you verify this solution? Are there any mistakes, gaps, etc.?

Thank you very much!

  • 2
    $\begingroup$ That is all good. The proof is quite tricky so don't hesitate to add a little bit more details between each steps (eg, "using the definition of $f$", "replacing $z$ with $f(x)$", etc...). It can be useful to give names to the meaningful equations for this purpose. Also, in step $2$, when you deduce $f(f(x))\leq 0$ from $f(x)\leq 0$, it seems to me that you are replacing $x$ with $f(x)$. The argument is correct, however in the above you only worked with $x<-\sqrt{\alpha}$. We may not have $f(x)\leq -\sqrt{\alpha}$ though. $\endgroup$ – Suzet Feb 25 '20 at 13:55
  • $\begingroup$ Actually for the first part of Case 2, you only need $x\leq 0$, and since we also have $f(x)\leq 0$, then we can replace $x$ by $f(x)$ and obtain the desired $f(f(x))$ without problem. $\endgroup$ – Suzet Feb 25 '20 at 13:55
  • $\begingroup$ @suzet Thank you for comment. I have difficulty understanding your last sentence.If $x<-\sqrt{\alpha}$ we have $f(x) <-\sqrt{\alpha}$ according our assumption. This is independent from $f(f(x)) \leq 0$ $\endgroup$ – lone student Feb 25 '20 at 14:19
  • $\begingroup$ @Suzet If I just apply $x \leq 0$, I can't finally get a contradiction. $\endgroup$ – lone student Feb 25 '20 at 14:39
  • 2
    $\begingroup$ Aaaah, I just got it, alright. Nevermind, it works fine ! $\endgroup$ – Suzet Feb 25 '20 at 14:58

I had trouble following your proof because it wasn't always clear what you were doing as you skipped some steps. Also, you sometimes introduced certain values or limitations well before they were used, making it seem they are needed earlier when they're not. In addition, there were some statements using $x$ which were only applicable to certain values or ranges instead of all of them, which could also be somewhat confusing. Nonetheless, although your proof might not be particularly pretty or elegant, it seems to all be correct.

I'm not sure what you consider to be a "canonical" answer, but I thought I would basically repeat your proof method here, filling in some details to possibly help make it easier for other people to follow along & also confirm it works.

First, the relation you're provided with is that

$$f(x + y) \leq yf(x) + f(f(x)) \tag{1}\label{eq1A}$$

for all real $x$ and $y$. You are trying to prove that $f(x) = 0$ for all $x \le 0$.

You are first trying to prove that $f(0) = 0$ by contradiction, where you assume it's either positive or negative, in the $2$ cases below.

$\color{black}{\large\text{Case} \thinspace 1:}$ $f(0) \gt 0$

First, have $\varepsilon$ be any real number satisfying

$$\varepsilon \gt \frac{f(f(0))}{f(0)} \implies \varepsilon f(0) \gt f(f(0)) \tag{2}\label{eq2A}$$

Next, consider a restricted region of $x_1$ where

$$x_1 \lt -\frac{\varepsilon + f(f(0))}{f(0)} \implies x_1 f(0) \lt -\varepsilon - f(f(0)) \tag{3}\label{eq3A}$$

Choosing $x = 0$ and $y = x$ in \eqref{eq1A} gives

$$f(x) \leq xf(0) + f(f(0)) \tag{4}\label{eq4A}$$

Using \eqref{eq3A} in \eqref{eq4A} with $x = x_1$ gives

$$f(x_1) \lt (-\varepsilon - f(f(0))) + f(f(0)) = -\varepsilon \tag{5}\label{eq5A}$$

Choosing $x = 0$ and $y = f(x)$ in \eqref{eq1A} gives

$$f(f(x)) \leq f(x)f(0) + f(f(0)) \tag{6}\label{eq6A}$$

Multiplying both sides of \eqref{eq5A} by $f(0)$, and using \eqref{eq2A}, gives

$$f(x_1)f(0) \lt -\varepsilon f(0) \lt -f(f(0)) \implies f(x_1)f(0) + f(f(0)) \lt 0 \tag{7}\label{eq7A}$$

Using this in \eqref{eq6A} with $x = x_1$ gives

$$f(f(x_1)) \lt 0 \tag{8}\label{eq8A}$$

From this case's assumption that $f(0) \gt 0$, plus using $x = y = 0$ in \eqref{eq1A} gives

$$0 \lt f(0) \le f(f(0)) \tag{9}\label{eq9A}$$

Leaving $x$ as is and using $y = -x$ in \eqref{eq1A} gives

$$f(0) \leq -xf(x) + f(f(x)) \tag{10}\label{eq10A}$$

From \eqref{eq9A}, you get in \eqref{eq2A} that $\varepsilon \gt 0$. Thus, from \eqref{eq5A}, you have $f(x_1) \lt 0$. Also, from \eqref{eq3A}, you have $x_1 \lt 0$. This means $x_1f(x_1) \gt 0 \implies -x_1f(x_1) \lt 0$. This, along with \eqref{eq8A} gives in \eqref{eq10A} using $x = x_1$ that

$$f(0) \lt 0 \tag{11}\label{eq11A}$$

This contradicts the assumption for the case, so it shows it can't be true.

$\color{black}{\large\text{Case} \thinspace 2:}$ $-\alpha \lt f(0) \lt 0$, where $\alpha\in\mathbb{R^+}.$

Leaving $x$ as is and using $y = 0$ in \eqref{eq1A} gives

$$f(x) \leq f(f(x)) \tag{12}\label{eq12A}$$

Using $x = f(x)$ in \eqref{eq4A} gives

$$f(f(x)) \leq f(x)f(0) + f(f(0)) \tag{13}\label{eq13A}$$

Using this with \eqref{eq12A} gives

$$f(x) \leq f(x)f(0) + f(f(0)) \implies f(x)(1 - f(0)) \leq f(f(0)) \tag{14}\label{eq14A}$$

Using $x = f(0)$ in \eqref{eq14A} gives

$$\begin{equation}\begin{aligned} f(f(0))(1 - f(0)) & \leq f(f(0)) \\ f(f(0)) - f(0)f(f(0)) & \leq f(f(0)) \\ - f(0)f(f(0)) & \leq 0 \\ f(f(0)) & \leq 0 \end{aligned}\end{equation}\tag{15}\label{eq15A}$$

Using this, along with $1 - f(0) \gt 0$ in \eqref{eq14A} gives

$$f(x) \leq 0 \implies f(f(x)) \leq 0 \tag{16}\label{eq16A}$$

where $x = f(x)$ was used to get the implied part. Note if $f(x) = 0$ for any $x$, using that in \eqref{eq12A} gives $0 \le f(0)$, which can't be, so \eqref{eq16A} shows $f(x) \lt 0$.

In \eqref{eq1A}, keep $x$ as is and have $y = z - x$ to get

$$f(z) \leq (z - x)f(x) + f(f(x)) \tag{17}\label{eq17A}$$

Have $z = f(x)$ to get

$$f(f(x)) \leq (f(x) - x)f(x) + f(f(x)) \implies (f(x) - x)f(x) \geq 0 \tag{18}\label{eq18A}$$

Since $f(x) \lt 0$, this means

$$f(x) - x \le 0 \implies f(x) \le x \tag{19}\label{eq19A}$$

If now consider the restricted region of

$$x_2 \lt -\sqrt{\alpha} \tag{20}\label{eq20A}$$

then with $x = x_2$, \eqref{eq19A} becomes

$$f(x_2) \le x_2 \lt -\sqrt{\alpha} \tag{21}\label{eq21A}$$

From \eqref{eq20A} and \eqref{eq21A}, you get

$$x_2f(x_2) \gt \alpha \implies -x_2f(x_2) \lt -\alpha \tag{22}\label{eq22A}$$

Using this, along with $x = x_2$, in \eqref{eq16A} in \eqref{eq10A} gives $f(0) \lt -\alpha$, which contradicts this case's assumption. Since $\alpha$ can be any positive real value, this shows there can't be any lower bound on the value of $f(0)$, i.e., it can't be negative.

Since neither case $1$ or $2$ are true, this means that $f(0) = 0$. Using this with \eqref{eq10A}, along with \eqref{eq16A}, gives

$$0 \leq -xf(x) + f(f(x)) \implies xf(x) \leq f(f(x)) \leq 0 \tag{23}\label{eq23A}$$

From \eqref{eq4A} and \eqref{eq15A}, you get

$$f(x) \leq f(f(0)) \leq 0 \tag{24}\label{eq24A}$$

From \eqref{eq23A}, for $x \leq 0$, you get

$$f(x) \geq 0 \tag{25}\label{eq25A}$$

This, together with \eqref{eq24A}, gives that

$$f(x) = 0, \; \forall \; x \leq 0 \tag{26}\label{eq26A}$$

  • 1
    $\begingroup$ @lonestudent You are welcome for my time, and thank you for the compliment. Yes, this really your proof, basically with the missing details filled in. I believe it should be easier now for anybody, including you as you've indicated, to follow along with what is going on. Note I have found for myself that if I don't include enough details in what I'm trying to prove, then sometime later even I myself often have trouble reading my own proofs. This is why I often prefer to ensure I add in many of the details even if the proof is only for my own use. $\endgroup$ – John Omielan Feb 25 '20 at 18:04
  • 1
    $\begingroup$ @lonestudent The one thing I'm not completely sure of is whether or not there's a considerably shorter way (e.g., use a theorem somebody has proven before) to prove this. However, with only the one relation given and nothing else (e.g., continuity, differentiability, boundedness, etc.) about the function, I doubt there's anything much, if any, shorter or simpler than what you did. Congratulations on solving it! $\endgroup$ – John Omielan Feb 25 '20 at 18:08
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    $\begingroup$ @lonestudent People, myself included, generally like short, elegant proofs. However, although they (including myself) don't like too long proofs, I at least would prefer a somewhat longer proof I can easily follow than a very terse proof where I have to fill in many of the missing steps myself. To be honest, I saw & started to read your proof shortly after you posted it, but found it was too difficult for me to follow at the time, so I put it aside to possibly look at again later. When I saw you had a bounty, I tried reading it again, this time finishing & being able to confirm it worked. $\endgroup$ – John Omielan Feb 25 '20 at 18:21
  • 1
    $\begingroup$ @lonestudent One other thing is about getting a downvote. Please don't read too much into that. I've got downvotes a few times for no apparent reason, sometimes for something I did months earlier. Although the answers have generally been longer proofs, it's also happened for shorter ones. I don't know why the person downvoted as I'm quite certain they're correct. Thus, don't try to guess the motive of the down-voters since, unless they leave a comment or you find a mistake that they perhaps noticed, you can't really tell what the reason for the down-vote is. $\endgroup$ – John Omielan Feb 25 '20 at 18:27
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    $\begingroup$ @lonestudent Thanks for checking & noticing. I made a mistake there. It should have been $f(x) - x \le 0$, which I've now fixed. However, it still does give that $f(x_2) \lt -\sqrt{\alpha}$. Thus, along with $x \lt -\sqrt{\alpha}$, you have $-xf(x) \lt -\alpha$. This still implies that $f(0) \lt -\alpha$ because since $f(f(x_2)) \leq 0$, this means $f(0) \leq -x_2f(x_2) + f(f(x_2)) \lt -\alpha \implies f(0) \lt -\alpha$. Of course, this means $f(0) \le -\alpha$ is obviously also true. $\endgroup$ – John Omielan Feb 25 '20 at 19:45

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