Let $f$ be any function that maps the set of real numbers into the set of real numbers. Prove that there exist real numbers $x$ and $y$ such that \[f\left(x-f(y)\right)>yf(x)+x\] Proposed by Igor Voronovich, Belarus
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Tags: function, inequalities, algebra, Functional inequality, IMO Shortlist
09.07.2010 11:49
I didn't like it too much, spent a couple hours just writing inequality after inequality until I found something that satisfied me as an adequate solution, so somebody please post a better one! As "expected" with this type of problem statement, we will assume that a function $f$ exists such that $f(x-f(y))\leq yf(x)+x$ for any two reals $x,y$, and reach a contradiction. We start with a few partial results:
Now we reach the following contradiction: $f(1)$ cannot exist:
15.07.2010 03:26
daniel73 wrote: I didn't like it too much, spent a couple hours just writing inequality after inequality until I found something that satisfied me as an adequate solution, so somebody please post a better one! After solving the problem myself, grading solutions found by US IMO team members and candidates on a mock test, and looking at the official solution in the shortlist packet, I have found no reason to believe an elegant solution or even one more motivated than "stumble around until a contradiction pops up" exists. I guess it's a little surprising that an innocuous looking inequality has no solutions, but as a solve this is maybe the worst shortlist problem I've done. Relatively speaking, your solution looks fairly slick to me.
15.07.2010 06:18
MellowMelon wrote: daniel73 wrote: I didn't like it too much, spent a couple hours just writing inequality after inequality until I found something that satisfied me as an adequate solution, so somebody please post a better one! After solving the problem myself, grading solutions found by US IMO team members and candidates on a mock test, and looking at the official solution in the shortlist packet, I have found no reason to believe an elegant solution or even one more motivated than "stumble around until a contradiction pops up" exists. I guess it's a little surprising that an innocuous looking inequality has no solutions, but as a solve this is maybe the worst shortlist problem I've done. Relatively speaking, your solution looks fairly slick to me. Thanks, Mellon, but what I did was exactly what you said, "stumble around (in my case for about four pages of scrap calculations) until a contradiction pops up"... I just hoped that a shortlisted problem would have a better solution...
11.08.2010 05:27
Suppose for the sake of contradiction that $f(x-f(y)) \leq yf(x) + x$ for all reals $x,y$. Replacing $x$ with $x+f(y)$ gives $f(x) \leq yf(x+f(y)) + x+f(y)$, and setting $y=0$ gives $f(x) \leq x + f(0)$. Consequently, if $x \leq -f(0)$, then $f(x) \leq 0$. If $s>0 \geq f(s)$, set $r = -1-f(0)$ to get $f(r) \leq sf(r+f(s)) + r + f(s) \leq s(r+f(s)+f(0)) + r \leq -s + r$. Hence, $s \leq r - f(r)$, so if $c = \max(r - f(r) + 1,1)$, $f(x) > 0$ whenever $x \geq c$. Setting $y = \min(-f(0),-c/f(c))$ and $x = c - f(y) \geq c$ gives $0 < f(x) \leq yf(c) + c \leq 0$, which is a contradiction.
11.08.2010 05:58
Zhero wrote: Suppose for the sake of contradiction that $f(x-f(y)) \leq yf(x) + x$ for all reals $x,y$. Replacing $x$ with $x+f(y)$ gives $f(x) \leq yf(x+f(y)) + x+f(y)$, and setting $y=0$ gives $f(x) \leq x + f(0)$. Call a real number $y$ weird if $y>0 \geq f(y)$. Let $x = -1-f(0)$. For any weird $y$, $f(x) \leq yf(x+f(y)) + x + f(y) \leq y(x+f(y)+f(0)) + x + f(y) \leq -y + x$. Hence, $y \leq x - f(x)$, so the set of weird $y$'s is bounded from above. Therefore, if $c = \max(x - f(x) + 1,1)$, $c > 0$ and $f(x) > 0$ whenever $x \geq c$. Setting $y = \min(-f(0),-c/f(c))$ and $x = c - f(y) \geq c$ (since $f(y) \leq -f(0) + f(0)$) gives $0 < f(x) \leq yf(c) + c \leq 0$, which is a contradiction.
I think the easy way to find the motivation for this is to ask why this is true when f(x) = x.
13.03.2011 16:05
daniel73 wrote: Finally, if $y>0$, then $f(0)+f(y)\geq f(f(y))\geq-1$, or $f(y)\geq-1-f(0)$ for all $y>0$. Now we reach the following contradiction: $f(1)$ cannot exist: Let $y=1$, $x=f(1)$, then $f(0)\leq f(f(1))+f(1)$, or $f(f(1))\leq f(1)+f(0)\leq f(f(1))+2f(1)$, and $f(1)\geq0$.[\quote] Can anyone explain this two sentences?
13.03.2011 20:06
Zhero wrote: Suppose for the sake of contradiction that $f(x-f(y)) \leq yf(x) + x$ for all reals $x,y$. Replacing $x$ with $x+f(y)$ gives $f(x) \leq yf(x+f(y)) + x+f(y)$, and setting $y=0$ gives $f(x) \leq x + f(0)$. Consequently, if $x \leq -f(0)$, then $f(x) \leq 0$. If $s>0 \geq f(s)$, set $r = -1-f(0)$ to get $f(r) \leq sf(r+f(s)) + r + f(s) \leq s(r+f(s)+f(0)) + r \leq -s + r$. Hence, $s \leq r - f(r)$, so if $c = \max(r - f(r) + 1,1)$, $f(x) > 0$ whenever $x \geq c$. Setting $y = \min(-f(0),-c/f(c))$ and $x = c - f(y) \geq c$ gives $0 < f(x) \leq yf(c) + c \leq 0$, which is a contradiction.
Are you sure $c = \max(r - f(r) + 1,1)$ is finite?
14.03.2011 05:46
Yes, since $r = -1 - f(0)$. My apologies if the original solution is unclear; I was attempting to cram it into as little space as possible (to contradict MellowMelon. ) I'll rewrite my solution more comprehensibly below: Suppose for the sake of contradiction that $f(x-f(y)) \leq yf(x) + x$ for all reals $x,y$. Replacing $x$ with $x+f(y)$ gives $f(x) \leq yf(x+f(y)) + x+f(y)$, and setting $y=0$ gives $f(x) \leq x + f(0)$. Consequently, if $x \leq -f(0)$, then $f(x) \leq 0$. We will show that $f(u)$ is positive for all sufficiently large $u$. Let $r = -1 - f(0)$. We claim that if $s > 0 \geq f(s)$, then $s \leq r - f(r)$. Substituting $x=r$ and $y = f(s)$ into $f(x) \leq y f(x+f(y)) + x + f(y)$ gives $f(r) \leq sf(r+f(s)) + r + f(s) \leq s(r+f(s)+f(0)) + r \leq s(r + f(0)) + r \leq -s + r$, which follows from $s > 0$ and $f(s) < 0$, so $s$ is indeed at most $r - f(r)$. Therefore, if we take some $c$ larger than both $r - f(r)$ and $0$, we must have $f(x) > 0$ whenever $x \geq c$. Take some $y$ smaller than $-f(0)$ and $-\frac{c}{f(c)}$ and set $x =c - f(y)$. Since $y \leq -f(0)$, $f(y) \leq 0$, so $x \geq c$. Hence, $0 < f(x)$. Returning to the original functional inequality, we have $f(x) = f(c - f(y)) \leq yf(c) + c \leq 0$, so $0 < 0$, contradiction. Edit: Fixed a typo.
14.03.2011 06:36
Zhero wrote: Yes, since $r = -1 - f(0)$. My apologies if the original solution is unclear; I was attempting to cram it into as little space as possible (to contradict MellowMelon. ) I'll rewrite my solution more comprehensibly below: Suppose for the sake of contradiction that $f(x-f(y)) \leq yf(x) + x$ for all reals $x,y$. Replacing $x$ with $x+f(y)$ gives $f(x) \leq yf(x+f(y)) + x+f(y)$, and setting $y=0$ gives $f(x) \leq x + f(0)$. Consequently, if $x \leq -f(0)$, then $f(x) \leq 0$. We will show that $f(u)$ is positive for all sufficiently large $u$. Let $r = -1 - f(0)$. We claim that if $s > 0 \geq f(s)$, then $s \leq r - f(r)$. Substituting $x=r$ and $y = f(s)$ into $f(x) \leq y f(x+f(y)) + x + f(y)$ gives $f(r) \leq sf(r+f(s)) + r + f(s) \leq s(r+f(s)+f(0)) + r \leq s(r + f(0)) + r \leq -s + r$, which follows from $s > 0$ and $f(s) < 0$, so $s$ is indeed at most $r - f(r)$. Therefore, if we take some $c$ larger than both $r - f(r)$ and $0$, we must have $f(x) > 0$ whenever $x \geq c$. Take some $y$ larger than $-f(0)$ and $-\frac{c}{f(c)}$ and set $x =c - f(y)$. Since $y \geq -f(0)$, $f(y) \leq 0$, so $x \geq c$. Hence, $0 < f(x)$. Returning to the original functional inequality, we have $f(x) = f(c - f(y)) \leq yf(c) + c \leq 0$, so $0 < 0$, contradiction. I only have problem with the last one: $yf(c) + c \leq 0$$ You told $y\ge -\frac{c}{f(c)}$ $yf(c) + c\ge -c+c=0$ hence I didn't have $<0$ but $>0$ Or don't I understand?
14.03.2011 20:40
No, you're right. I meant to take a $y$ smaller than both $-f(0)$ and $\frac{-c}{f(c)}$. I fixed this in my post above. Hope it works now.
05.04.2014 13:24
The problem is for all reals $x,y$ \[f(x-f(y))\le yf(x)+x\] Plugging $y=0$ \[f(x-f(0))\le x\implies f(x)\le x+f(0)\] Plugging $x=f(y)$ \[f(0)\le yf(f(y))+f(y)\le y(y+2f(0))+(y+f(0))\implies y(y+2f(0)+1)\ge 0\] As $y$ is any real number, hence $f(0)=-1/2$. Thus $f(x)\le x-\dfrac12$. Notice that $f(x) < 0$ for all $x<1/2$. We assume there is $x>0$ such that $f(x)\ge 0$ Take some $x\ge 0$ and $y\le 0$ such that $f(x)\ge 0$ , $x-f(y)>0$ , $f(x-f(y))\ge 0$ and $x+yf(x) < 0$ Thus $0>x+yf(x)\ge f(x-f(y))\ge 0$. Hence for any $x\ge 0$ and $y\le 0$ we have $f(x-f(y))<0$ or $x+f(x)y\ge 0$ but not both. We get $f(x-f(y))(x+f(x)y)\ge 0$. Plugging $y=0$ \[f\left(x+\dfrac12\right)x\ge 0\] Thus $f(x)\ge 0$ for $x>\dfrac12$. Take a pair of $x>0$ and $y<0$ such that $x-f(y)\ge\dfrac12$ Hence $yf(x)+x\ge 0\implies y\left(x-\dfrac12\right)+x\ge 0\implies (y+1)\left(x-\dfrac12\right)\ge\dfrac12$. But this can't be true for all such $x,y$. Hence we have a contradiction. Thus $f(x)\le 0$ for all $x$. Again take $x>0$ and $y<0$ such that $f(x-f(y))-x>0$. Then we get $f(x)\ge 0$. Thus $f(x-f(y))\le x\implies f(x)\le x+f(y)$. Therefore $f(x)\le x+y-\dfrac12$. Hence we arrive at a contradiction and no such $f(x)$ exists.
16.06.2015 23:42
Assume the contrary, that $f(x-f(y))\leq yf(x)+x$ for any $x, y\in\mathbb{R}$. Let $P(x, y)$ denote this assertion. $P(x+f(0), 0)$ gives $$f(x)\leq x+f(0)$$ for any $x\in\mathbb{R}$. Now we prove the following double claim: Claim: We have (i) $\lim_{y\rightarrow -\infty}f(y)=-\infty$; (ii) $\lim_{y\rightarrow +\infty}f(y)=+\infty$. Proof: Part (i) is an immediate corollary of $f(y)\leq y+f(0)$, so let's focus on the latter. Let $y\in\mathbb{R}^+$; $P(x+f(y), y)$ yields $$f(x)\leq yf(x+f(y))+x+f(y)\leq y(x+f(y)+f(0))+x+f(y)=(y+1)x+(y+1)f(y)+yf(0)$$ for every real $x$. Our aim is to prove that for any real $t$ there is an $y_0$ such that for any $y\geq y_0$, $f(y)\geq t$. If we can find arbitrarily large $y$ such that $f(y)<t$, then for arbitrarily large $y$ we have $$(y+1)f(y)+yf(0)\leq (y+1)t+yf(0)\leq (y+1)(t+|f(0)|)$$ and, by fixing $x<-t-|f(0)|$, for arbitrarily large $y$ we have the inequality $$f(x)\leq (y+1)(x+t+|f(0)|)$$ however the last factor is negative, implying that this expression may attain arbitrarily small values when we increase $y$. This is a contradiction. $\square$ Now it's easy to devise a contradiction. Choose $x$ such that $f(x)>0$ (it trivially exists by our claim). Then $\lim_{y\rightarrow -\infty}(yf(x)+x)=-\infty$, implying that $\lim_{y\rightarrow -\infty}f(x-f(y))=-\infty$. But $\lim_{y\rightarrow -\infty}(x-f(y))=+\infty$, therefore $\lim_{y\rightarrow -\infty}f(x-f(y))=+\infty$. This is a contradiction, establishing the problem statement.
19.03.2016 01:51
My Solution: Assume the opposite for the sake of contradiction. Let $I(x,y)$ be assertion of the $f(x-f(y))\le yf(x)+x$ $\forall x,y\in \mathbb R$. $I(x,0)$ implies $f(x-f(0))\le x$, substituting $x'=x-f(0)$ we get $f(x)\le x+f(0)$ -- $(1)$ $I(f(y),y)$ implies $f(0)\le yf(f(y))+f(y)$ --$(2)$implies $0\le yf(f(y))+f(y)-f(0)\le yf(f(y))+y$ implies $f(f(y))\le -1$ for $y<0$--$(3)$ $(3)+(1)$ implies $-1\le f(f(y))\le f(y)+f(0)\le y+2f(0)$ for $y>0$ implies $f(0)\ge -\frac{1}{2}$--$(4)$. Now we have $yf(f(y))+f(y)\ge f(0)\ge -\frac{1}{2}$ and for $y<0$ $f(y)\ge -\frac{1}{2}-yf(f(y))\ge y-\frac{1}{2}$.--$(5)$ Now from $(2)+(1)$ we get $f(0)\le yf(f(y))+f(y)\le y(y+2f(0))+y+f(0)$ which is equvalent to $0\le y(y+2f(0)+1)$ $\implies$ $f(0)=-\frac{1}{2}$. From $(1)$ : $f(y)\le y+f(0)=y-\frac{1}{2}$ but combined with $(5)$ implies $f(y)=y-\frac{1}{2}$. Now we have $I(x,y)$ equivalent to $f(x-f(y))=x-f(y)-\frac{1}{2}=x-y\le yf(x)+x$ $0\le y(f(x)+1)$ which is in contradiction with $I(1,-1)$.
03.06.2017 02:38
It's a nice problem in my opinion. It's sad though, as there is no elegant proof without contradiction. IMO SL 2009 A5 wrote: Let $f$ be any function that maps the set of real numbers into the set of real numbers. Prove that there exist real numbers $x$ and $y$ such that \[f\left(x-f(y)\right)>yf(x)+x\] Proposed by Igor Voronovich, Belarus Solution: For the sake of contradiction, suppose there exists a function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying $$f(x-f(y)) \le yf(x)+x,$$for all $x, y \in \mathbb{R}$. Denote the above property of $f$ by $P(x, y)$; then, $$P(x+f(0),0) \implies f(x) \le x+f(0),$$and $$P(f(y), y) \implies y\cdot \left(f(f(y))+1\right) \ge 0,$$for all $x, y \in \mathbb{R}$. Call the previous two relations $(1)$ and $(2)$ respectively. Let $C=-(1+f(0))$ and consider the subsequent claim. Claim 1: For all $t \in \mathbb{R}$ we have $t>0 \implies f(t) \ge C$. (Proof) Evidently, $(2)$ implies for all $t>0$ we have $f(f(t)) \ge -1$ so $(1)$ yields $f(t)+f(0) \ge -1 \implies f(t) \ge C$ as stated. $\square$ Now, we define the set $$\mathcal{S} \overset{\text{def}}{:=} \left \{ x>0: f(x)<0 \right \}.$$ Claim 2: $\mathcal{S}$ is bounded above. (Proof) Assume to the contrary; let $s \in \mathcal{S}$. By our assumption, there exists $t \in \mathcal{S}$ such that $$t>\frac{C-s}{f(s)},$$then we have $$f(s-f(t)) \le tf(s)+s<C$$so by the previous claim, we conclude $0>f(t) \ge s>0$, a contradiction! Hence, the claim must hold. $\square$ Thus, there exists an $N>0$ for which $x>N \implies f(x) \ge 0$. Now, we consider the next crucial step; define the set $$\mathcal{P} \overset{\text{def}}{:=} \left \{ x>N: f(x)>0 \right \}.$$ Claim 3: $\mathcal{P}$ is unbounded. (Proof) Else, we have $M>N>0$ such that $x>M \implies f(x)=0$. If there exists $x_0 \in \mathbb{R}$ such that $f(x_0)<0$ then plug in $x=x_0, y>M$ to get $$P(x_0, y) \implies f(x_0)=f(x_0-f(y)) \le yf(x_0)+x_0,$$which fails for $y \rightarrow \infty$. Therefore, $f(x) \ge 0$ always holds. However, $$P(x, y) \implies 0 \le f(x-f(y)) \le yf(x)+x,$$which fails if for any fixed $x$ we have $f(x)>0$; by setting $y \rightarrow -\infty$. Since $f \not \equiv 0$, we get the desired contradiction! $\square$ Finally, we fix an $x$ sufficiently large, so that $f(x)>0$ and take any $y$ with $$y<\frac{C-x}{f(x)},$$to get $$f(x-f(y)) \le yf(x)+x<C \implies f(y) \ge x.$$Notice that this means $\lim_{y \rightarrow -\infty} f(y) \rightarrow \infty$; so let $y \rightarrow -\infty$ in $P(x, y)$. Evidently, for fixed $x$, we have $(x-f(y)) \rightarrow -\infty$ so $$f(x-f(y)) \rightarrow \infty.$$However, $yf(x)+x \rightarrow -\infty$ so we get the required contradiction! $\blacksquare$
29.04.2020 09:31
We go by proof by contradiction: suppose $f(x-f(y)) \le yf(x)+x$ for all $x,y\in \mathbb{R}$. We will prove that there are no solutions. Call this $P(x,y)$. Claim 1: $f(x) \le x-1/2$ for all $x$. Proof: $P(x,0)$ gives $f(x-f(0)) \le x$, so $f(x) \le x+f(0)$. We repeatedly abuse this condition on $P(f(y),y)$ to unwrap $f$'s. We get \begin{align*} f(0) &\le yf(f(y)) + f(y) \\ &\le y(f(y)+f(0)) + (y+f(0)) \\ &\le y(y+2f(0)) + (y+f(0)) \\ &= y^2 + (2f(0)+1)y + f(0) \\ \implies 0&\le y^2+(2f(0)+1)y = (y+f(0)+1/2)^2 - (f(0)+1/2)^2. \end{align*}Plug in $y=-f(0)-1/2$, so $0\le -(f(0)+1/2)^2$. So $f(0)=-1/2$. Hence $f(x) \le x-1/2$; this is an upper bound on $f$. $\square$ Claim 2: $f(y) > y-1/2$ for $y>0$. Proof: Now we want to lower bound $f(x)$. We use the FE to do this: \[ f(x) \ge \frac{f(x-f(y))-x}{y}.\]We have $f(f(y)) \le f(y)-1/2 \le y-1 < -1$ for $y>0$. Plug in $x=f(y)$, since then $f(x-f(y)$ is tractable; we know $f(0)$. We get \[ -1>f(f(y)) \ge \frac{-1/2-f(y)}{y} \implies -y>-1/2-f(y) \implies f(y) > y-1/2. \ \square\]Now, $f(42) \le 42-1/2$ by Claim 1 but $f(42) > 42-1/2$ by Claim 2. Contradiction. Hence, there are no solutions.
09.08.2020 20:46
As in every other solution, look for functions that satisfy $P(x,y)\iff f(x-f(y))\leq yf(x)+x$ for all real $x,y$, use $P(x+f(0),0)$ to get $f(x)\leq x+f(0)\implies f(f(x))\leq f(x)+f(0)$. Use that for the following: $P(f(x),x) \implies f(0)\leq xf(f(x))+f(x)\leq xf(f(x))+x+f(0) \implies x(f(f(x))+1)\geq 0$ $\implies f(f(0))\geq -1 \text{ for } x>0 \implies f(x)\geq -1-f(0) \text{ for } x\geq 0$ Now suppose the exists some $x$ with $f(x)> 0$. Then $x-f(y)>0$ when $y<x-f(0)$, which means $yf(x)+x\geq f(x-f(y))\geq -1-f(0)$ when $y\rightarrow-\infty$, but also $yf(x)\rightarrow-\infty$, a contradiction. So $f(x)\leq 0$ for all real $x$. With this in mind, take $P(x,y)$ where $y\rightarrow\infty$ and $x$ arbitrarily large, we have $f(x-f(y))\leq yf(x)+x\rightarrow-\infty$, and $x-f(y)\geq x\geq 0$ because $f(y)\leq 0$, but that contradicts our first conclusion, that $f$ is bounded below on the positive reals. So there is no such function.
20.02.2021 23:12
Storage: Assume the contrary, i.e., $$f(x-f(y)) \leq yf(x)+x$$for all real $x,y$. Call the above $P(x,y)$. Let $c=-1-f(0)$ Claim 1: $x>0$ $\implies$ $f(x) \geq c$. Proof: $P(x+f(0),0)$ gives $f(x) \leq x+f(0)$ for all $x$. Now $P(f(x),x)$ for any $x>0$ gives $$f(0) \leq xf(f(x))+f(x) \leq xf(f(x))+x+f(0)$$$\implies$ $f(f(x)) \geq -1$ $\implies$ $f(x) \geq f(f(x))-f(0) \geq -1-f(0)$ as required. $\blacksquare$ Claim 2: $f(x) \leq 0$ for all real $x$. Proof: Assume $f(t)>0$ for some $t$. If $y$ is large, then $-yf(t)+t <c$ $$\implies f(t-f(-y)) \leq -yf(t)+t<c$$which gives $f(-y) \geq t$ for all large $y$. Now note that $P(-x,-1)$ gives $$f(-x-f(-1))+f(-x) \leq -x$$Now if we take $x$ very large, then by the above we get $-x \geq 2t$, which is absurd. $\blacksquare$ From the first two claims and $P(0,y)$, we get $$yf(0) \geq f(-f(y)) \geq \min \{f(0),c \}$$for all real $y$. Clearly this is only possible if $f(0)=0$. Now $P(f(1),1)$ gives $f(1)+f(f(1)) \geq 0$. If $f(1)<0$, then $$f(1)+f(f(1)) < f(f(1)) \leq 0$$Contradiction! So $f(1)=0$. Now $P(-2021,1)$ gives $0 \leq -2021$, contradiction! $\blacksquare$
15.03.2021 06:12
I agree it is not a particularly instructive problem, but the solution does have some interesting and nonstandard ideas. Suppose for contradiction that $f(x-f(y)) \le yf(x) + x$ for any reals $x,y$. Let $(x,y) = (x+f(0),0)$: $f(x)\le x+f(0)$ for all $x$. Let $(x,y) = (f(x),x)$: $f(0)\le y=xf(f(x)) + f(x) \le xf(f(x)) + x + f(0)$, so $x(f(f(x))+1) \ge 0$. This means that $f(f(x))\ge -1$ when $x>0$, and $f(f(x))\le -1$ when $x\le 0$. Then we have $f(x)+f(0)\ge f(f(x)) \ge -1$, i.e. $f(x)\ge -1-f(0)$ for all positive real $x$. Let $(x,y) = (x,1)$: $f(x-f(1)) \le f(x) + x \le 2x + f(0)$. Putting $x = f(1)$ gives $f(1) \ge 0$. If $f(1) = 0$, then $f(x) \le f(x) + x$ for all $x$, which is clearly false. Therefore, $f(1) > 0$. Claim. $f(x)\ge 1$ for all $x\in (-\infty, \frac{-2-f(0)}{f(1)})$. Proof. Let $(x,y) = (f(x)+\epsilon, x)$ for some $\epsilon>0$: \[-1-f(0)\le f(\epsilon) \le xf(f(x)+\epsilon) + f(x) + \epsilon.\]If $f(x) < 1$ for some $x$, then taking $\epsilon = 1-f(x) > 0$, we have \[-1-f(0) \le xf(1) + 1 \implies x \ge \frac{-2-f(0)}{f(1)}. \square\] On the other hand, using $f(x-f(1))\le f(x) + x$ recursively (and fixing $x<0$), we have \[f(x-nf(1)) \le f(x) + nx - \frac{n(n-1)}{2}f(1)\]which can get arbitrarily small as $n$ gets large. This is a contradiction.
09.05.2022 14:56
Assume the contrary. Let $P(x,y)$ denote the assertion $yf(x)+x\geq f(x-f(y)).$ $P(x,0)\implies f(x)\leq x+f(0).$ $P(f(y),y)\implies 0\leq y(f(f(y))+1)\implies f(f(y))\geq -1~~\forall y\in \mathbb{R^+}.$ It follows that $f(y)\geq -1-f(0) ~~\forall y\in \mathbb{R^+}.$ Claim: $f(x)\leq 0,$ for all $x.$ Proof. Assume the contrary. If $u<v-f(0),$ then $v-f(u)>0.$ It follows that $uf(v)+v\geq -1-f(0).$ But this is a contradiction for sufficiently large negative values of $u,$ thus the claim. $\blacksquare$ Thus $f(0)\leq 0\implies f(x)\leq x.$ Notice that for $z>0$ and $z>-f(-1)-1,$ setting $x=f(z)-1$ easily gives a contradiction. Thus $f(x-f(y))>yf(x)+x~~\forall x,y\in \mathbb{R}.$
02.11.2022 19:57
Can someone check this?
31.12.2022 02:03
Hopefully this works. Suppose that $f(x-f(y)) \leq yf(x)+x$. Setting $y=0$ implies that $f(x) \leq x+f(0)$ (shift $x$ with $f(0)$) and $x=f(y)$ combined with the previous inequality gives that $yf(f(y))+y+f(0) \geq yf(f(y))+f(y) \geq f(0)$, so $f(f(y)) \geq -1$ for $y>0$. Combining with the first inequality again, we obtain $f(y)+f(0) \geq -1$, so $f(x) \geq c$ for all $x>0$ (for some constant $c=-1-f(0)$). If there is some $t$, such that $f(t)>0$, take $x=t$ and $y \rightarrow -\infty$ to get that $c \leq f(t-f(y)) \leq yf(t)+t \rightarrow -\infty$, contradiction (obviously $f(y) \leq y+f(0) \leq t$ would be satisfied for small enough $y$). Hence $f(x) \leq 0$ for all $x$. Now fix some $x>0$ and take $y \rightarrow +\infty$ in the original inequality to get that $c \leq f(x-f(y)) \leq yf(x)+x$ and the RHS again tends to $-\infty$, contradiction (we used that $x-f(y)>0$).
25.01.2023 17:45
We show that there are no functions $f\colon \mathbb{R} \to \mathbb{R}$ satisfying \[f(x - f(y)) \le yf(x) + x \]Let $P(x,y)$ denote the given assertion. Claim: $f(0) = -\frac{1}{2}$ Proof: $P(x+f(0),0): f(x)\le x + f(0)$. $P(f(x),x): f(0)\le xf(f(x)) + f(x)$. This implies that $f(0)\le x(x + 2f(0) + 1) + f(0)\implies x(x + 2f(0) + 1) \ge 0$ since $f(f(x))\le x + 2f(0)$. Taking $x>0$ gives that $x + 2f(0) + 1 \ge 0$, so $f(0)\ge -\frac{1}{2}$. Taking $x<0$ gives that $x + 2f(0) + 1\le 0$, so $f(0)\le - \frac{1}{2}$. Thus, $f(0) = -\frac{1}{2}$. $\square$ So $f(x)\le x - 1/2$. We get \[f(0)\le xf(f(x)) + f(x)\le xf(f(x)) + x + f(0),\]so $x(f(f(x)) + 1) \ge 0$, which implies $f(f(x))\ge -1\forall x>0$, so $f(x) + f(0)\ge -1\implies f(x) \ge -1/2$. Claim: $f(x) \le 0\forall x\in \mathbb{R}$. Proof: Clearly we only need to prove this for $x>1/2$. Suppose some $k>1/2$ satisfied $f(k)>0$. $P(k,x): f(k - f(x)) \le xf(k) + k$. Fix $x$ arbitrarily small. Since $f(x) \le 0$, we have $k-f(x)>0$, so the LHS is at least $-1/2$. However, the RHS approaches $-\infty$, contradiction. $\square$ $P(1434,x): f(1434 - f(x)) \le xf(1434) + 1434$. Fix $x$ arbitrarily large. Since $f(x)\le 0$, the LHS is at least $-1/2$. If $f(1434) = 0$, $P(f(1434),1434)$ gives that $-1/2 \le -517$, which is absurd. Thus, $f(1434) < 0$, which implies that the RHS gets arbitrarily small, contradiction.
12.04.2023 03:16
Assume otherwise that $f(x-f(y))\le yf(x)+x$ for all $x$ and $y$. Then, let this be $P(x,y)$. Let $c=f(0)$. $P(x,0)$ gives $f(x-c)\le x$ so $f(x)\le x+c$ for all $c$. $P(f(y),y)$ gives \[c\le yf(f(y))+f(y)\le y(y+2c)+y+c\implies 0\le y(y+(2c+1)).\]If we pick $y=-\tfrac{2c+1}{2}$ then this becomes $0\le -\tfrac{(2c+1)^2}{4}$ which implies $c=-\tfrac12$. But also, $y(f(f(y))+1)\ge 0$ so for $y\ge 0$ we have $f(f(y))\ge -1$ and so \[-y\ge yf(f(y))\ge c-f(y)\]so $y\le f(y)-c$ so $f(y)\ge y+c$. Therefore, $f(x)=x+c$ for all $x$. In the original inequality this becomes $-y\le y(x+c)$ which is clearly false.