# Finding all positive real functions satisfying $xf(y)+f(f(y))\leq f(x+y)$

Find function $$f: \mathbb{R}_{> 0}\rightarrow \mathbb{R}_{> 0}$$ such that: $$xf(y)+f(f(y))\leq f(x+y)$$ for all positive $$x$$ and $$y$$?

That problem made me think a lot. This is the first time I solve functional inequality. Please show me the way to solve such problems. From the inequality, can we prove that $$f$$ is injective or surjective?

• I suspect that no such function exists. Is that a possibility that you would be interested in discussing ? – WW1 Jun 18 at 1:28
• Why you know that? I can not find the way... – user628755 Jun 18 at 1:31
• I got it! It took me a while, but such a function doesn't exist. I don't want to post an answer without giving some hints first. The first step that helped me was to show that $f(y) \le y$ for all $y$. I proved it by contradiction. Let me know if you want me to post a full answer. – Theo Bendit Jun 18 at 2:20
• Theo Bendit, give me the answer pls! – user628755 Jun 18 at 4:02
• I have cast a "close as duplicate" vote to this question but now I think my vote was wrong. This is not a duplicate of the above-linked question. In that question, the domain/codomain of $f$ is $\mathbb R$, while in this question, the domain/codomain the positive real line. It is not clear how the answers to that question can be adapted to this one. – user1551 Jun 18 at 8:46

## 2 Answers

$$f(f(y))-f(x+y)\leq -xf(y)$$.

As $$xf(y)>0$$, $$f(f(y))-f(x+y)<0$$.

Now, if $$f(y_0)>y_0$$ for some $$y_0>0$$, then we can put $$x=f(y_0)-y_0$$ and $$y=y_0$$ which yields $$f(f(y_0))-f(f(y_0))<0$$ which is clearly wrong.

Therefore, $$f(y)\leq y$$

Now, as $$f(f(y))>0$$, $$xf(y)

$$x(f(y)-1) $$\forall$$ $$x,y>0$$.

If $$f(y)>1$$ for any finite $$y>0$$, by making $$x$$ arbitrarily large, we obtain a contradiction.

Thus, $$f(y)\leq 1$$

As $$f(f(y))>0$$, $$f(x+y)\leq 1$$ and $$xf(y)+f(f(y))\leq f(x+y)$$, $$xf(y)<1$$

As $$f(y)>0$$, by making $$x$$ arbitrarily large, we obtain another contradiction.

Therefore, there exists no function $$f:\mathbb{R}_{>0}\to \mathbb{R}_{>0}$$ such that $$xf(y)+f(f(y))\leq f(x+y)$$.

We first prove that $$f(x)\leq x$$. Suppose that for some $$k$$, $$f(k)>k$$. Then letting $$x=f(k)-k$$ and $$y=k$$ in the original equation shows that $$(f(k)-k)f(k)+f(f(k))\leq f(f(k)) \implies (f(k)-k)f(k)\leq 0$$ which is clearly a contradiction since we supposed $$f(k)-k$$ was positive and $$f$$ attains positive real values. Hence $$f(x)\leq x$$ for all $$x\in\mathbb{R^+}$$.

Now fix a constant $$c>0$$. Note that for sufficiently large $$N$$, we have that \begin{align*} N(f(c)-1)&>c-f(f(c))\qquad \text{ (true for sufficienty large N)} \\ \implies Nf(c)-N&>c-f(f(c)) \\ \implies Nf(c)+f(f(c))&>N+c\tag{1} \end{align*}

But if we put $$x=N$$ and $$y=c$$ into our original equation, we find that, on the contrary, $$Nf(c)+f(f(c))\leq f(N+c)\leq N+c$$ where the last inequality comes from the fact that $$f(x)\leq x$$. This clearly contradicts $$(1)$$, hence no such functions exist.