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Does there exist the function $f:\mathbb R^+\rightarrow \mathbb R^+$, such that $$f^2(x)\ge f(x+y)\left(f(x)+y \right) \forall x,y \in \mathbb R^+$$

My work so far:

Assume that a function exists. Then $$f(x+y)\le \frac{f^2(x)}{f(x)+y}<f(x).$$ Then this function is strictly decreasing. (And this function is injective).

Roman83
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2 Answers2

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Taking logarithms gives $$ 2\log f(x)\geq\log f(x+y)+\log(1+y/f(x))+\log f(x) $$ After rearranging, $$ \frac{\log f(x)-\log f(x+y)}{y}\geq\frac{\log(1+y/f(x))}{y}, $$ so after taking limit as $y\rightarrow0$ we get $$ \frac{-f'(x)}{f(x)}=(-\log f)'(x)\geq\frac{1}{f(x)}. $$ and so $f'(x)\leq-1$, which contradicts $f:\mathbb{R}^+\rightarrow\mathbb{R}^+$. (As $f$ is decreasing, it is almost everywhere differentiable)

m7e
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  • $f$ may not derivative; – partofsha May 05 '16 at 13:27
  • see the last sentence – m7e May 05 '16 at 13:29
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    You both could combine your answers and simplify mge's idea a little: There is no need to take logarithms if you use partofsha's first inequality. Then you immediately get $f'(x)\leq -1$ a.e. – sranthrop May 05 '16 at 13:31
  • @sranthrop That's right. – m7e May 05 '16 at 13:32
  • Why $f'(x)\leq-1$ contradicts $f:\mathbb{R}^+\rightarrow\mathbb{R}^+$? Let $f(x)=\frac 1 x$ –  May 05 '16 at 15:28
  • @EugenCovaci Are you sure $-1/x^2\leq -1$? – m7e May 05 '16 at 15:52
  • @mge for $x \in (0,1)$. But it doesn't matter, how do you justify the contradiction? –  May 05 '16 at 16:01
  • It does matter, because if you don't read the statement (the domain of $f$ is $\mathbb{R}^+$, not $(0,1)$) don't expect detailed explanations – m7e May 05 '16 at 16:13
  • @mge just ignore $f(x)= \frac 1 x$. How do you justify the contradiction? –  May 05 '16 at 16:20
  • @EugenCovaci: Integration over non-degenerate intervals preserves inequalities. In particular, $f'(x)\le-1$ implies, for any $a>0$, $$\int_a^x f'(t)dt=f(x)-f(a)<-x+a.$$ – Vincenzo Oliva May 06 '16 at 11:29
  • @VincenzoOliva Your inequality only proves $f(x)<f(a)$. Still I don't see any contradiction. –  May 06 '16 at 13:43
  • @EugenCovaci: Mh? Take $x\to+\infty$. – Vincenzo Oliva May 06 '16 at 13:56
  • @EugenCovaci: Let $a=1$. His inequality shows that $f(x)$ becomes negative for large enough $x$. More intuitively: if the slope is $<-1$ almost everywhere, then the function must decrease at least as fast as $-x$, and in particular it must eventually cross the x-axis. – Samuel May 06 '16 at 13:59
  • @Samuel I knew that from the beginning, still I want to see that in the proof. I also can prove that without using integrals. –  May 06 '16 at 14:03
  • @EugenCovaci: But you said you still didn't see any contradiction. – Vincenzo Oliva May 06 '16 at 14:25
  • @VincenzoOliva I don't see it in the proof. –  May 06 '16 at 19:03
  • @EugenCovaci: Lol. True, but we were talking about my inequality, not mge's answer itself. Anyway, it was apparently left to the OP, and apparently he didn't need to ask for clarification. Cheers. – Vincenzo Oliva May 06 '16 at 19:30
  • @VincenzoOliva About your inequality: How do you know $f'$ is Riemann integrable? –  May 07 '16 at 03:45
  • @Samuel I genuinely do not know why $f′$ is Riemann integrable. Please tell me. –  May 07 '16 at 14:00
  • @EugenCovaci: Perhaps I was a bit sloppy about that. $f'$ is Lebesgue integrable.

    Firstly, let us note that if $f$ exists then $\lim\limits_{x\to\infty}f(x)=0$. Indeed, we have $$0<f(x+y)\le\frac{f^2(x)}{f(x)+y}$$so, taking $y\to+\infty$, the result follows from the squeeze theorem. We already know $f'(x)\le-1$. For a contradiction, suppose $$\liminf_{x\to+\infty}f'(x)=\liminf_{x\to+\infty}\left(\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\right)=-\infty.$$

     – Vincenzo Oliva May 18 '16 at 09:11
  • Then for any $M\in\mathbb{N}$ there exists $N\in\mathbb{N}$ such that, for infinitely many $x$, $$\frac{f(x+1/N)-f(x)}{1/N}<-M \ f(x+1/N)<f(x)-M/N,$$which contradicts $f$ being always positive, as the RHS converges to $-M/N$. Therefore $\lvert f'(x)\rvert$ is bounded, which makes $f$ Lipschitz and thus absolutely continuous. By the Fundamental Theorem of Calculus for Lebesgue Integration, this is equivalent to $f$ being almost everywhere differentiable (which in this case we already knew), $f'$ being integrable and $\int_a^bf'(x)dx=f(b)-f(a)$. – Vincenzo Oliva May 18 '16 at 09:12
  • @EugenCovaci: Yet, a similar argument applied just to $f'(x)\le-1$ is obviously enough to derive that $f$ ought to become negative, eventually. Integration was just the first thing that naively came to my mind, I guess. – Vincenzo Oliva May 19 '16 at 21:10
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Suppose that there is a function $f$.then we have since $$f^2(x)\ge f(x+y)(f(x)+y)$$ we have $$f(x)-f(x+y)\ge\dfrac{f(x)y}{f(x)+y}>0$$ which shows that $f$is a strictly decreasing function.Give an $x\in R^{+}$, we choose an $n\in N^{+}$ such $nf(x+1)\ge 1$, then $$f\left(x+\dfrac{k}{n}\right)-f\left(x+\dfrac{k+1}{n}\right)\ge\dfrac{f(x+k/n)\cdot\frac{1}{n}}{f(x+\frac{k}{n})+\frac{1}{n}}>\dfrac{1}{2n}$$ summing up these inequalities for $k=0,1,\cdots,n-1$,we have $$f(x)-f(x+1)>\dfrac{1}{2}$$ take an $m$ such $m\ge 2f(x)$,then $$f(x)-f(x+m)>\dfrac{m}{2}\ge f(x)$$a contradiction

partofsha
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