4
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ How did you get the second inequality? $\endgroup$ – sranthrop May 5 '16 at 12:50
  • $\begingroup$ @sranthrop: I edited. $\endgroup$ – Roman83 May 5 '16 at 12:54
2
$\begingroup$

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)

$\endgroup$
  • $\begingroup$ $f$ may not derivative; $\endgroup$ – partofsha May 5 '16 at 13:27
  • $\begingroup$ see the last sentence $\endgroup$ – m7e May 5 '16 at 13:29
  • 1
    $\begingroup$ 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. $\endgroup$ – sranthrop May 5 '16 at 13:31
  • $\begingroup$ @sranthrop That's right. $\endgroup$ – m7e May 5 '16 at 13:32
  • $\begingroup$ Why $f'(x)\leq-1$ contradicts $f:\mathbb{R}^+\rightarrow\mathbb{R}^+$? Let $f(x)=\frac 1 x$ $\endgroup$ – user261263 May 5 '16 at 15:28
0
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ how do you get the first inequality? $\endgroup$ – Samuel May 5 '16 at 13:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.