Show that there is no function $f:{{\mathbb{R}}^{+}}\to {{\mathbb{R}}^{+}}$ such that $f(x+y)>f(x)(1+yf(x))$ for all $x,y\in {{\mathbb{R}}^{+}}$.
Problem
Source: Turkish NMO 1996, 6. Problem
Tags: function, induction, algebra proposed, algebra
31.07.2011 20:42
for $y=0$ you get $f(x)>f(x)$ for all$x\in\mathbb{R^{+}}$ am i missing something?
31.07.2011 20:44
$0 \notin \mathbb{R}^+ \ !$
31.07.2011 20:50
costantin07 wrote: for $y=0$ you get $f(x)\geqq(x)$ for all$x\in\mathbb{R}^+\ $ am i missing something? Yes you do, exactly you did three things wrong: First of all the function is defined for all positive real $x$ to all positive real $x$: $f(x):\mathbb{R^{+}}\to \mathbb{R^{+}}$, so you can't say that $y=0$, second in the problem the expression $\ge$ is never used (only $>$) and third, if you fill $y=0$ (this isn't even allowed) then you get: $f(x)>f(x)$ (so you even messed up the basic algebra). Read the problem more careful the next time!
31.07.2011 21:17
lol ..three stuffs..i get it now ..sorry anyway it s just because i am morrocan and we follow the french notations so for us $0\in\mathbb{R^{+}}$ and the set of positif reals without $0$ is : $\mathbb{R_{+}^{*}}$ sorry again! i will try to be carefull with the notations used in the forum next time Let's return to the foundamental problem now !
04.08.2011 14:28
Clearly $f(x+y)>{yf(x)^2}$. We can even prove that no function satisfies this inequallity in the following way: Fixing x and taking y large enough, we see that there exists a $x_0$ such that $f(x_0)>3$ Inductively $\displaystyle f(x+\frac{k}{n})>\frac{f(x)^{2^k}}{n^{2k-1}}$ (induction on k). For $k=n$ and $x=x_0$, $\displaystyle f(x_0+1)>\frac{f(x_0)^{2^k}}{n^{2n-1}}>\frac{3^{2^n}}{n^{2n-1}}$, which tends to infinity, contradiction.
05.04.2014 19:05
panos_lo wrote: Inductively $\displaystyle f(x+\frac{k}{n})>\frac{f(x)^{2^k}}{n^{2k-1}}$ (induction on k). Typo. It should be $\displaystyle f(x+\frac{k}{n})>\frac{f(x)^{2^k}}{n^{2^k-1}}$. panos_lo wrote: For $k=n$ and $x=x_0$, $\displaystyle f(x_0+1)>\frac{f(x_0)^{2^k}}{n^{2n-1}}>\frac{3^{2^n}}{n^{2n-1}}$, which tends to infinity, contradiction. Again, here it should be $\displaystyle f(x_0+1)>\frac{f(x_0)^{2^k}}{n^{2n-1}}>\frac{3^{2^n}}{n^{2^n-1}}$. But I don't think $\frac{3^{2^n}}{n^{2^n-1}}$ actually tends to infinity, so the solution does not make any sense.
08.04.2014 06:08
efoski1687 wrote: Show that there is no function $f:{{\mathbb{R}}^{+}}\to {{\mathbb{R}}^{+}}$ such that $f(x+y)>f(x)(1+yf(x))\quad (1)$ for all $x,y\in {{\mathbb{R}}^{+}}$. Clearly, $f$ is strictly increasing on $\mathbb R^+.$ Replacing $y=\frac{1}{f(x)}$ in $(1),$ we get \[f\left(x+\frac{1}{f(x)}\right)>2\cdot f(x),\quad \forall x>0.\quad (2)\] Next, replacing $x$ by $x+\frac{1}{f(x)}$ in $(2),$ we get \[f\left(x+\frac{1}{f(x)}+\frac{1}{f\left(x+\frac{1}{f(x)}\right)}\right)>2\cdot f\left(x+\frac{1}{f(x)}\right)>2^2\cdot f(x).\] Since $f$ is strictly increasing, we have \[\begin{aligned} f\left(x+\frac{1}{f(x)}+\frac{1}{f\left(x+\frac{1}{f(x)}\right)}\right)&< f\left(x+\frac{1}{f(x)}+\frac{1}{2\cdot f(x)}\right)\\ &=f\left(x+\left(1+\frac{1}{2}\right)\frac{1}{f(x)}\right)\end{aligned}\] and hence, it follows that \[f\left(x+\left(1+\frac{1}{2}\right)\frac{1}{f(x)}\right)>2^2\cdot f(x),\quad \forall x>0.\quad (2)\] From this, using induction, we can easily prove that \[f\left(x+\left(1+\frac{1}{2}+\cdots +\frac{1}{2^n}\right)\frac{1}{f(x)}\right)>2^{n+1}\cdot f(x),\quad \forall x>0. \quad (3)\] Since $1+\frac{1}{2}+\cdots +\frac{1}{2^n}<2,$ it follows that \[f\left(x+\frac{2}{f(x)}\right)>2^{n+1}\cdot f(x),\quad \forall x \in \mathbb R^+,\, n \in \mathbb N^*,\] which is impossible. So no function exists.