Prove that there exist two functions $f,g \colon \mathbb{R} \to \mathbb{R}$, such that $f\circ g$ is strictly decreasing and $g\circ f$ is strictly increasing. (Poland) Andrzej Komisarski and Marcin Kuczma
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Tags: function, algebra proposed, algebra
26.02.2011 21:48
27.01.2015 18:00
isn't f(x)={x^2 for x>0 and -x^2 for x<=0 g(x)={-x^2 for x>0 and x^2 for x<=0 a solution
27.01.2015 18:10
If you take $x>0$: $g(f(x))=g(x^2)=-x^4$ which is not strictly increasing .
27.01.2015 18:11
lggl wrote: isn't f(x)={x^2 for x>0 and -x^2 for x<=0 g(x)={-x^2 for x>0 and x^2 for x<=0 a solution No, with your example $f$ is increasing and $g$ is decreasing, so both $f\circ g$ and $g\circ f$ are decreasing (in fact they're identical).
10.12.2017 13:49
is not a solution
10.12.2017 16:07
MATH1945 wrote:
a solution? $-2^{x^2}$ is not strictly decreasing.
23.09.2019 08:09
As a notational convenience, we let $h(a\mapsto b)=c\mapsto d$ indicate that $h(x)=\frac{d-c}{b-a}(x-a)+c$ for $x\in(a,b)$. Let $\alpha,\beta:\mathbb{Z}\mapsto\mathbb{Z}$ be functions defined as \[\alpha(n)=\begin{cases}-9n-1 & \text{if }n\text{ even} \\ 9n+1 & \text{if }n\text{ odd}\end{cases}\]and \[\beta(n)=\begin{cases}9n & \text{if }n\text{ even} \\ -9n & \text{if }n\text{ odd}\end{cases}.\]Note that $\alpha(n)$ has the opposite parity as $n$ and $\beta(n)$ has the same parity as $n$. Furthermore, we compute that \[\beta\circ\alpha(n)=81n+9\]and \[\alpha\circ\beta(n)=\begin{cases}-81n-1 & \text{if }n\text{ even} \\ -81n+1 & \text{if }n\text{ odd}\end{cases}.\]We see that $\beta\circ\alpha$ increases by $81$ every time $n$ increases by $1$ and $\alpha\circ\beta$ decreases by at least $79$ each time $n$ increases by $1$. Now, we define $f$ and $g$ to be \[f(n\mapsto n+1)=\begin{cases}\alpha(n)\mapsto\alpha(n)+1 & \text{if }n\text{ even} \\ \alpha(n)+1\mapsto\alpha(n) & \text{if }n\text{ odd}\end{cases}\]and \[g(n\mapsto n+1)=\begin{cases} \beta(n)+1\mapsto\beta(n) & \text{if }n\text{ even} \\ \beta(n)\mapsto\beta(n)+1 & \text{if }n\text{ odd}.\end{cases}\]Note that this uniquely determines $f$ and $g$ over $\mathbb{R}\setminus\mathbb{Z}$ is the disjoint union of all the intervals $[n,n+1)$ over all $n\in\mathbb{Z}$. We compute \[f\circ g(n\mapsto n+1)=\alpha\circ\beta(n)+1\mapsto\alpha\circ\beta(n)\]and \[g\circ f(n\mapsto n+1)=\beta\circ\alpha(n)\mapsto\beta\circ\alpha(n)+1.\]The fact that $\beta\circ\alpha$ is very strongly increasing and $\alpha\circ\beta$ is very strongly decreasing gives that $g\circ f$ and $f\circ g$ are strictly increasing and decreasing respectively. It remains to assign values to $f(n)$ and $g(n)$ for integral $n$ such that the increasing/decreasing conditions still hold. This is quite tricky, but I think this construction works. For integral $n$, define \[f(n)=\begin{cases}-27n & \text{if }n\text{ even} \\ 27n+2 & \text{if }n\text{ odd}\end{cases}\]and \[g(n)=\begin{cases}3n & \text{if }n\text{ even} \\ -3n & \text{if }n\text{ odd}\end{cases}.\]We have then that \[g\circ f(n)=81n+6\]and \[f\circ g(n)=\begin{cases}-81n & \text{if }n\text{ even} \\ -81n+2 & \text{if }n\text{ odd}\end{cases}.\]Note that \[\beta\circ\alpha(n-1)+1\le g\circ f(n)\le\beta\circ\alpha(n)\]and \[\alpha\circ\beta(n-1)\ge f\circ g(n)\ge\alpha\circ\beta(n)+1,\]so $g\circ f$ is strictly increasing and $f\circ g$ is strictly decreasing.
01.03.2020 18:11
24.12.2022 16:59
Take two bijections $a,b:\mathbb{R}\to\mathbb{R}$, such that the directed graphs they induce are isomorphic (bijection between cycles of certain length, and chains), and $a$ is strictly increasing, while $b$ is strictly decreasing. The functions mentioned above $a(x)=2x$, $\forall x\in \mathbb{R}$ and $b(x)=-2x$, $ \forall x\in \mathbb{R}$ work, since they have a bunch of chains and one cycle of length 1 (or loop, whichever way you want to call it). Now we take matching components of $a$ and $b$, let the component of $a$ be blue and the component of $b$ red. We make a new component like this: take a blue element, then a red element, then the next blue element, then the next red element, and so on, while also doing this backwards. Notice, we're essentially "mixing" the 2 components in an alternating fashion. We will let $f(x)$ be the red element after $x$ in the component that contains blue $x$, and $g(x)$ be the blue element after $x$ in the component that contains red $x$. Essentially, if in the new graph we color red the edges from the red elements and color blue the edges from the blue elements, we let $f$ be the blue function and $g$ be the red function. Now it is clear $g\circ f=a$ and $f\circ g =b$, so our functions work.
04.08.2023 16:42
I have no moral qualms whatsoever about this solution. I will exhibit $f$ and $g$ such that $f(g(x))=-2x$ and $g(f(x))=2x$. Partition the real numbers into equivalence classes where $x$ and $y$ are in the same class iff $\{\log_2 |x|\}=\{\log_2 |y|\}$, or equivalently there exists some $k \in \mathbb{Z}$ with $|x|=2^k|y|$. Then pair these equivalence classes arbitrarily (one way to do this explicitly, without the axiom of choice or something like that, is to pair the class with $\{\log_2 |x|\}=t$ with the class with $\{\log_2|x|\}=1-t$ for $t \in (0,1) \setminus \{\tfrac{1}{2}\}$, and then pair the class with $\{\log_2 |x|\}=0$ with the one with $\{\log_2 |x|\}=\tfrac{1}{2}$). Then for each constructed pair of equivalence classes, pick one representative from each. Suppose some pair of equivalence classes has representatives $a$ and $b$. Then for any $k \in \mathbb{Z}$, let $$f(\varepsilon 2^ka)=\varepsilon(-2)^kb,f(\varepsilon 2^kb)=\varepsilon(-2)^{k+1}a,g(\varepsilon 2^k a)=\varepsilon(-2)^kb,g(\varepsilon 2^kb)=-\varepsilon(-2)^{k+1}a,$$where $\varepsilon \in \{-1,1\}$. Then $$f(g(\varepsilon 2^ka))=f((-1)^k\varepsilon\cdot 2^kb)=(-1)^k \varepsilon \cdot (-2)^{k+1}a=-\varepsilon 2^{k+1}a\text{ and } f(g(\varepsilon 2^kb))=f(-(-1)^{k+1}\varepsilon\cdot 2^{k+1}a)=-(-1)^{k+1}\varepsilon \cdot (-2)^{k+1}a=-\varepsilon 2^{k+1}a$$and $$g(f(\varepsilon 2^ka))=g((-1)^k\varepsilon \cdot 2^kb)=-(-1)^k \varepsilon \cdot (-2)^{k+1}a=\varepsilon 2^{k+1}a\text{ and }g(f(\varepsilon 2^kb))=g((-1)^{k+1}\varepsilon \cdot 2^{k+1}a)=(-1)^{k+1}\varepsilon\cdot(-2)^{k+1}b=\varepsilon 2^{k+1}b$$so $f(g(x))=-2x$ and $g(f(x))=2x$ whenever $x$ is in the equivalence class of $a$ or $b$, hence $f(g(x))=-2x$ and $g(f(x))=2x$ for all $x \in \mathbb{R}$ since our pair of equivalence classes was picked arbitrarily. $\blacksquare$ Remark: This solution is strongly motivated by considering the arrows graph. The well-known (?) problem to find functions $f : \mathbb{Z}^+ \to \mathbb{Z}^+$ such that $f(f(n))=2n$ results in a similar structure.
01.03.2024 14:41
Conside the solution $f(x)=-x, g(x)=x^2\forall x\geq 0, -x^2 \forall x\leq 0$ Obviously $g(x)$ is strictly increasing (yes its derivative at 0 is 0 but you need 2 points where it does not increase to make it not strictly increasing, this function is like $x^3$ but for $x^2$ $f(g(x))=-g(x)$ which is strictly decreasing, and $g(f(x))=g(x)$ which is strictly increasing
04.04.2024 00:30
I think the motivation behind this problem is a way of "flagging" some property through the construction, either through intervals or equivalence classes. Most solutions feel roughly isomorphic, and this problem should easily generalize to letting you specify what $f \circ g$ and $g \circ f$ are more or less. We claim that such functions exist. Note that $\tan$ and $\tan^{-1}$ are bijective order preserving maps between $(-\infty, \infty)$ and $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. As such, by considering $f = \tan \circ f' \circ \tan^{-1}, g = \tan \circ g' \tan^{-1}$, it remains to solve the problem for $f'$ and $g'$ over $(0, 1)$. We define $f'(x) = x + 1434$ and \[ g'(x) = \begin{cases} x - 1434 & x > 1400 \\ -x - 1434 & x < 1400 \\ \end{cases} \]Then $g' \circ f'$ is the identity and $f' \circ g'$ is inverting. The result follows.