A functional equation with strange domain

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A functional equation with strange domain

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Source: 2022 Taiwan TST Round 3 Mock P4

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Li4

#1 h Apr 30, 2022, 6:11 AM • 2 Y

Y by Ainulindale, Mango247

Let $\mathcal{X}$ be the collection of all non-empty subsets (not necessarily finite) of the positive integer set $\mathbb{N}$ . Determine all functions $f: \mathcal{X} \to \mathbb{R}^+$ satisfying the following properties:

(i) For all $S$ , $T \in \mathcal{X}$ with $S\subseteq T$ , there holds $f(T) \le f(S)$ .
(ii) For all $S$ , $T \in \mathcal{X}$ , there hold
$f(S) + f(T) \le f(S + T),\quad f(S)f(T) = f(S\cdot T),$ where $S + T = \{s + t\mid s\in S, t\in T\}$ and $S \cdot T = \{s\cdot t\mid s\in S, t\in T\}$ .

Proposed by Li4, Untro368, and Ming Hsiao.

This post has been edited 4 times. Last edited by Li4, May 1, 2022, 9:58 AM

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gghx

#3 h Apr 30, 2022, 7:24 AM • 1 Y

Y by PRMOisTheHardestExam

Chef's kiss :love:

:love:

The only solution is $f(S)=(\min_{s\in S} s)^c$ where $c\ge 1$ is some constant real number. The first condition holds because since $T$ contains $S$ , the smallest element of $T$ is at least as small as the smallest element of $S$ . The second condition holds because $x^c$ is convex and multiplicative.

We now show it is the only solution.

Let $x,y$ be natural numbers unless otherwise specified. For simplicity, define $f(\{x\})=f(x)$ where $x$ is any positive integer.
Note that from condition (ii), $f(x)+f(y)\le f(x+y)$ and $f(x)f(y)=f(xy)$ .

Obviously, $f(1)=1$ , and $f$ is strictly increasing.

Consider any $y>2$ .

Let $r,s$ be positive integers such that $\frac{r}{s}>\frac{\log y}{\log 2}$ . Then, $2^r>y^s\implies f(2^r)>f(y^s)\implies f(2)^r>f(y)^s\implies \frac{r}{s}>\frac{\log f(y)}{\log f(2)}$ . Similarly, if $\frac{r}{s}<\frac{\log y}{\log 2}$ , then $\frac{r}{s}<\frac{\log f(y)}{\log f(2)}$ .

Since rationals are dense over $\mathbb{R}$ , $\frac{\log y}{\log 2} = \frac{\log f(y)}{\log f(2)}\implies \frac{\log f(y)}{\log y}$ is constant.
Hence, $\frac{\log f(y)}{\log y}=c\implies f(y)=y^c$ for some constant $c$ . Since $f(x)+f(y)\le f(x+y)$ , we get $c\ge 1$ .

Let $S$ be a set containing $1$ . Then, $f(S)f(\mathbb{N})=f(\mathbb{N})$ implies $f(S)=1$ .

Let $S$ be a set with minimal element $a$ .
Then, from (i) we get $f(S)\le a^c$ and from (ii) we get $f(S)\ge f(S-(a-1))+f(a-1)=1+(a-1)^c$ .
Hence, $a^c\ge f(S)\ge 1+(a-1)^c$ .

Take a positive integer $k$ , $f(S)^k=f(S^k)\ge 1+(a^k-1)^c$ .
Thus, $f(S)>\sqrt[k]{(a^k-1)^c}$ . The limit of this is $a^c$ , and so taking $k$ to infinity we get $f(S)=a^c$ .
This concludes the proof.

wow.

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#4 h Apr 30, 2022, 7:40 AM • 2 Y

Y by Mango247, Mango247

The only function is $f(S)=(\min S)^c$ where $c\ge1$ is constant. We prove the following claims.

Claim 1: $f(\mathbb N)=1$ .
Proof: Setting $S=T=\mathbb N$ in the latter equation of condition (ii) gives
$f(\mathbb N)f(\mathbb N)=f(\mathbb N)\implies f(\mathbb N)=1$
Claim 2: For any set $S$ , if $m=\min S$ , then we have $f(S)=f(m\mathbb N)$ .
Proof: Since $m$ is the minimum element of $S$ , we have $S\cdot\mathbb N=m\mathbb N$ .
Setting $T=\mathbb N$ in the latter of condition (ii) gives
$f(S)f(\mathbb N)=f(m\mathbb N)\implies f(S)=f(m\mathbb N)$ since $f(\mathbb N)=1$ from Claim 1.

Because of Claim 2, we can define a function $g:\mathbb N\to\mathbb R^+$ such that $f(S)=g(\min S)$ for any subset $S$ . In terms of $g$ , the new conditions are:
(i) For all $s,t\in\mathbb N$ with $t\le s$ , we have $g(t)\le g(s)$
(ii) For all $s,t\in\mathbb N$ , we have
$g(s)+g(t)\le g(s+t),\quad g(s)g(t)=g(st)$ The proof that $g(n)=n^c$ for some constant $c$ follows similarly as @gghx's solution above.

EDIT: Whoops. @below, yes you are right, my solution is incorrect. My apologies.

This post has been edited 2 times. Last edited by P1kachu, May 4, 2022, 1:46 PM

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gghx

#5 h Apr 30, 2022, 7:42 AM • 3 Y

Y by Mango247, Mango247, Mango247

P1kachu wrote:

Proof: Since $m$ is the minimum element of $S$ , we have $S\cdot\mathbb N=m\mathbb N$ .

I dont think so? for example $S=\{2,3\}$ .

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