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LSpice
LSpice
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No$\newcommand\N{\mathbb N}$No, it is not possible to have $\mu_{[\N]^2}\big(\text{rev}(f)\big) = 1$$\mu_{[\N]^2}\big({\operatorname{rev}(f)}\big) = 1$.

Given the function $f$, we will say that $n\in\mathbb{N}$ is good if $f(n)<f(k)$ for more than $\frac{3n}{4}$ values of $k\in\{1,\dots,n-1\}$. If $\mu_{[\N]^2}\big(\text{rev}(f)\big) = 1$$\mu_{[\N]^2}\big({\operatorname{rev}(f)}\big) = 1$, then the set of good numbers necessarily has density $1$.

So for big enough $N$ there will be more than $2^{N-1}+1$ good numbers in $[2^N,2^{N+1}-1]$. But that poses a problem: let $N$ be big and let $L_N$ be the median of the set of values $\{f(1),\dots,f(2^N-1)\}$. Then, for all good numbers $n\in[2^N,2^{N+1}-1]$ we have $f(n)<L_N$. Thus, $f(n)\leq L$ for at least $2^N$ values in $\{1,\dots,2^{N+1}\}$. Thus $L_{N+1}\leq L_N$. This cannot happen for all $N$, of course, as $\lim_{N\to\infty}L_N=\infty$, so we have a contradiction.

No, it is not possible to have $\mu_{[\N]^2}\big(\text{rev}(f)\big) = 1$.

Given the function $f$, we will say that $n\in\mathbb{N}$ is good if $f(n)<f(k)$ for more than $\frac{3n}{4}$ values of $k\in\{1,\dots,n-1\}$. If $\mu_{[\N]^2}\big(\text{rev}(f)\big) = 1$, then the set of good numbers necessarily has density $1$.

So for big enough $N$ there will be more than $2^{N-1}+1$ good numbers in $[2^N,2^{N+1}-1]$. But that poses a problem: let $N$ be big and let $L_N$ be the median of the set of values $\{f(1),\dots,f(2^N-1)\}$. Then, for all good numbers $n\in[2^N,2^{N+1}-1]$ we have $f(n)<L_N$. Thus, $f(n)\leq L$ for at least $2^N$ values in $\{1,\dots,2^{N+1}\}$. Thus $L_{N+1}\leq L_N$. This cannot happen for all $N$, of course, as $\lim_{N\to\infty}L_N=\infty$, so we have a contradiction.

$\newcommand\N{\mathbb N}$No, it is not possible to have $\mu_{[\N]^2}\big({\operatorname{rev}(f)}\big) = 1$.

Given the function $f$, we will say that $n\in\mathbb{N}$ is good if $f(n)<f(k)$ for more than $\frac{3n}{4}$ values of $k\in\{1,\dots,n-1\}$. If $\mu_{[\N]^2}\big({\operatorname{rev}(f)}\big) = 1$, then the set of good numbers necessarily has density $1$.

So for big enough $N$ there will be more than $2^{N-1}+1$ good numbers in $[2^N,2^{N+1}-1]$. But that poses a problem: let $N$ be big and let $L_N$ be the median of the set of values $\{f(1),\dots,f(2^N-1)\}$. Then, for all good numbers $n\in[2^N,2^{N+1}-1]$ we have $f(n)<L_N$. Thus, $f(n)\leq L$ for at least $2^N$ values in $\{1,\dots,2^{N+1}\}$. Thus $L_{N+1}\leq L_N$. This cannot happen for all $N$, of course, as $\lim_{N\to\infty}L_N=\infty$, so we have a contradiction.

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Saúl RM
Saúl RM
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No, it is not possible to have $\mu\big(\text{rev}(f)\big) = 1$$\mu_{[\N]^2}\big(\text{rev}(f)\big) = 1$.

Given the function $f$, we will say that $n\in\mathbb{N}$ is good if $f(n)<f(k)$ for more than $\frac{3n}{4}$ values of $k\in\{1,\dots,n-1\}$. If $\mu\big(\text{rev}(f)\big) = 1$$\mu_{[\N]^2}\big(\text{rev}(f)\big) = 1$, then the set of good numbers necessarily has density $1$.

So for big enough $N$ there will be more than $2^{N-1}+1$ good numbers in $[2^N,2^{N+1}-1]$. But that poses a problem: let $N$ be big and let $L_N$ be the median of the set of values $\{f(1),\dots,f(2^N-1)\}$. Then, for all good numbers $n\in[2^N,2^{N+1}-1]$ we have $f(n)<L_N$. Thus, $f(n)\leq L$ for at least $2^N$ values in $\{1,\dots,2^{N+1}\}$. Thus $L_{N+1}\leq L_N$. This cannot happen for all $N$, of course, as $\lim_{N\to\infty}L_N=\infty$, so we have a contradiction.

No, it is not possible to have $\mu\big(\text{rev}(f)\big) = 1$.

Given the function $f$, we will say that $n\in\mathbb{N}$ is good if $f(n)<f(k)$ for more than $\frac{3n}{4}$ values of $k\in\{1,\dots,n-1\}$. If $\mu\big(\text{rev}(f)\big) = 1$, then the set of good numbers necessarily has density $1$.

So for big enough $N$ there will be more than $2^{N-1}+1$ good numbers in $[2^N,2^{N+1}-1]$. But that poses a problem: let $N$ be big and let $L_N$ be the median of the set of values $\{f(1),\dots,f(2^N-1)\}$. Then, for all good numbers $n\in[2^N,2^{N+1}-1]$ we have $f(n)<L_N$. Thus, $f(n)\leq L$ for at least $2^N$ values in $\{1,\dots,2^{N+1}\}$. Thus $L_{N+1}\leq L_N$. This cannot happen for all $N$, of course, as $\lim_{N\to\infty}L_N=\infty$, so we have a contradiction.

No, it is not possible to have $\mu_{[\N]^2}\big(\text{rev}(f)\big) = 1$.

Given the function $f$, we will say that $n\in\mathbb{N}$ is good if $f(n)<f(k)$ for more than $\frac{3n}{4}$ values of $k\in\{1,\dots,n-1\}$. If $\mu_{[\N]^2}\big(\text{rev}(f)\big) = 1$, then the set of good numbers necessarily has density $1$.

So for big enough $N$ there will be more than $2^{N-1}+1$ good numbers in $[2^N,2^{N+1}-1]$. But that poses a problem: let $N$ be big and let $L_N$ be the median of the set of values $\{f(1),\dots,f(2^N-1)\}$. Then, for all good numbers $n\in[2^N,2^{N+1}-1]$ we have $f(n)<L_N$. Thus, $f(n)\leq L$ for at least $2^N$ values in $\{1,\dots,2^{N+1}\}$. Thus $L_{N+1}\leq L_N$. This cannot happen for all $N$, of course, as $\lim_{N\to\infty}L_N=\infty$, so we have a contradiction.

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Saúl RM
Saúl RM
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No, it is not possible to have $\mu\big(\text{rev}(f)\big) = 1$.

Given the function $f$, we will say that $n\in\mathbb{N}$ is good if $f(n)<f(k)$ for more than $\frac{3n}{4}$ values of $k\in\{1,\dots,n-1\}$. If $\mu\big(\text{rev}(f)\big) = 1$, then the set of good numbers necessarily has density $1$.

So for big enough $N$ there will be more than $2^{N-1}+1$ good numbers in $[2^N,2^{N+1}-1]$. But that poses a problem: let $N$ be big and let $L_N$ be the median of the set of values $\{f(1),\dots,f(2^N-1)\}$. Then, for all good numbers $n\in[2^N,2^{N+1}-1]$ we have $f(n)<L_N$. Thus, $f(n)\leq L$ for at least $2^N$ values in $\{1,\dots,2^{N+1}\}$. Thus $L_{N+1}\leq L_N$. This cannot happen for all $N$, of course, as $\lim_{N\to\infty}L_N=\infty$, so we have a contradiction.