Are all linear functions monotonic?












1














I have never come across a statement linking linearity and monotony - but it seems that for each linear function (positive, negative or even constant slope), the function is monotonic:



I.e. for y $geq$ x it follows that f(y) $geq$ f(x).



Is this correct?






analysis monotone-functions







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  • Yes, it's monotonic, but not necessarily increasing. You could have $yge ximplies f(y)le f(x)$
    – saulspatz
    Nov 25 at 14:08










  • Exactly, so positive slope = increasing; negative slope = decreasing, in both cases strictly monotonic as well -- constant slope = monotonic, but not strictly. Is this correct?
    – TestGuest
    Nov 25 at 14:10












  • Note that your question only makes sense for 1-dimensional functions, if you want to talk about linear functions in general, say $f:mathbb{R}^nrightarrow mathbb{R}^m$ or any vector spaces then you need to define what you mean by $xleq y$
    – Yanko
    Nov 25 at 14:11










  • @TestGuest yes this is the point.
    – Yanko
    Nov 25 at 14:12










  • @TestGuest Yes, that's right.
    – saulspatz
    Nov 25 at 14:12
















1














I have never come across a statement linking linearity and monotony - but it seems that for each linear function (positive, negative or even constant slope), the function is monotonic:



I.e. for y $geq$ x it follows that f(y) $geq$ f(x).



Is this correct?






analysis monotone-functions







share|cite|improve this question






















  • Yes, it's monotonic, but not necessarily increasing. You could have $yge ximplies f(y)le f(x)$
    – saulspatz
    Nov 25 at 14:08










  • Exactly, so positive slope = increasing; negative slope = decreasing, in both cases strictly monotonic as well -- constant slope = monotonic, but not strictly. Is this correct?
    – TestGuest
    Nov 25 at 14:10












  • Note that your question only makes sense for 1-dimensional functions, if you want to talk about linear functions in general, say $f:mathbb{R}^nrightarrow mathbb{R}^m$ or any vector spaces then you need to define what you mean by $xleq y$
    – Yanko
    Nov 25 at 14:11










  • @TestGuest yes this is the point.
    – Yanko
    Nov 25 at 14:12










  • @TestGuest Yes, that's right.
    – saulspatz
    Nov 25 at 14:12














1












1








1







I have never come across a statement linking linearity and monotony - but it seems that for each linear function (positive, negative or even constant slope), the function is monotonic:



I.e. for y $geq$ x it follows that f(y) $geq$ f(x).



Is this correct?






analysis monotone-functions







share|cite|improve this question













I have never come across a statement linking linearity and monotony - but it seems that for each linear function (positive, negative or even constant slope), the function is monotonic:



I.e. for y $geq$ x it follows that f(y) $geq$ f(x).



Is this correct?






analysis monotone-functions




analysis monotone-functions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 25 at 14:04









TestGuest

405821




405821












  • Yes, it's monotonic, but not necessarily increasing. You could have $yge ximplies f(y)le f(x)$
    – saulspatz
    Nov 25 at 14:08










  • Exactly, so positive slope = increasing; negative slope = decreasing, in both cases strictly monotonic as well -- constant slope = monotonic, but not strictly. Is this correct?
    – TestGuest
    Nov 25 at 14:10












  • Note that your question only makes sense for 1-dimensional functions, if you want to talk about linear functions in general, say $f:mathbb{R}^nrightarrow mathbb{R}^m$ or any vector spaces then you need to define what you mean by $xleq y$
    – Yanko
    Nov 25 at 14:11










  • @TestGuest yes this is the point.
    – Yanko
    Nov 25 at 14:12










  • @TestGuest Yes, that's right.
    – saulspatz
    Nov 25 at 14:12


















  • Yes, it's monotonic, but not necessarily increasing. You could have $yge ximplies f(y)le f(x)$
    – saulspatz
    Nov 25 at 14:08










  • Exactly, so positive slope = increasing; negative slope = decreasing, in both cases strictly monotonic as well -- constant slope = monotonic, but not strictly. Is this correct?
    – TestGuest
    Nov 25 at 14:10












  • Note that your question only makes sense for 1-dimensional functions, if you want to talk about linear functions in general, say $f:mathbb{R}^nrightarrow mathbb{R}^m$ or any vector spaces then you need to define what you mean by $xleq y$
    – Yanko
    Nov 25 at 14:11










  • @TestGuest yes this is the point.
    – Yanko
    Nov 25 at 14:12










  • @TestGuest Yes, that's right.
    – saulspatz
    Nov 25 at 14:12
















Yes, it's monotonic, but not necessarily increasing. You could have $yge ximplies f(y)le f(x)$
– saulspatz
Nov 25 at 14:08




Yes, it's monotonic, but not necessarily increasing. You could have $yge ximplies f(y)le f(x)$
– saulspatz
Nov 25 at 14:08












Exactly, so positive slope = increasing; negative slope = decreasing, in both cases strictly monotonic as well -- constant slope = monotonic, but not strictly. Is this correct?
– TestGuest
Nov 25 at 14:10






Exactly, so positive slope = increasing; negative slope = decreasing, in both cases strictly monotonic as well -- constant slope = monotonic, but not strictly. Is this correct?
– TestGuest
Nov 25 at 14:10














Note that your question only makes sense for 1-dimensional functions, if you want to talk about linear functions in general, say $f:mathbb{R}^nrightarrow mathbb{R}^m$ or any vector spaces then you need to define what you mean by $xleq y$
– Yanko
Nov 25 at 14:11




Note that your question only makes sense for 1-dimensional functions, if you want to talk about linear functions in general, say $f:mathbb{R}^nrightarrow mathbb{R}^m$ or any vector spaces then you need to define what you mean by $xleq y$
– Yanko
Nov 25 at 14:11












@TestGuest yes this is the point.
– Yanko
Nov 25 at 14:12




@TestGuest yes this is the point.
– Yanko
Nov 25 at 14:12












@TestGuest Yes, that's right.
– saulspatz
Nov 25 at 14:12




@TestGuest Yes, that's right.
– saulspatz
Nov 25 at 14:12










1 Answer
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1














Not necessarily monotonic increasing, consider $f:mathbb{R}rightarrowmathbb{R}$ with $f(x)=-x$



clearly $f$ is linear but $1geq 0$ and $f(1)=-1<0=f(0)$.



In general if $f$ is linear then so is $-f$ and it is impossible that both are monotonic increasing.



It is true however that every linear function is monotonic because every linear function from $mathbb{R}rightarrowmathbb{R}$ takes the form $f(x)=ax$ where $a=f(1)$.






share|cite|improve this answer





















  • Thanks! How is a = f(1) a proof of monotonicity, however?
    – TestGuest
    Nov 25 at 14:14






  • 1




    @TestGuest I just meant that $f(x)=ax$ is necessarily monotone. If $a>0$ it is increasing, if $a=0$ well it's constant and if $a<0$ then decreasing. The fact that $a=f(1)$ is unnecessary.
    – Yanko
    Nov 25 at 14:15













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1 Answer
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1














Not necessarily monotonic increasing, consider $f:mathbb{R}rightarrowmathbb{R}$ with $f(x)=-x$



clearly $f$ is linear but $1geq 0$ and $f(1)=-1<0=f(0)$.



In general if $f$ is linear then so is $-f$ and it is impossible that both are monotonic increasing.



It is true however that every linear function is monotonic because every linear function from $mathbb{R}rightarrowmathbb{R}$ takes the form $f(x)=ax$ where $a=f(1)$.






share|cite|improve this answer





















  • Thanks! How is a = f(1) a proof of monotonicity, however?
    – TestGuest
    Nov 25 at 14:14






  • 1




    @TestGuest I just meant that $f(x)=ax$ is necessarily monotone. If $a>0$ it is increasing, if $a=0$ well it's constant and if $a<0$ then decreasing. The fact that $a=f(1)$ is unnecessary.
    – Yanko
    Nov 25 at 14:15


















1














Not necessarily monotonic increasing, consider $f:mathbb{R}rightarrowmathbb{R}$ with $f(x)=-x$



clearly $f$ is linear but $1geq 0$ and $f(1)=-1<0=f(0)$.



In general if $f$ is linear then so is $-f$ and it is impossible that both are monotonic increasing.



It is true however that every linear function is monotonic because every linear function from $mathbb{R}rightarrowmathbb{R}$ takes the form $f(x)=ax$ where $a=f(1)$.






share|cite|improve this answer





















  • Thanks! How is a = f(1) a proof of monotonicity, however?
    – TestGuest
    Nov 25 at 14:14






  • 1




    @TestGuest I just meant that $f(x)=ax$ is necessarily monotone. If $a>0$ it is increasing, if $a=0$ well it's constant and if $a<0$ then decreasing. The fact that $a=f(1)$ is unnecessary.
    – Yanko
    Nov 25 at 14:15
















1












1








1






Not necessarily monotonic increasing, consider $f:mathbb{R}rightarrowmathbb{R}$ with $f(x)=-x$



clearly $f$ is linear but $1geq 0$ and $f(1)=-1<0=f(0)$.



In general if $f$ is linear then so is $-f$ and it is impossible that both are monotonic increasing.



It is true however that every linear function is monotonic because every linear function from $mathbb{R}rightarrowmathbb{R}$ takes the form $f(x)=ax$ where $a=f(1)$.






share|cite|improve this answer












Not necessarily monotonic increasing, consider $f:mathbb{R}rightarrowmathbb{R}$ with $f(x)=-x$



clearly $f$ is linear but $1geq 0$ and $f(1)=-1<0=f(0)$.



In general if $f$ is linear then so is $-f$ and it is impossible that both are monotonic increasing.



It is true however that every linear function is monotonic because every linear function from $mathbb{R}rightarrowmathbb{R}$ takes the form $f(x)=ax$ where $a=f(1)$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 25 at 14:08









Yanko

5,829723




5,829723












  • Thanks! How is a = f(1) a proof of monotonicity, however?
    – TestGuest
    Nov 25 at 14:14






  • 1




    @TestGuest I just meant that $f(x)=ax$ is necessarily monotone. If $a>0$ it is increasing, if $a=0$ well it's constant and if $a<0$ then decreasing. The fact that $a=f(1)$ is unnecessary.
    – Yanko
    Nov 25 at 14:15




















  • Thanks! How is a = f(1) a proof of monotonicity, however?
    – TestGuest
    Nov 25 at 14:14






  • 1




    @TestGuest I just meant that $f(x)=ax$ is necessarily monotone. If $a>0$ it is increasing, if $a=0$ well it's constant and if $a<0$ then decreasing. The fact that $a=f(1)$ is unnecessary.
    – Yanko
    Nov 25 at 14:15


















Thanks! How is a = f(1) a proof of monotonicity, however?
– TestGuest
Nov 25 at 14:14




Thanks! How is a = f(1) a proof of monotonicity, however?
– TestGuest
Nov 25 at 14:14




1




1




@TestGuest I just meant that $f(x)=ax$ is necessarily monotone. If $a>0$ it is increasing, if $a=0$ well it's constant and if $a<0$ then decreasing. The fact that $a=f(1)$ is unnecessary.
– Yanko
Nov 25 at 14:15






@TestGuest I just meant that $f(x)=ax$ is necessarily monotone. If $a>0$ it is increasing, if $a=0$ well it's constant and if $a<0$ then decreasing. The fact that $a=f(1)$ is unnecessary.
– Yanko
Nov 25 at 14:15




















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