Are all linear functions monotonic?
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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
asked Nov 25 at 14:04
TestGuest
400721
add a comment |
up vote
1
down vote
favorite
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
asked Nov 25 at 14:04
TestGuest
400721
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
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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
asked Nov 25 at 14:04
TestGuest
400721
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
asked Nov 25 at 14:04
TestGuest
400721
asked Nov 25 at 14:04
TestGuest
400721
asked Nov 25 at 14:04
TestGuest
400721
asked Nov 25 at 14:04
TestGuest
400721
asked Nov 25 at 14:04
TestGuest
400721
400721
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
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accepted
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 $a=f(1)$.
answered Nov 25 at 14:08
Yanko
5,123722
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
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
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 $a=f(1)$.
answered Nov 25 at 14:08
Yanko
5,123722
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
add a comment |
up vote
1
down vote
accepted
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 $a=f(1)$.
answered Nov 25 at 14:08
Yanko
5,123722
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
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
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 $a=f(1)$.
answered Nov 25 at 14:08
Yanko
5,123722
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 $a=f(1)$.
answered Nov 25 at 14:08
Yanko
5,123722
answered Nov 25 at 14:08
Yanko
5,123722
answered Nov 25 at 14:08
Yanko
5,123722
answered Nov 25 at 14:08
Yanko
5,123722
5,123722
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
add a comment |
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
add a comment |
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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