Test if vector valued functions is increasing?












0












$begingroup$


Edited in response to @Randall's question, although I'm still not sure if this question makes sense.



I am trying to find out whether there exists a simple way to tell if a function from $quad{mathbb R}^nto{mathbb R}^n$ is strictly increasing. For example, with functions from $quad{mathbb R}to{mathbb R}$, I would take the first derivative and try to show that it is positive on the interval I'm interested in.



In higher dimensions, the function would have to be from $quad{mathbb R}^nto{mathbb R}^n$ so that the input and output vectors would be of the same dimensions. Then (as I understand it) the function $f : mathbb{R}^n to
mathbb{R}^n$ is strictly increasing if for all $mathbf{x_1}, mathbf{x_2} in mathbb{R}^n, mathbf{x_1} gtgt mathbf{x_2}$, we have $ mathbf{y_1}=f(mathbf{x_1}) gtgt mathbf{y_2}=f(mathbf{x_2})$, where $mathbf{x_1} gtgt mathbf{x_2}$ means that each component of $ mathbf{x_1}$ is greater than than the corresponding component of $mathbf{x_2}$.



Is there a test like looking at the first derivative, but in higher dimensions? Thanks in advance for the help.






multivariable-calculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What would this even mean? What's the ordering on points in the plane?
    $endgroup$
    – Randall
    Jan 15 at 20:24










  • $begingroup$
    @Randall I'm sorry, I don't understand your question.
    $endgroup$
    – Julia B
    Jan 15 at 20:31






  • 1




    $begingroup$
    There is no such notion of increasing function in higher dimensions because there is no sensible $leq$.
    $endgroup$
    – Randall
    Jan 15 at 20:32
















0












$begingroup$


Edited in response to @Randall's question, although I'm still not sure if this question makes sense.



I am trying to find out whether there exists a simple way to tell if a function from $quad{mathbb R}^nto{mathbb R}^n$ is strictly increasing. For example, with functions from $quad{mathbb R}to{mathbb R}$, I would take the first derivative and try to show that it is positive on the interval I'm interested in.



In higher dimensions, the function would have to be from $quad{mathbb R}^nto{mathbb R}^n$ so that the input and output vectors would be of the same dimensions. Then (as I understand it) the function $f : mathbb{R}^n to
mathbb{R}^n$ is strictly increasing if for all $mathbf{x_1}, mathbf{x_2} in mathbb{R}^n, mathbf{x_1} gtgt mathbf{x_2}$, we have $ mathbf{y_1}=f(mathbf{x_1}) gtgt mathbf{y_2}=f(mathbf{x_2})$, where $mathbf{x_1} gtgt mathbf{x_2}$ means that each component of $ mathbf{x_1}$ is greater than than the corresponding component of $mathbf{x_2}$.



Is there a test like looking at the first derivative, but in higher dimensions? Thanks in advance for the help.






multivariable-calculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What would this even mean? What's the ordering on points in the plane?
    $endgroup$
    – Randall
    Jan 15 at 20:24










  • $begingroup$
    @Randall I'm sorry, I don't understand your question.
    $endgroup$
    – Julia B
    Jan 15 at 20:31






  • 1




    $begingroup$
    There is no such notion of increasing function in higher dimensions because there is no sensible $leq$.
    $endgroup$
    – Randall
    Jan 15 at 20:32














0












0








0





$begingroup$


Edited in response to @Randall's question, although I'm still not sure if this question makes sense.



I am trying to find out whether there exists a simple way to tell if a function from $quad{mathbb R}^nto{mathbb R}^n$ is strictly increasing. For example, with functions from $quad{mathbb R}to{mathbb R}$, I would take the first derivative and try to show that it is positive on the interval I'm interested in.



In higher dimensions, the function would have to be from $quad{mathbb R}^nto{mathbb R}^n$ so that the input and output vectors would be of the same dimensions. Then (as I understand it) the function $f : mathbb{R}^n to
mathbb{R}^n$ is strictly increasing if for all $mathbf{x_1}, mathbf{x_2} in mathbb{R}^n, mathbf{x_1} gtgt mathbf{x_2}$, we have $ mathbf{y_1}=f(mathbf{x_1}) gtgt mathbf{y_2}=f(mathbf{x_2})$, where $mathbf{x_1} gtgt mathbf{x_2}$ means that each component of $ mathbf{x_1}$ is greater than than the corresponding component of $mathbf{x_2}$.



Is there a test like looking at the first derivative, but in higher dimensions? Thanks in advance for the help.






multivariable-calculus







share|cite|improve this question











$endgroup$




Edited in response to @Randall's question, although I'm still not sure if this question makes sense.



I am trying to find out whether there exists a simple way to tell if a function from $quad{mathbb R}^nto{mathbb R}^n$ is strictly increasing. For example, with functions from $quad{mathbb R}to{mathbb R}$, I would take the first derivative and try to show that it is positive on the interval I'm interested in.



In higher dimensions, the function would have to be from $quad{mathbb R}^nto{mathbb R}^n$ so that the input and output vectors would be of the same dimensions. Then (as I understand it) the function $f : mathbb{R}^n to
mathbb{R}^n$ is strictly increasing if for all $mathbf{x_1}, mathbf{x_2} in mathbb{R}^n, mathbf{x_1} gtgt mathbf{x_2}$, we have $ mathbf{y_1}=f(mathbf{x_1}) gtgt mathbf{y_2}=f(mathbf{x_2})$, where $mathbf{x_1} gtgt mathbf{x_2}$ means that each component of $ mathbf{x_1}$ is greater than than the corresponding component of $mathbf{x_2}$.



Is there a test like looking at the first derivative, but in higher dimensions? Thanks in advance for the help.






multivariable-calculus




multivariable-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 at 21:10







Julia B

















asked Jan 15 at 20:22









Julia BJulia B

64




64








  • 1




    $begingroup$
    What would this even mean? What's the ordering on points in the plane?
    $endgroup$
    – Randall
    Jan 15 at 20:24










  • $begingroup$
    @Randall I'm sorry, I don't understand your question.
    $endgroup$
    – Julia B
    Jan 15 at 20:31






  • 1




    $begingroup$
    There is no such notion of increasing function in higher dimensions because there is no sensible $leq$.
    $endgroup$
    – Randall
    Jan 15 at 20:32














  • 1




    $begingroup$
    What would this even mean? What's the ordering on points in the plane?
    $endgroup$
    – Randall
    Jan 15 at 20:24










  • $begingroup$
    @Randall I'm sorry, I don't understand your question.
    $endgroup$
    – Julia B
    Jan 15 at 20:31






  • 1




    $begingroup$
    There is no such notion of increasing function in higher dimensions because there is no sensible $leq$.
    $endgroup$
    – Randall
    Jan 15 at 20:32








1




1




$begingroup$
What would this even mean? What's the ordering on points in the plane?
$endgroup$
– Randall
Jan 15 at 20:24




$begingroup$
What would this even mean? What's the ordering on points in the plane?
$endgroup$
– Randall
Jan 15 at 20:24












$begingroup$
@Randall I'm sorry, I don't understand your question.
$endgroup$
– Julia B
Jan 15 at 20:31




$begingroup$
@Randall I'm sorry, I don't understand your question.
$endgroup$
– Julia B
Jan 15 at 20:31




1




1




$begingroup$
There is no such notion of increasing function in higher dimensions because there is no sensible $leq$.
$endgroup$
– Randall
Jan 15 at 20:32




$begingroup$
There is no such notion of increasing function in higher dimensions because there is no sensible $leq$.
$endgroup$
– Randall
Jan 15 at 20:32










1 Answer
1






active

oldest

votes


















0












$begingroup$

If I understand you question correctly, my answer should provide a sufficient condition. If $f:Bbb R^ntoBbb R^n$ is differentiable, then it is increasing if
$$
frac {partial f_j(xi)}{partial x_i} ge 0 quadtext{for all $i,j=1,2,dots,n$}
$$
and all vectors $xiin Bbb R^n$. You can replace the $ge$ sign with $>$ if you want your function to be strictly increasing.



The proof that this works is as follows: Suppose that $mathbf x<<mathbf y$ then each component of $mathbf v=mathbf y -mathbf x$ is greater than zero. By the fundamental theorem of calculus,
$$begin{align}
f(mathbf y) &= f(mathbf x) + left(int_0^1 Df(mathbf x + tmathbf v) ,dtright)mathbf v.
end{align}$$
The above means that for each $j$ we have
$$begin{align}
f_j(mathbf y) - f_j(mathbf x) = sum_{i=1}^n int_0^1frac {partial f(mathbf x + tmathbf v)}{partial x_i}, v_i ,dt ge 0
end{align}$$
by our assumption on the derivatives of $f$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This was my intuition - thank you very much for the response and particularly for the proof! I'll go through it in detail to make sure I understand it.
    $endgroup$
    – Julia B
    Jan 15 at 21:48










  • $begingroup$
    Sorry I was being careless with the MVT, I changed it to FTC instead.
    $endgroup$
    – BigbearZzz
    Jan 15 at 21:54














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1 Answer
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1 Answer
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active

oldest

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active

oldest

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0












$begingroup$

If I understand you question correctly, my answer should provide a sufficient condition. If $f:Bbb R^ntoBbb R^n$ is differentiable, then it is increasing if
$$
frac {partial f_j(xi)}{partial x_i} ge 0 quadtext{for all $i,j=1,2,dots,n$}
$$
and all vectors $xiin Bbb R^n$. You can replace the $ge$ sign with $>$ if you want your function to be strictly increasing.



The proof that this works is as follows: Suppose that $mathbf x<<mathbf y$ then each component of $mathbf v=mathbf y -mathbf x$ is greater than zero. By the fundamental theorem of calculus,
$$begin{align}
f(mathbf y) &= f(mathbf x) + left(int_0^1 Df(mathbf x + tmathbf v) ,dtright)mathbf v.
end{align}$$
The above means that for each $j$ we have
$$begin{align}
f_j(mathbf y) - f_j(mathbf x) = sum_{i=1}^n int_0^1frac {partial f(mathbf x + tmathbf v)}{partial x_i}, v_i ,dt ge 0
end{align}$$
by our assumption on the derivatives of $f$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This was my intuition - thank you very much for the response and particularly for the proof! I'll go through it in detail to make sure I understand it.
    $endgroup$
    – Julia B
    Jan 15 at 21:48










  • $begingroup$
    Sorry I was being careless with the MVT, I changed it to FTC instead.
    $endgroup$
    – BigbearZzz
    Jan 15 at 21:54


















0












$begingroup$

If I understand you question correctly, my answer should provide a sufficient condition. If $f:Bbb R^ntoBbb R^n$ is differentiable, then it is increasing if
$$
frac {partial f_j(xi)}{partial x_i} ge 0 quadtext{for all $i,j=1,2,dots,n$}
$$
and all vectors $xiin Bbb R^n$. You can replace the $ge$ sign with $>$ if you want your function to be strictly increasing.



The proof that this works is as follows: Suppose that $mathbf x<<mathbf y$ then each component of $mathbf v=mathbf y -mathbf x$ is greater than zero. By the fundamental theorem of calculus,
$$begin{align}
f(mathbf y) &= f(mathbf x) + left(int_0^1 Df(mathbf x + tmathbf v) ,dtright)mathbf v.
end{align}$$
The above means that for each $j$ we have
$$begin{align}
f_j(mathbf y) - f_j(mathbf x) = sum_{i=1}^n int_0^1frac {partial f(mathbf x + tmathbf v)}{partial x_i}, v_i ,dt ge 0
end{align}$$
by our assumption on the derivatives of $f$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This was my intuition - thank you very much for the response and particularly for the proof! I'll go through it in detail to make sure I understand it.
    $endgroup$
    – Julia B
    Jan 15 at 21:48










  • $begingroup$
    Sorry I was being careless with the MVT, I changed it to FTC instead.
    $endgroup$
    – BigbearZzz
    Jan 15 at 21:54
















0












0








0





$begingroup$

If I understand you question correctly, my answer should provide a sufficient condition. If $f:Bbb R^ntoBbb R^n$ is differentiable, then it is increasing if
$$
frac {partial f_j(xi)}{partial x_i} ge 0 quadtext{for all $i,j=1,2,dots,n$}
$$
and all vectors $xiin Bbb R^n$. You can replace the $ge$ sign with $>$ if you want your function to be strictly increasing.



The proof that this works is as follows: Suppose that $mathbf x<<mathbf y$ then each component of $mathbf v=mathbf y -mathbf x$ is greater than zero. By the fundamental theorem of calculus,
$$begin{align}
f(mathbf y) &= f(mathbf x) + left(int_0^1 Df(mathbf x + tmathbf v) ,dtright)mathbf v.
end{align}$$
The above means that for each $j$ we have
$$begin{align}
f_j(mathbf y) - f_j(mathbf x) = sum_{i=1}^n int_0^1frac {partial f(mathbf x + tmathbf v)}{partial x_i}, v_i ,dt ge 0
end{align}$$
by our assumption on the derivatives of $f$.






share|cite|improve this answer











$endgroup$



If I understand you question correctly, my answer should provide a sufficient condition. If $f:Bbb R^ntoBbb R^n$ is differentiable, then it is increasing if
$$
frac {partial f_j(xi)}{partial x_i} ge 0 quadtext{for all $i,j=1,2,dots,n$}
$$
and all vectors $xiin Bbb R^n$. You can replace the $ge$ sign with $>$ if you want your function to be strictly increasing.



The proof that this works is as follows: Suppose that $mathbf x<<mathbf y$ then each component of $mathbf v=mathbf y -mathbf x$ is greater than zero. By the fundamental theorem of calculus,
$$begin{align}
f(mathbf y) &= f(mathbf x) + left(int_0^1 Df(mathbf x + tmathbf v) ,dtright)mathbf v.
end{align}$$
The above means that for each $j$ we have
$$begin{align}
f_j(mathbf y) - f_j(mathbf x) = sum_{i=1}^n int_0^1frac {partial f(mathbf x + tmathbf v)}{partial x_i}, v_i ,dt ge 0
end{align}$$
by our assumption on the derivatives of $f$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 15 at 21:54

























answered Jan 15 at 21:41









BigbearZzzBigbearZzz

8,96021652




8,96021652












  • $begingroup$
    This was my intuition - thank you very much for the response and particularly for the proof! I'll go through it in detail to make sure I understand it.
    $endgroup$
    – Julia B
    Jan 15 at 21:48










  • $begingroup$
    Sorry I was being careless with the MVT, I changed it to FTC instead.
    $endgroup$
    – BigbearZzz
    Jan 15 at 21:54




















  • $begingroup$
    This was my intuition - thank you very much for the response and particularly for the proof! I'll go through it in detail to make sure I understand it.
    $endgroup$
    – Julia B
    Jan 15 at 21:48










  • $begingroup$
    Sorry I was being careless with the MVT, I changed it to FTC instead.
    $endgroup$
    – BigbearZzz
    Jan 15 at 21:54


















$begingroup$
This was my intuition - thank you very much for the response and particularly for the proof! I'll go through it in detail to make sure I understand it.
$endgroup$
– Julia B
Jan 15 at 21:48




$begingroup$
This was my intuition - thank you very much for the response and particularly for the proof! I'll go through it in detail to make sure I understand it.
$endgroup$
– Julia B
Jan 15 at 21:48












$begingroup$
Sorry I was being careless with the MVT, I changed it to FTC instead.
$endgroup$
– BigbearZzz
Jan 15 at 21:54






$begingroup$
Sorry I was being careless with the MVT, I changed it to FTC instead.
$endgroup$
– BigbearZzz
Jan 15 at 21:54




















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