Prove increasing convex function has increasing differences
Let $v: mathbb{R} to mathbb{R}$ be an increasing, convex function.
For any $t>0$ I want to show that for all $x_{1} leq x_{2}$ we have: $$v(x_{1}+t) - v(x_{1}) leq v(x_{2} +t) - v(x_{2})$$
This of course can be illustrated heuristically if $v$ is twice differentiable. But I am trying to show this from the definition of a convex function and by the fact that $v$ is increasing, but I am just moving in circles. I am pretty sure this result is true, and I need it to finish a proof I am working on.
Any suggestions will help.
real-analysis convex-analysis
asked Oct 9 '15 at 18:03
möbius
849721
add a comment |
Let $v: mathbb{R} to mathbb{R}$ be an increasing, convex function.
For any $t>0$ I want to show that for all $x_{1} leq x_{2}$ we have: $$v(x_{1}+t) - v(x_{1}) leq v(x_{2} +t) - v(x_{2})$$
This of course can be illustrated heuristically if $v$ is twice differentiable. But I am trying to show this from the definition of a convex function and by the fact that $v$ is increasing, but I am just moving in circles. I am pretty sure this result is true, and I need it to finish a proof I am working on.
Any suggestions will help.
real-analysis convex-analysis
asked Oct 9 '15 at 18:03
möbius
849721
add a comment |
Let $v: mathbb{R} to mathbb{R}$ be an increasing, convex function.
For any $t>0$ I want to show that for all $x_{1} leq x_{2}$ we have: $$v(x_{1}+t) - v(x_{1}) leq v(x_{2} +t) - v(x_{2})$$
This of course can be illustrated heuristically if $v$ is twice differentiable. But I am trying to show this from the definition of a convex function and by the fact that $v$ is increasing, but I am just moving in circles. I am pretty sure this result is true, and I need it to finish a proof I am working on.
Any suggestions will help.
real-analysis convex-analysis
asked Oct 9 '15 at 18:03
möbius
849721
Let $v: mathbb{R} to mathbb{R}$ be an increasing, convex function.
For any $t>0$ I want to show that for all $x_{1} leq x_{2}$ we have: $$v(x_{1}+t) - v(x_{1}) leq v(x_{2} +t) - v(x_{2})$$
This of course can be illustrated heuristically if $v$ is twice differentiable. But I am trying to show this from the definition of a convex function and by the fact that $v$ is increasing, but I am just moving in circles. I am pretty sure this result is true, and I need it to finish a proof I am working on.
Any suggestions will help.
real-analysis convex-analysis
real-analysis convex-analysis
asked Oct 9 '15 at 18:03
möbius
849721
asked Oct 9 '15 at 18:03
möbius
849721
asked Oct 9 '15 at 18:03
möbius
849721
asked Oct 9 '15 at 18:03
möbius
849721
asked Oct 9 '15 at 18:03
möbius
849721
849721
add a comment |
add a comment |
2 Answers
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Since $x_1 le x_1+t,x_2 le x_2+t$, from the definition of convexity we have
$$ v(x_1+t) le left(frac{x_2-x_1}{x_2-x_1+t}right)v(x_1) + left(frac{t}{x_2-x_1+t}right)v(x_2+t) $$
and
$$ v(x_2) le left(frac{t}{x_2-x_1+t}right)v(x_1) + left(frac{x_2-x_1}{x_2-x_1+t}right)v(x_2+t).$$
Adding up the inequalities gives the desired inequality.
answered Oct 9 '15 at 18:20
Joey Zou
7,47721131
I think this is a great and succinct answer. However, before I accept and upvote, I am wondering if you can explain the steps you took to get to this solution? I worked for a while on this problem and could not get here, so I want to understand how you reached this point.
– möbius
Oct 9 '15 at 18:33
Nevermind, I see the logic. Thanks for your help.
– möbius
Oct 9 '15 at 18:42
add a comment |
[This inequality is true for any convex function, increasing or not.]
I'm going to assume that $x_2le x_1+t$. The reasoning in the remaining case is similar. First consider the three points $x_1le x_2le x_1+t$. We have
$$
x_2={x_1+t-x_2over t}x_1+{x_2-x_1over t}(x_1+t),
$$
so by the convexity inequality for $nu$,
$$
nu(x_2)le {x_1+t-x_2over t}nu(x_1)+{x_2-x_1over t}nu(x_1+t).
$$
Similarly,
$$
nu(x_1+t)le {x_2-x_1over t}nu(x_2)+{x_1-x_2+tover t}nu(x_2+t).
$$
Now add these two inequalities, clear $t$ from the denominator, and simplify.
answered Oct 9 '15 at 18:22
John Dawkins
13k11017
add a comment |
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2 Answers
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2 Answers
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Since $x_1 le x_1+t,x_2 le x_2+t$, from the definition of convexity we have
$$ v(x_1+t) le left(frac{x_2-x_1}{x_2-x_1+t}right)v(x_1) + left(frac{t}{x_2-x_1+t}right)v(x_2+t) $$
and
$$ v(x_2) le left(frac{t}{x_2-x_1+t}right)v(x_1) + left(frac{x_2-x_1}{x_2-x_1+t}right)v(x_2+t).$$
Adding up the inequalities gives the desired inequality.
answered Oct 9 '15 at 18:20
Joey Zou
7,47721131
I think this is a great and succinct answer. However, before I accept and upvote, I am wondering if you can explain the steps you took to get to this solution? I worked for a while on this problem and could not get here, so I want to understand how you reached this point.
– möbius
Oct 9 '15 at 18:33
Nevermind, I see the logic. Thanks for your help.
– möbius
Oct 9 '15 at 18:42
add a comment |
Since $x_1 le x_1+t,x_2 le x_2+t$, from the definition of convexity we have
$$ v(x_1+t) le left(frac{x_2-x_1}{x_2-x_1+t}right)v(x_1) + left(frac{t}{x_2-x_1+t}right)v(x_2+t) $$
and
$$ v(x_2) le left(frac{t}{x_2-x_1+t}right)v(x_1) + left(frac{x_2-x_1}{x_2-x_1+t}right)v(x_2+t).$$
Adding up the inequalities gives the desired inequality.
answered Oct 9 '15 at 18:20
Joey Zou
7,47721131
I think this is a great and succinct answer. However, before I accept and upvote, I am wondering if you can explain the steps you took to get to this solution? I worked for a while on this problem and could not get here, so I want to understand how you reached this point.
– möbius
Oct 9 '15 at 18:33
Nevermind, I see the logic. Thanks for your help.
– möbius
Oct 9 '15 at 18:42
add a comment |
Since $x_1 le x_1+t,x_2 le x_2+t$, from the definition of convexity we have
$$ v(x_1+t) le left(frac{x_2-x_1}{x_2-x_1+t}right)v(x_1) + left(frac{t}{x_2-x_1+t}right)v(x_2+t) $$
and
$$ v(x_2) le left(frac{t}{x_2-x_1+t}right)v(x_1) + left(frac{x_2-x_1}{x_2-x_1+t}right)v(x_2+t).$$
Adding up the inequalities gives the desired inequality.
answered Oct 9 '15 at 18:20
Joey Zou
7,47721131
Since $x_1 le x_1+t,x_2 le x_2+t$, from the definition of convexity we have
$$ v(x_1+t) le left(frac{x_2-x_1}{x_2-x_1+t}right)v(x_1) + left(frac{t}{x_2-x_1+t}right)v(x_2+t) $$
and
$$ v(x_2) le left(frac{t}{x_2-x_1+t}right)v(x_1) + left(frac{x_2-x_1}{x_2-x_1+t}right)v(x_2+t).$$
Adding up the inequalities gives the desired inequality.
answered Oct 9 '15 at 18:20
Joey Zou
7,47721131
answered Oct 9 '15 at 18:20
Joey Zou
7,47721131
answered Oct 9 '15 at 18:20
Joey Zou
7,47721131
answered Oct 9 '15 at 18:20
Joey Zou
7,47721131
7,47721131
I think this is a great and succinct answer. However, before I accept and upvote, I am wondering if you can explain the steps you took to get to this solution? I worked for a while on this problem and could not get here, so I want to understand how you reached this point.
– möbius
Oct 9 '15 at 18:33
Nevermind, I see the logic. Thanks for your help.
– möbius
Oct 9 '15 at 18:42
add a comment |
I think this is a great and succinct answer. However, before I accept and upvote, I am wondering if you can explain the steps you took to get to this solution? I worked for a while on this problem and could not get here, so I want to understand how you reached this point.
– möbius
Oct 9 '15 at 18:33
Nevermind, I see the logic. Thanks for your help.
– möbius
Oct 9 '15 at 18:42
I think this is a great and succinct answer. However, before I accept and upvote, I am wondering if you can explain the steps you took to get to this solution? I worked for a while on this problem and could not get here, so I want to understand how you reached this point.
– möbius
Oct 9 '15 at 18:33
I think this is a great and succinct answer. However, before I accept and upvote, I am wondering if you can explain the steps you took to get to this solution? I worked for a while on this problem and could not get here, so I want to understand how you reached this point.
– möbius
Oct 9 '15 at 18:33
Nevermind, I see the logic. Thanks for your help.
– möbius
Oct 9 '15 at 18:42
Nevermind, I see the logic. Thanks for your help.
– möbius
Oct 9 '15 at 18:42
add a comment |
[This inequality is true for any convex function, increasing or not.]
I'm going to assume that $x_2le x_1+t$. The reasoning in the remaining case is similar. First consider the three points $x_1le x_2le x_1+t$. We have
$$
x_2={x_1+t-x_2over t}x_1+{x_2-x_1over t}(x_1+t),
$$
so by the convexity inequality for $nu$,
$$
nu(x_2)le {x_1+t-x_2over t}nu(x_1)+{x_2-x_1over t}nu(x_1+t).
$$
Similarly,
$$
nu(x_1+t)le {x_2-x_1over t}nu(x_2)+{x_1-x_2+tover t}nu(x_2+t).
$$
Now add these two inequalities, clear $t$ from the denominator, and simplify.
answered Oct 9 '15 at 18:22
John Dawkins
13k11017
add a comment |
[This inequality is true for any convex function, increasing or not.]
I'm going to assume that $x_2le x_1+t$. The reasoning in the remaining case is similar. First consider the three points $x_1le x_2le x_1+t$. We have
$$
x_2={x_1+t-x_2over t}x_1+{x_2-x_1over t}(x_1+t),
$$
so by the convexity inequality for $nu$,
$$
nu(x_2)le {x_1+t-x_2over t}nu(x_1)+{x_2-x_1over t}nu(x_1+t).
$$
Similarly,
$$
nu(x_1+t)le {x_2-x_1over t}nu(x_2)+{x_1-x_2+tover t}nu(x_2+t).
$$
Now add these two inequalities, clear $t$ from the denominator, and simplify.
answered Oct 9 '15 at 18:22
John Dawkins
13k11017
add a comment |
[This inequality is true for any convex function, increasing or not.]
I'm going to assume that $x_2le x_1+t$. The reasoning in the remaining case is similar. First consider the three points $x_1le x_2le x_1+t$. We have
$$
x_2={x_1+t-x_2over t}x_1+{x_2-x_1over t}(x_1+t),
$$
so by the convexity inequality for $nu$,
$$
nu(x_2)le {x_1+t-x_2over t}nu(x_1)+{x_2-x_1over t}nu(x_1+t).
$$
Similarly,
$$
nu(x_1+t)le {x_2-x_1over t}nu(x_2)+{x_1-x_2+tover t}nu(x_2+t).
$$
Now add these two inequalities, clear $t$ from the denominator, and simplify.
answered Oct 9 '15 at 18:22
John Dawkins
13k11017
[This inequality is true for any convex function, increasing or not.]
I'm going to assume that $x_2le x_1+t$. The reasoning in the remaining case is similar. First consider the three points $x_1le x_2le x_1+t$. We have
$$
x_2={x_1+t-x_2over t}x_1+{x_2-x_1over t}(x_1+t),
$$
so by the convexity inequality for $nu$,
$$
nu(x_2)le {x_1+t-x_2over t}nu(x_1)+{x_2-x_1over t}nu(x_1+t).
$$
Similarly,
$$
nu(x_1+t)le {x_2-x_1over t}nu(x_2)+{x_1-x_2+tover t}nu(x_2+t).
$$
Now add these two inequalities, clear $t$ from the denominator, and simplify.
answered Oct 9 '15 at 18:22
John Dawkins
13k11017
answered Oct 9 '15 at 18:22
John Dawkins
13k11017
answered Oct 9 '15 at 18:22
John Dawkins
13k11017
answered Oct 9 '15 at 18:22
John Dawkins
13k11017
13k11017
add a comment |
add a comment |
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