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Showing that $x mapsto |x|^p$ is strictly mid-point convex for $2 leq p < infty$ [duplicate]
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Elementary proof that $|x|^p$ is convex.
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Let $2 leq p < infty$ and consider the function $f : mathbbR^n to mathbbR$ defined by $$f(x) := |x|^p.$$ Then this function is mid-point convex (in fact strictly), i.e. we have that $$fleft(fracx + y2right) leq fracf(x) + f(y)2$$ holds for all $x,y in mathbbR^n$. Is there a nice way of showing this?
real-analysis convexity-inequality
asked Jul 18 at 8:28
TheGeekGreek
5,01831033
marked as duplicate by John Ma, Mostafa Ayaz, Shailesh, Adrian Keister, Parcly Taxel Jul 18 at 13:52
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
0
down vote
favorite
This question already has an answer here:
Elementary proof that $|x|^p$ is convex.
4 answers
Let $2 leq p < infty$ and consider the function $f : mathbbR^n to mathbbR$ defined by $$f(x) := |x|^p.$$ Then this function is mid-point convex (in fact strictly), i.e. we have that $$fleft(fracx + y2right) leq fracf(x) + f(y)2$$ holds for all $x,y in mathbbR^n$. Is there a nice way of showing this?
real-analysis convexity-inequality
asked Jul 18 at 8:28
TheGeekGreek
5,01831033
marked as duplicate by John Ma, Mostafa Ayaz, Shailesh, Adrian Keister, Parcly Taxel Jul 18 at 13:52
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
@JohnMa My answer uses an old geometric idea. It is different from all the answers in the post you have referred to. I hope it has some merit, so this question need not be closed.
– Kavi Rama Murthy
Jul 18 at 9:11
Then you should post your answer to that question instead of this one. The goal is to put all good answers in a same place! @KaviRamaMurthy
– John Ma
Jul 18 at 9:12
I am doing that but that post is 1 year and 4 months old. It is unlikely that anyone would look at my answer.
– Kavi Rama Murthy
Jul 18 at 9:14
That question was found when I put "|x|^p is convex" is google. That's the first post I find and has 1k views. I don't see why this question will be more popular in the long run? @KaviRamaMurthy
– John Ma
Jul 18 at 9:16
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question already has an answer here:
Elementary proof that $|x|^p$ is convex.
4 answers
Let $2 leq p < infty$ and consider the function $f : mathbbR^n to mathbbR$ defined by $$f(x) := |x|^p.$$ Then this function is mid-point convex (in fact strictly), i.e. we have that $$fleft(fracx + y2right) leq fracf(x) + f(y)2$$ holds for all $x,y in mathbbR^n$. Is there a nice way of showing this?
real-analysis convexity-inequality
asked Jul 18 at 8:28
TheGeekGreek
5,01831033
This question already has an answer here:
Elementary proof that $|x|^p$ is convex.
4 answers
Let $2 leq p < infty$ and consider the function $f : mathbbR^n to mathbbR$ defined by $$f(x) := |x|^p.$$ Then this function is mid-point convex (in fact strictly), i.e. we have that $$fleft(fracx + y2right) leq fracf(x) + f(y)2$$ holds for all $x,y in mathbbR^n$. Is there a nice way of showing this?
This question already has an answer here:
Elementary proof that $|x|^p$ is convex.
4 answers
real-analysis convexity-inequality
asked Jul 18 at 8:28
TheGeekGreek
5,01831033
asked Jul 18 at 8:28
TheGeekGreek
5,01831033
asked Jul 18 at 8:28
TheGeekGreek
5,01831033
asked Jul 18 at 8:28
TheGeekGreek
5,01831033
5,01831033
marked as duplicate by John Ma, Mostafa Ayaz, Shailesh, Adrian Keister, Parcly Taxel Jul 18 at 13:52
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by John Ma, Mostafa Ayaz, Shailesh, Adrian Keister, Parcly Taxel Jul 18 at 13:52
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
@JohnMa My answer uses an old geometric idea. It is different from all the answers in the post you have referred to. I hope it has some merit, so this question need not be closed.
– Kavi Rama Murthy
Jul 18 at 9:11
Then you should post your answer to that question instead of this one. The goal is to put all good answers in a same place! @KaviRamaMurthy
– John Ma
Jul 18 at 9:12
I am doing that but that post is 1 year and 4 months old. It is unlikely that anyone would look at my answer.
– Kavi Rama Murthy
Jul 18 at 9:14
That question was found when I put "|x|^p is convex" is google. That's the first post I find and has 1k views. I don't see why this question will be more popular in the long run? @KaviRamaMurthy
– John Ma
Jul 18 at 9:16
add a comment |Â
@JohnMa My answer uses an old geometric idea. It is different from all the answers in the post you have referred to. I hope it has some merit, so this question need not be closed.
– Kavi Rama Murthy
Jul 18 at 9:11
Then you should post your answer to that question instead of this one. The goal is to put all good answers in a same place! @KaviRamaMurthy
– John Ma
Jul 18 at 9:12
I am doing that but that post is 1 year and 4 months old. It is unlikely that anyone would look at my answer.
– Kavi Rama Murthy
Jul 18 at 9:14
That question was found when I put "|x|^p is convex" is google. That's the first post I find and has 1k views. I don't see why this question will be more popular in the long run? @KaviRamaMurthy
– John Ma
Jul 18 at 9:16
@JohnMa My answer uses an old geometric idea. It is different from all the answers in the post you have referred to. I hope it has some merit, so this question need not be closed.
– Kavi Rama Murthy
Jul 18 at 9:11
@JohnMa My answer uses an old geometric idea. It is different from all the answers in the post you have referred to. I hope it has some merit, so this question need not be closed.
– Kavi Rama Murthy
Jul 18 at 9:11
Then you should post your answer to that question instead of this one. The goal is to put all good answers in a same place! @KaviRamaMurthy
– John Ma
Jul 18 at 9:12
Then you should post your answer to that question instead of this one. The goal is to put all good answers in a same place! @KaviRamaMurthy
– John Ma
Jul 18 at 9:12
I am doing that but that post is 1 year and 4 months old. It is unlikely that anyone would look at my answer.
– Kavi Rama Murthy
Jul 18 at 9:14
I am doing that but that post is 1 year and 4 months old. It is unlikely that anyone would look at my answer.
– Kavi Rama Murthy
Jul 18 at 9:14
That question was found when I put "|x|^p is convex" is google. That's the first post I find and has 1k views. I don't see why this question will be more popular in the long run? @KaviRamaMurthy
– John Ma
Jul 18 at 9:16
That question was found when I put "|x|^p is convex" is google. That's the first post I find and has 1k views. I don't see why this question will be more popular in the long run? @KaviRamaMurthy
– John Ma
Jul 18 at 9:16
add a comment |Â
2 Answers
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Here is an elementary argument which uses only MVT: Claim: for $x geq 0$ we have $x^p=sup a^p+(x-a)pa^p-1: ageq 0$. To prove that apply MVT to $x^p-a^p$ and consider the cases $a<x$ and $a geq x$ separately. [ The idea for this comes from the geometric fact that a convex function is the upper envelope of the tangent lines]. Now define $f_a(x)=a^p+(x-a)pa^p-1$. [MVT gives $a^p+(x-a)pa^p-1leq x^p$. To show that the supremum of the left side equals $x^p$ you just have to note that $a^p+(x-a)pa^p-1=x^p$ when $a=x$]. Then $f_a(frac x+y 2)=frac f_a(x)+f_a(y) 2$ (since $f_a$ is an affine function ). Just take supremum over $a geq 0$ to get the desired inequality.
answered Jul 18 at 8:51
Kavi Rama Murthy
20.8k2829
add a comment |Â
up vote
0
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After finishing my answer I realized I assumed $p$ to be an integer. I'll post my answer anyway for completeness:
The following inequality trivially holds for all $x, y in mathbbR$, $0 leq n leq p$:
$$0 leq (|x|^n - |y|^n)(|x|^p-n - |y|^p-n)$$
Multiplying all crossing terms and moving negative terms to the left, we get:
$$|x|^n|y|^p-n +|x|^p-n|y|^n leq |x|^p + |y|^p$$
We sum over all $S subseteq [1..p]$, substituting $n=|S|$:
$$2sum_S subseteq [1..p] |x|^|y|^p- leq sum_S subseteq [1..p] |x|^p + |y|^p$$
Both sides can be simplified:
$$2(|x|+|y|)^p leq 2^p(|x|^p+|y|^p)$$
By the triangle inequality the left hand side bounds $2|x+y|^p$ from above. We divide by $2^p+1$, getting:
$$fracx+y2^p leq fracx2$$
answered Jul 18 at 9:20
Timon Knigge
484
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Here is an elementary argument which uses only MVT: Claim: for $x geq 0$ we have $x^p=sup a^p+(x-a)pa^p-1: ageq 0$. To prove that apply MVT to $x^p-a^p$ and consider the cases $a<x$ and $a geq x$ separately. [ The idea for this comes from the geometric fact that a convex function is the upper envelope of the tangent lines]. Now define $f_a(x)=a^p+(x-a)pa^p-1$. [MVT gives $a^p+(x-a)pa^p-1leq x^p$. To show that the supremum of the left side equals $x^p$ you just have to note that $a^p+(x-a)pa^p-1=x^p$ when $a=x$]. Then $f_a(frac x+y 2)=frac f_a(x)+f_a(y) 2$ (since $f_a$ is an affine function ). Just take supremum over $a geq 0$ to get the desired inequality.
answered Jul 18 at 8:51
Kavi Rama Murthy
20.8k2829
add a comment |Â
up vote
0
down vote
accepted
Here is an elementary argument which uses only MVT: Claim: for $x geq 0$ we have $x^p=sup a^p+(x-a)pa^p-1: ageq 0$. To prove that apply MVT to $x^p-a^p$ and consider the cases $a<x$ and $a geq x$ separately. [ The idea for this comes from the geometric fact that a convex function is the upper envelope of the tangent lines]. Now define $f_a(x)=a^p+(x-a)pa^p-1$. [MVT gives $a^p+(x-a)pa^p-1leq x^p$. To show that the supremum of the left side equals $x^p$ you just have to note that $a^p+(x-a)pa^p-1=x^p$ when $a=x$]. Then $f_a(frac x+y 2)=frac f_a(x)+f_a(y) 2$ (since $f_a$ is an affine function ). Just take supremum over $a geq 0$ to get the desired inequality.
answered Jul 18 at 8:51
Kavi Rama Murthy
20.8k2829
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Here is an elementary argument which uses only MVT: Claim: for $x geq 0$ we have $x^p=sup a^p+(x-a)pa^p-1: ageq 0$. To prove that apply MVT to $x^p-a^p$ and consider the cases $a<x$ and $a geq x$ separately. [ The idea for this comes from the geometric fact that a convex function is the upper envelope of the tangent lines]. Now define $f_a(x)=a^p+(x-a)pa^p-1$. [MVT gives $a^p+(x-a)pa^p-1leq x^p$. To show that the supremum of the left side equals $x^p$ you just have to note that $a^p+(x-a)pa^p-1=x^p$ when $a=x$]. Then $f_a(frac x+y 2)=frac f_a(x)+f_a(y) 2$ (since $f_a$ is an affine function ). Just take supremum over $a geq 0$ to get the desired inequality.
answered Jul 18 at 8:51
Kavi Rama Murthy
20.8k2829
Here is an elementary argument which uses only MVT: Claim: for $x geq 0$ we have $x^p=sup a^p+(x-a)pa^p-1: ageq 0$. To prove that apply MVT to $x^p-a^p$ and consider the cases $a<x$ and $a geq x$ separately. [ The idea for this comes from the geometric fact that a convex function is the upper envelope of the tangent lines]. Now define $f_a(x)=a^p+(x-a)pa^p-1$. [MVT gives $a^p+(x-a)pa^p-1leq x^p$. To show that the supremum of the left side equals $x^p$ you just have to note that $a^p+(x-a)pa^p-1=x^p$ when $a=x$]. Then $f_a(frac x+y 2)=frac f_a(x)+f_a(y) 2$ (since $f_a$ is an affine function ). Just take supremum over $a geq 0$ to get the desired inequality.
answered Jul 18 at 8:51
Kavi Rama Murthy
20.8k2829
edited Jul 18 at 9:07
answered Jul 18 at 8:51
Kavi Rama Murthy
20.8k2829
answered Jul 18 at 8:51
Kavi Rama Murthy
20.8k2829
answered Jul 18 at 8:51
Kavi Rama Murthy
20.8k2829
20.8k2829
add a comment |Â
add a comment |Â
up vote
0
down vote
After finishing my answer I realized I assumed $p$ to be an integer. I'll post my answer anyway for completeness:
The following inequality trivially holds for all $x, y in mathbbR$, $0 leq n leq p$:
$$0 leq (|x|^n - |y|^n)(|x|^p-n - |y|^p-n)$$
Multiplying all crossing terms and moving negative terms to the left, we get:
$$|x|^n|y|^p-n +|x|^p-n|y|^n leq |x|^p + |y|^p$$
We sum over all $S subseteq [1..p]$, substituting $n=|S|$:
$$2sum_S subseteq [1..p] |x|^|y|^p- leq sum_S subseteq [1..p] |x|^p + |y|^p$$
Both sides can be simplified:
$$2(|x|+|y|)^p leq 2^p(|x|^p+|y|^p)$$
By the triangle inequality the left hand side bounds $2|x+y|^p$ from above. We divide by $2^p+1$, getting:
$$fracx+y2^p leq fracx2$$
answered Jul 18 at 9:20
Timon Knigge
484
add a comment |Â
up vote
0
down vote
After finishing my answer I realized I assumed $p$ to be an integer. I'll post my answer anyway for completeness:
The following inequality trivially holds for all $x, y in mathbbR$, $0 leq n leq p$:
$$0 leq (|x|^n - |y|^n)(|x|^p-n - |y|^p-n)$$
Multiplying all crossing terms and moving negative terms to the left, we get:
$$|x|^n|y|^p-n +|x|^p-n|y|^n leq |x|^p + |y|^p$$
We sum over all $S subseteq [1..p]$, substituting $n=|S|$:
$$2sum_S subseteq [1..p] |x|^|y|^p- leq sum_S subseteq [1..p] |x|^p + |y|^p$$
Both sides can be simplified:
$$2(|x|+|y|)^p leq 2^p(|x|^p+|y|^p)$$
By the triangle inequality the left hand side bounds $2|x+y|^p$ from above. We divide by $2^p+1$, getting:
$$fracx+y2^p leq fracx2$$
answered Jul 18 at 9:20
Timon Knigge
484
add a comment |Â
up vote
0
down vote
up vote
0
down vote
After finishing my answer I realized I assumed $p$ to be an integer. I'll post my answer anyway for completeness:
The following inequality trivially holds for all $x, y in mathbbR$, $0 leq n leq p$:
$$0 leq (|x|^n - |y|^n)(|x|^p-n - |y|^p-n)$$
Multiplying all crossing terms and moving negative terms to the left, we get:
$$|x|^n|y|^p-n +|x|^p-n|y|^n leq |x|^p + |y|^p$$
We sum over all $S subseteq [1..p]$, substituting $n=|S|$:
$$2sum_S subseteq [1..p] |x|^|y|^p- leq sum_S subseteq [1..p] |x|^p + |y|^p$$
Both sides can be simplified:
$$2(|x|+|y|)^p leq 2^p(|x|^p+|y|^p)$$
By the triangle inequality the left hand side bounds $2|x+y|^p$ from above. We divide by $2^p+1$, getting:
$$fracx+y2^p leq fracx2$$
answered Jul 18 at 9:20
Timon Knigge
484
After finishing my answer I realized I assumed $p$ to be an integer. I'll post my answer anyway for completeness:
The following inequality trivially holds for all $x, y in mathbbR$, $0 leq n leq p$:
$$0 leq (|x|^n - |y|^n)(|x|^p-n - |y|^p-n)$$
Multiplying all crossing terms and moving negative terms to the left, we get:
$$|x|^n|y|^p-n +|x|^p-n|y|^n leq |x|^p + |y|^p$$
We sum over all $S subseteq [1..p]$, substituting $n=|S|$:
$$2sum_S subseteq [1..p] |x|^|y|^p- leq sum_S subseteq [1..p] |x|^p + |y|^p$$
Both sides can be simplified:
$$2(|x|+|y|)^p leq 2^p(|x|^p+|y|^p)$$
By the triangle inequality the left hand side bounds $2|x+y|^p$ from above. We divide by $2^p+1$, getting:
$$fracx+y2^p leq fracx2$$
answered Jul 18 at 9:20
Timon Knigge
484
answered Jul 18 at 9:20
Timon Knigge
484
answered Jul 18 at 9:20
Timon Knigge
484
answered Jul 18 at 9:20
Timon Knigge
484
484
add a comment |Â
add a comment |Â
@JohnMa My answer uses an old geometric idea. It is different from all the answers in the post you have referred to. I hope it has some merit, so this question need not be closed.
– Kavi Rama Murthy
Jul 18 at 9:11
Then you should post your answer to that question instead of this one. The goal is to put all good answers in a same place! @KaviRamaMurthy
– John Ma
Jul 18 at 9:12
I am doing that but that post is 1 year and 4 months old. It is unlikely that anyone would look at my answer.
– Kavi Rama Murthy
Jul 18 at 9:14
That question was found when I put "|x|^p is convex" is google. That's the first post I find and has 1k views. I don't see why this question will be more popular in the long run? @KaviRamaMurthy
– John Ma
Jul 18 at 9:16