Mixing
Please explain inequality $|x^{p}-y^{p}| leq |x-y|^p$
$begingroup$
Suppose $x geq 0$, $y geq 0$ and $0<p<1$. Why is the following inequality true?
$|x^{p}-y^{p}| leq |x-y|^p$
real-analysis
edited Nov 20 '10 at 0:21
Ross Millikan
299k24200374
asked Nov 19 '10 at 19:27
studentstudent
505414
$endgroup$
add a comment |
$begingroup$
Suppose $x geq 0$, $y geq 0$ and $0<p<1$. Why is the following inequality true?
$|x^{p}-y^{p}| leq |x-y|^p$
real-analysis
edited Nov 20 '10 at 0:21
Ross Millikan
299k24200374
asked Nov 19 '10 at 19:27
studentstudent
505414
$endgroup$
$begingroup$
The inequality $|x^p-y^p|leq|x-y|^p$ holds even when $x$ and $y$ are positive operators on Hilbert space. This follows from a more general result proved as Theorem 1.5 in the paper [1] by J. Phillips, which gives the inequality $|f(x)-f(y)|leq f(|x-y|)$ whenever $f$ is an operator monotone function on $[0,infty)$ such that $f(0)=0$. The fact that $tmapsto t^p$ is operator monotone if $0lt pleq 1$ is proven as Proposition 1.3.8 in Pedersen's C*-algebras book. Your inequality is the $1$-dimensional case. [1]: dspace.library.uvic.ca:8443/dspace/handle/1828/1506
$endgroup$
– Jonas Meyer
Nov 20 '10 at 7:07
add a comment |
$begingroup$
Suppose $x geq 0$, $y geq 0$ and $0<p<1$. Why is the following inequality true?
$|x^{p}-y^{p}| leq |x-y|^p$
real-analysis
edited Nov 20 '10 at 0:21
Ross Millikan
299k24200374
asked Nov 19 '10 at 19:27
studentstudent
505414
$endgroup$
Suppose $x geq 0$, $y geq 0$ and $0<p<1$. Why is the following inequality true?
$|x^{p}-y^{p}| leq |x-y|^p$
real-analysis
real-analysis
edited Nov 20 '10 at 0:21
Ross Millikan
299k24200374
asked Nov 19 '10 at 19:27
studentstudent
505414
edited Nov 20 '10 at 0:21
Ross Millikan
299k24200374
asked Nov 19 '10 at 19:27
studentstudent
505414
edited Nov 20 '10 at 0:21
Ross Millikan
299k24200374
edited Nov 20 '10 at 0:21
Ross Millikan
299k24200374
edited Nov 20 '10 at 0:21
Ross Millikan
299k24200374
299k24200374
asked Nov 19 '10 at 19:27
studentstudent
505414
asked Nov 19 '10 at 19:27
studentstudent
505414
asked Nov 19 '10 at 19:27
studentstudent
505414
505414
$begingroup$
The inequality $|x^p-y^p|leq|x-y|^p$ holds even when $x$ and $y$ are positive operators on Hilbert space. This follows from a more general result proved as Theorem 1.5 in the paper [1] by J. Phillips, which gives the inequality $|f(x)-f(y)|leq f(|x-y|)$ whenever $f$ is an operator monotone function on $[0,infty)$ such that $f(0)=0$. The fact that $tmapsto t^p$ is operator monotone if $0lt pleq 1$ is proven as Proposition 1.3.8 in Pedersen's C*-algebras book. Your inequality is the $1$-dimensional case. [1]: dspace.library.uvic.ca:8443/dspace/handle/1828/1506
$endgroup$
– Jonas Meyer
Nov 20 '10 at 7:07
add a comment |
$begingroup$
The inequality $|x^p-y^p|leq|x-y|^p$ holds even when $x$ and $y$ are positive operators on Hilbert space. This follows from a more general result proved as Theorem 1.5 in the paper [1] by J. Phillips, which gives the inequality $|f(x)-f(y)|leq f(|x-y|)$ whenever $f$ is an operator monotone function on $[0,infty)$ such that $f(0)=0$. The fact that $tmapsto t^p$ is operator monotone if $0lt pleq 1$ is proven as Proposition 1.3.8 in Pedersen's C*-algebras book. Your inequality is the $1$-dimensional case. [1]: dspace.library.uvic.ca:8443/dspace/handle/1828/1506
$endgroup$
– Jonas Meyer
Nov 20 '10 at 7:07
$begingroup$
The inequality $|x^p-y^p|leq|x-y|^p$ holds even when $x$ and $y$ are positive operators on Hilbert space. This follows from a more general result proved as Theorem 1.5 in the paper [1] by J. Phillips, which gives the inequality $|f(x)-f(y)|leq f(|x-y|)$ whenever $f$ is an operator monotone function on $[0,infty)$ such that $f(0)=0$. The fact that $tmapsto t^p$ is operator monotone if $0lt pleq 1$ is proven as Proposition 1.3.8 in Pedersen's C*-algebras book. Your inequality is the $1$-dimensional case. [1]: dspace.library.uvic.ca:8443/dspace/handle/1828/1506
$endgroup$
– Jonas Meyer
Nov 20 '10 at 7:07
$begingroup$
The inequality $|x^p-y^p|leq|x-y|^p$ holds even when $x$ and $y$ are positive operators on Hilbert space. This follows from a more general result proved as Theorem 1.5 in the paper [1] by J. Phillips, which gives the inequality $|f(x)-f(y)|leq f(|x-y|)$ whenever $f$ is an operator monotone function on $[0,infty)$ such that $f(0)=0$. The fact that $tmapsto t^p$ is operator monotone if $0lt pleq 1$ is proven as Proposition 1.3.8 in Pedersen's C*-algebras book. Your inequality is the $1$-dimensional case. [1]: dspace.library.uvic.ca:8443/dspace/handle/1828/1506
$endgroup$
– Jonas Meyer
Nov 20 '10 at 7:07
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Assume w.lo.g $x>y$. Then the statement becomes $x^p - y^p leq (x-y)^p$. Since $y > 0$, divide through out by $y^p$. So we need to show $(frac{x}{y})^p - 1 leq (frac{x}{y}-1)^p$ whenever $x > y >0$ and $0<p<1$. Let $t = frac{x}{y}$. So we need to show $t^p - 1 leq (t-1)^p$ whenever $t > 1$ and $0<p<1$.
This is a calculus problem.
Let $f(t) = (t-1)^p - (t^p -1) $ where $0<p<1$.
Show that the function $f(t)$ is increasing for $t geq 1$ and when $0 < p < 1$.
So $f(t) geq f(1)$ and $f(1) = 0$. So, we get the desired result.
$textbf{EDIT:}$ To show $f(t)$ is increasing for $t geq 1$ and when $0 < p < 1$.
We need to show $0 < f'(t) = p (t-1)^{p-1} - pt^{p-1}$ and since $p>0$, all we need to show is that $(t-1)^{p-1}>t^{p-1}$, $forall t > 1$ and $0<p<1$.
Since $0<p<1$, we need to show $t^{1-p} > (t-1)^{1-p}$ which is true since $p<1$ and $t>t-1>0$.
edited Sep 20 '11 at 13:35
Davide Giraudo
127k17154268
$endgroup$
11
$begingroup$
@user10: I see you're a new user to this site. Just so you know, it's considered polite on this site both to upvote the answers to your questions that you find helpful (click on the little up arrow next to the answer) and to formally accept the answer to each of your questions that you think is the best (click on the little check mark by the answer you want to accept).
$endgroup$
– Mike Spivey
Nov 20 '10 at 0:08
add a comment |
$begingroup$
This follows from the more general inequality
begin{equation}(1)qquadqquad|x+y|^ple |x|^p+|y|^pqquadqquadtext{(for $x,yinmathbb{C}$ and $0lt ple1$)}end{equation}
Indeed, if we replace $x$ by $x-y$ in (1) we get
$$|x|^ple |x-y|^p+|y|^p$$
which imply
$$|x|^p-|y|^ple |x-y|^p$$
To prove (1), first note
$$|x+y|^ple(|x|+|y|)^p$$
Hence it is sufficient to prove (1) for $x,yge0$ in which case we may apply Sivaram's argument in the previous answer.
edited Nov 20 '10 at 5:32
Arturo Magidin
264k34590917
answered Nov 19 '10 at 22:50
AD.AD.
8,76383163
$endgroup$
$begingroup$
@ AD: You should have $0<p<1$ in your argument.
$endgroup$
– user17762
Nov 19 '10 at 23:21
$begingroup$
@Sivaram: Of course, I add that. Thanks.
$endgroup$
– AD.
Nov 20 '10 at 5:27
$begingroup$
@Arturo Magidin: I know about the problem with <. I was just fixing that.
$endgroup$
– AD.
Nov 20 '10 at 5:38
$begingroup$
Sorry about that; saw the problem, so I tried to fix it. Sometimes it happens to me and I navigate away before realizing it.
$endgroup$
– Arturo Magidin
Nov 20 '10 at 5:41
$begingroup$
@Arturo Magidin: Ok.
$endgroup$
– AD.
Nov 20 '10 at 5:46
|
show 6 more comments
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2 Answers
2
active
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2 Answers
2
active
oldest
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$begingroup$
Assume w.lo.g $x>y$. Then the statement becomes $x^p - y^p leq (x-y)^p$. Since $y > 0$, divide through out by $y^p$. So we need to show $(frac{x}{y})^p - 1 leq (frac{x}{y}-1)^p$ whenever $x > y >0$ and $0<p<1$. Let $t = frac{x}{y}$. So we need to show $t^p - 1 leq (t-1)^p$ whenever $t > 1$ and $0<p<1$.
This is a calculus problem.
Let $f(t) = (t-1)^p - (t^p -1) $ where $0<p<1$.
Show that the function $f(t)$ is increasing for $t geq 1$ and when $0 < p < 1$.
So $f(t) geq f(1)$ and $f(1) = 0$. So, we get the desired result.
$textbf{EDIT:}$ To show $f(t)$ is increasing for $t geq 1$ and when $0 < p < 1$.
We need to show $0 < f'(t) = p (t-1)^{p-1} - pt^{p-1}$ and since $p>0$, all we need to show is that $(t-1)^{p-1}>t^{p-1}$, $forall t > 1$ and $0<p<1$.
Since $0<p<1$, we need to show $t^{1-p} > (t-1)^{1-p}$ which is true since $p<1$ and $t>t-1>0$.
edited Sep 20 '11 at 13:35
Davide Giraudo
127k17154268
$endgroup$
11
$begingroup$
@user10: I see you're a new user to this site. Just so you know, it's considered polite on this site both to upvote the answers to your questions that you find helpful (click on the little up arrow next to the answer) and to formally accept the answer to each of your questions that you think is the best (click on the little check mark by the answer you want to accept).
$endgroup$
– Mike Spivey
Nov 20 '10 at 0:08
add a comment |
$begingroup$
Assume w.lo.g $x>y$. Then the statement becomes $x^p - y^p leq (x-y)^p$. Since $y > 0$, divide through out by $y^p$. So we need to show $(frac{x}{y})^p - 1 leq (frac{x}{y}-1)^p$ whenever $x > y >0$ and $0<p<1$. Let $t = frac{x}{y}$. So we need to show $t^p - 1 leq (t-1)^p$ whenever $t > 1$ and $0<p<1$.
This is a calculus problem.
Let $f(t) = (t-1)^p - (t^p -1) $ where $0<p<1$.
Show that the function $f(t)$ is increasing for $t geq 1$ and when $0 < p < 1$.
So $f(t) geq f(1)$ and $f(1) = 0$. So, we get the desired result.
$textbf{EDIT:}$ To show $f(t)$ is increasing for $t geq 1$ and when $0 < p < 1$.
We need to show $0 < f'(t) = p (t-1)^{p-1} - pt^{p-1}$ and since $p>0$, all we need to show is that $(t-1)^{p-1}>t^{p-1}$, $forall t > 1$ and $0<p<1$.
Since $0<p<1$, we need to show $t^{1-p} > (t-1)^{1-p}$ which is true since $p<1$ and $t>t-1>0$.
edited Sep 20 '11 at 13:35
Davide Giraudo
127k17154268
$endgroup$
11
$begingroup$
@user10: I see you're a new user to this site. Just so you know, it's considered polite on this site both to upvote the answers to your questions that you find helpful (click on the little up arrow next to the answer) and to formally accept the answer to each of your questions that you think is the best (click on the little check mark by the answer you want to accept).
$endgroup$
– Mike Spivey
Nov 20 '10 at 0:08
add a comment |
$begingroup$
Assume w.lo.g $x>y$. Then the statement becomes $x^p - y^p leq (x-y)^p$. Since $y > 0$, divide through out by $y^p$. So we need to show $(frac{x}{y})^p - 1 leq (frac{x}{y}-1)^p$ whenever $x > y >0$ and $0<p<1$. Let $t = frac{x}{y}$. So we need to show $t^p - 1 leq (t-1)^p$ whenever $t > 1$ and $0<p<1$.
This is a calculus problem.
Let $f(t) = (t-1)^p - (t^p -1) $ where $0<p<1$.
Show that the function $f(t)$ is increasing for $t geq 1$ and when $0 < p < 1$.
So $f(t) geq f(1)$ and $f(1) = 0$. So, we get the desired result.
$textbf{EDIT:}$ To show $f(t)$ is increasing for $t geq 1$ and when $0 < p < 1$.
We need to show $0 < f'(t) = p (t-1)^{p-1} - pt^{p-1}$ and since $p>0$, all we need to show is that $(t-1)^{p-1}>t^{p-1}$, $forall t > 1$ and $0<p<1$.
Since $0<p<1$, we need to show $t^{1-p} > (t-1)^{1-p}$ which is true since $p<1$ and $t>t-1>0$.
edited Sep 20 '11 at 13:35
Davide Giraudo
127k17154268
$endgroup$
Assume w.lo.g $x>y$. Then the statement becomes $x^p - y^p leq (x-y)^p$. Since $y > 0$, divide through out by $y^p$. So we need to show $(frac{x}{y})^p - 1 leq (frac{x}{y}-1)^p$ whenever $x > y >0$ and $0<p<1$. Let $t = frac{x}{y}$. So we need to show $t^p - 1 leq (t-1)^p$ whenever $t > 1$ and $0<p<1$.
This is a calculus problem.
Let $f(t) = (t-1)^p - (t^p -1) $ where $0<p<1$.
Show that the function $f(t)$ is increasing for $t geq 1$ and when $0 < p < 1$.
So $f(t) geq f(1)$ and $f(1) = 0$. So, we get the desired result.
$textbf{EDIT:}$ To show $f(t)$ is increasing for $t geq 1$ and when $0 < p < 1$.
We need to show $0 < f'(t) = p (t-1)^{p-1} - pt^{p-1}$ and since $p>0$, all we need to show is that $(t-1)^{p-1}>t^{p-1}$, $forall t > 1$ and $0<p<1$.
Since $0<p<1$, we need to show $t^{1-p} > (t-1)^{1-p}$ which is true since $p<1$ and $t>t-1>0$.
edited Sep 20 '11 at 13:35
Davide Giraudo
127k17154268
edited Sep 20 '11 at 13:35
Davide Giraudo
127k17154268
edited Sep 20 '11 at 13:35
Davide Giraudo
127k17154268
edited Sep 20 '11 at 13:35
Davide Giraudo
127k17154268
127k17154268
answered Nov 19 '10 at 19:53
user17762
11
$begingroup$
@user10: I see you're a new user to this site. Just so you know, it's considered polite on this site both to upvote the answers to your questions that you find helpful (click on the little up arrow next to the answer) and to formally accept the answer to each of your questions that you think is the best (click on the little check mark by the answer you want to accept).
$endgroup$
– Mike Spivey
Nov 20 '10 at 0:08
add a comment |
11
$begingroup$
@user10: I see you're a new user to this site. Just so you know, it's considered polite on this site both to upvote the answers to your questions that you find helpful (click on the little up arrow next to the answer) and to formally accept the answer to each of your questions that you think is the best (click on the little check mark by the answer you want to accept).
$endgroup$
– Mike Spivey
Nov 20 '10 at 0:08
11
11
$begingroup$
@user10: I see you're a new user to this site. Just so you know, it's considered polite on this site both to upvote the answers to your questions that you find helpful (click on the little up arrow next to the answer) and to formally accept the answer to each of your questions that you think is the best (click on the little check mark by the answer you want to accept).
$endgroup$
– Mike Spivey
Nov 20 '10 at 0:08
$begingroup$
@user10: I see you're a new user to this site. Just so you know, it's considered polite on this site both to upvote the answers to your questions that you find helpful (click on the little up arrow next to the answer) and to formally accept the answer to each of your questions that you think is the best (click on the little check mark by the answer you want to accept).
$endgroup$
– Mike Spivey
Nov 20 '10 at 0:08
add a comment |
$begingroup$
This follows from the more general inequality
begin{equation}(1)qquadqquad|x+y|^ple |x|^p+|y|^pqquadqquadtext{(for $x,yinmathbb{C}$ and $0lt ple1$)}end{equation}
Indeed, if we replace $x$ by $x-y$ in (1) we get
$$|x|^ple |x-y|^p+|y|^p$$
which imply
$$|x|^p-|y|^ple |x-y|^p$$
To prove (1), first note
$$|x+y|^ple(|x|+|y|)^p$$
Hence it is sufficient to prove (1) for $x,yge0$ in which case we may apply Sivaram's argument in the previous answer.
edited Nov 20 '10 at 5:32
Arturo Magidin
264k34590917
answered Nov 19 '10 at 22:50
AD.AD.
8,76383163
$endgroup$
$begingroup$
@ AD: You should have $0<p<1$ in your argument.
$endgroup$
– user17762
Nov 19 '10 at 23:21
$begingroup$
@Sivaram: Of course, I add that. Thanks.
$endgroup$
– AD.
Nov 20 '10 at 5:27
$begingroup$
@Arturo Magidin: I know about the problem with <. I was just fixing that.
$endgroup$
– AD.
Nov 20 '10 at 5:38
$begingroup$
Sorry about that; saw the problem, so I tried to fix it. Sometimes it happens to me and I navigate away before realizing it.
$endgroup$
– Arturo Magidin
Nov 20 '10 at 5:41
$begingroup$
@Arturo Magidin: Ok.
$endgroup$
– AD.
Nov 20 '10 at 5:46
|
show 6 more comments
$begingroup$
This follows from the more general inequality
begin{equation}(1)qquadqquad|x+y|^ple |x|^p+|y|^pqquadqquadtext{(for $x,yinmathbb{C}$ and $0lt ple1$)}end{equation}
Indeed, if we replace $x$ by $x-y$ in (1) we get
$$|x|^ple |x-y|^p+|y|^p$$
which imply
$$|x|^p-|y|^ple |x-y|^p$$
To prove (1), first note
$$|x+y|^ple(|x|+|y|)^p$$
Hence it is sufficient to prove (1) for $x,yge0$ in which case we may apply Sivaram's argument in the previous answer.
edited Nov 20 '10 at 5:32
Arturo Magidin
264k34590917
answered Nov 19 '10 at 22:50
AD.AD.
8,76383163
$endgroup$
$begingroup$
@ AD: You should have $0<p<1$ in your argument.
$endgroup$
– user17762
Nov 19 '10 at 23:21
$begingroup$
@Sivaram: Of course, I add that. Thanks.
$endgroup$
– AD.
Nov 20 '10 at 5:27
$begingroup$
@Arturo Magidin: I know about the problem with <. I was just fixing that.
$endgroup$
– AD.
Nov 20 '10 at 5:38
$begingroup$
Sorry about that; saw the problem, so I tried to fix it. Sometimes it happens to me and I navigate away before realizing it.
$endgroup$
– Arturo Magidin
Nov 20 '10 at 5:41
$begingroup$
@Arturo Magidin: Ok.
$endgroup$
– AD.
Nov 20 '10 at 5:46
|
show 6 more comments
$begingroup$
This follows from the more general inequality
begin{equation}(1)qquadqquad|x+y|^ple |x|^p+|y|^pqquadqquadtext{(for $x,yinmathbb{C}$ and $0lt ple1$)}end{equation}
Indeed, if we replace $x$ by $x-y$ in (1) we get
$$|x|^ple |x-y|^p+|y|^p$$
which imply
$$|x|^p-|y|^ple |x-y|^p$$
To prove (1), first note
$$|x+y|^ple(|x|+|y|)^p$$
Hence it is sufficient to prove (1) for $x,yge0$ in which case we may apply Sivaram's argument in the previous answer.
edited Nov 20 '10 at 5:32
Arturo Magidin
264k34590917
answered Nov 19 '10 at 22:50
AD.AD.
8,76383163
$endgroup$
This follows from the more general inequality
begin{equation}(1)qquadqquad|x+y|^ple |x|^p+|y|^pqquadqquadtext{(for $x,yinmathbb{C}$ and $0lt ple1$)}end{equation}
Indeed, if we replace $x$ by $x-y$ in (1) we get
$$|x|^ple |x-y|^p+|y|^p$$
which imply
$$|x|^p-|y|^ple |x-y|^p$$
To prove (1), first note
$$|x+y|^ple(|x|+|y|)^p$$
Hence it is sufficient to prove (1) for $x,yge0$ in which case we may apply Sivaram's argument in the previous answer.
edited Nov 20 '10 at 5:32
Arturo Magidin
264k34590917
answered Nov 19 '10 at 22:50
AD.AD.
8,76383163
edited Nov 20 '10 at 5:32
Arturo Magidin
264k34590917
edited Nov 20 '10 at 5:32
Arturo Magidin
264k34590917
edited Nov 20 '10 at 5:32
Arturo Magidin
264k34590917
264k34590917
answered Nov 19 '10 at 22:50
AD.AD.
8,76383163
answered Nov 19 '10 at 22:50
AD.AD.
8,76383163
answered Nov 19 '10 at 22:50
AD.AD.
8,76383163
8,76383163
$begingroup$
@ AD: You should have $0<p<1$ in your argument.
$endgroup$
– user17762
Nov 19 '10 at 23:21
$begingroup$
@Sivaram: Of course, I add that. Thanks.
$endgroup$
– AD.
Nov 20 '10 at 5:27
$begingroup$
@Arturo Magidin: I know about the problem with <. I was just fixing that.
$endgroup$
– AD.
Nov 20 '10 at 5:38
$begingroup$
Sorry about that; saw the problem, so I tried to fix it. Sometimes it happens to me and I navigate away before realizing it.
$endgroup$
– Arturo Magidin
Nov 20 '10 at 5:41
$begingroup$
@Arturo Magidin: Ok.
$endgroup$
– AD.
Nov 20 '10 at 5:46
|
show 6 more comments
$begingroup$
@ AD: You should have $0<p<1$ in your argument.
$endgroup$
– user17762
Nov 19 '10 at 23:21
$begingroup$
@Sivaram: Of course, I add that. Thanks.
$endgroup$
– AD.
Nov 20 '10 at 5:27
$begingroup$
@Arturo Magidin: I know about the problem with <. I was just fixing that.
$endgroup$
– AD.
Nov 20 '10 at 5:38
$begingroup$
Sorry about that; saw the problem, so I tried to fix it. Sometimes it happens to me and I navigate away before realizing it.
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– Arturo Magidin
Nov 20 '10 at 5:41
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@Arturo Magidin: Ok.
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– AD.
Nov 20 '10 at 5:46
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@ AD: You should have $0<p<1$ in your argument.
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– user17762
Nov 19 '10 at 23:21
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@ AD: You should have $0<p<1$ in your argument.
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– user17762
Nov 19 '10 at 23:21
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@Sivaram: Of course, I add that. Thanks.
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– AD.
Nov 20 '10 at 5:27
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@Sivaram: Of course, I add that. Thanks.
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– AD.
Nov 20 '10 at 5:27
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@Arturo Magidin: I know about the problem with <. I was just fixing that.
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– AD.
Nov 20 '10 at 5:38
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@Arturo Magidin: I know about the problem with <. I was just fixing that.
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– AD.
Nov 20 '10 at 5:38
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Sorry about that; saw the problem, so I tried to fix it. Sometimes it happens to me and I navigate away before realizing it.
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– Arturo Magidin
Nov 20 '10 at 5:41
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Sorry about that; saw the problem, so I tried to fix it. Sometimes it happens to me and I navigate away before realizing it.
$endgroup$
– Arturo Magidin
Nov 20 '10 at 5:41
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@Arturo Magidin: Ok.
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– AD.
Nov 20 '10 at 5:46
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@Arturo Magidin: Ok.
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– AD.
Nov 20 '10 at 5:46
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The inequality $|x^p-y^p|leq|x-y|^p$ holds even when $x$ and $y$ are positive operators on Hilbert space. This follows from a more general result proved as Theorem 1.5 in the paper [1] by J. Phillips, which gives the inequality $|f(x)-f(y)|leq f(|x-y|)$ whenever $f$ is an operator monotone function on $[0,infty)$ such that $f(0)=0$. The fact that $tmapsto t^p$ is operator monotone if $0lt pleq 1$ is proven as Proposition 1.3.8 in Pedersen's C*-algebras book. Your inequality is the $1$-dimensional case. [1]: dspace.library.uvic.ca:8443/dspace/handle/1828/1506
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– Jonas Meyer
Nov 20 '10 at 7:07