Proof of the inequality $||a+b|^p - |a|^p| leq epsilon |a|^p + C_epsilon |b|^p$

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In this paper, the following theorem is used implicitly (see Example (a) on page 488):

Let $0<p<infty$. For every $epsilon > 0$, there exists some $C_epsilon > 0$ such that $forall a,b in mathbbC$:
$$
left||a+b|^p - |a|^p right| leq epsilon |a|^p + C_epsilon |b|^p
$$

Is this a well-known result? I would like to know its proof, and its name if it has one. Can someone point me towards a reference to this result which includes a proof, or give me a proof outline?

I've tried using Jensen's Inequality for the case $p>1, a,b>0$, but it doesn't seem to work out.

real-analysis inequality

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asked Jul 16 at 20:21

Sambo

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up vote
4
down vote

favorite

In this paper, the following theorem is used implicitly (see Example (a) on page 488):

Let $0<p<infty$. For every $epsilon > 0$, there exists some $C_epsilon > 0$ such that $forall a,b in mathbbC$:
$$
left||a+b|^p - |a|^p right| leq epsilon |a|^p + C_epsilon |b|^p
$$

Is this a well-known result? I would like to know its proof, and its name if it has one. Can someone point me towards a reference to this result which includes a proof, or give me a proof outline?

I've tried using Jensen's Inequality for the case $p>1, a,b>0$, but it doesn't seem to work out.

real-analysis inequality

share|cite|improve this question

asked Jul 16 at 20:21

Sambo

1,2551427

add a comment | 

up vote
4
down vote

favorite

up vote
4
down vote

favorite

In this paper, the following theorem is used implicitly (see Example (a) on page 488):

Let $0<p<infty$. For every $epsilon > 0$, there exists some $C_epsilon > 0$ such that $forall a,b in mathbbC$:
$$
left||a+b|^p - |a|^p right| leq epsilon |a|^p + C_epsilon |b|^p
$$

Is this a well-known result? I would like to know its proof, and its name if it has one. Can someone point me towards a reference to this result which includes a proof, or give me a proof outline?

I've tried using Jensen's Inequality for the case $p>1, a,b>0$, but it doesn't seem to work out.

real-analysis inequality

share|cite|improve this question

asked Jul 16 at 20:21

Sambo

1,2551427

In this paper, the following theorem is used implicitly (see Example (a) on page 488):

Let $0<p<infty$. For every $epsilon > 0$, there exists some $C_epsilon > 0$ such that $forall a,b in mathbbC$:
$$
left||a+b|^p - |a|^p right| leq epsilon |a|^p + C_epsilon |b|^p
$$

Is this a well-known result? I would like to know its proof, and its name if it has one. Can someone point me towards a reference to this result which includes a proof, or give me a proof outline?

I've tried using Jensen's Inequality for the case $p>1, a,b>0$, but it doesn't seem to work out.

real-analysis inequality

share|cite|improve this question

asked Jul 16 at 20:21

Sambo

1,2551427

share|cite|improve this question

share|cite|improve this question

asked Jul 16 at 20:21

Sambo

1,2551427

asked Jul 16 at 20:21

Sambo

1,2551427

asked Jul 16 at 20:21

Sambo

1,2551427

1,2551427

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
2
down vote



accepted

Define $k=dfracba$. This theorem is equivalent to show that $$left||1+k|^p-1right|leepsilon+C_epsilon|k|^p$$First let $|1+k|ge 1$, then we must have$$|1+k|^p-1leepsilon+C_epsilon|k|^p$$or equivalently$$dfracle C_epsilon$$if $|1+k|^ple1+epsilon$ the theorem holds with any $C_epsilon>0$ then assume that $|1+k|^p>1+epsilon$. By triangle inequality we have $$(1+|k|)^pge|1+k|^p>1+epsilon$$which yields to $$|k|>(1+epsilon)^frac1p-1$$or$$|dfrac1k|<dfrac1(1+epsilon)^frac1p-1$$therefore $$dfrac<(1+|dfrac1k|)^p<dfrac1+epsilon((1+epsilon)^frac1p-1)^p$$then by choosing $C_epsilon=dfrac1+epsilon((1+epsilon)^frac1p-1)^p$ we complete our proof in this case.

The other case is where $|1+k|<1$. Here we need to show that $$dfrac1+kle C_epsilon$$ for $|1+k|^pge 1-epsilon$ this automatically holds and for $|1+k|^p<1-epsilon$ we have $$|1+k|<(1-epsilon)^dfrac1p$$this is inside a circle of radius $(1-epsilon)^dfrac1p$ centered at $-1$ which geometrically means that $$|k|>1-(1-epsilon)^dfrac1p$$. Here taking $C_epsilon=dfrac1left(1-(1-epsilon)^dfrac1pright)^p$ fulfills our condition. Therefore the general $C_epsilon$ would be $$Large C_epsilon=maxleftdfrac1left(1-(1-epsilon)^frac1pright)^p,dfrac1+epsilon((1+epsilon)^frac1p-1)^pright$$

share|cite|improve this answer

answered Jul 16 at 22:19

Mostafa Ayaz

8,6023630

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for the detailed explanation! Did you come up with this on your own, or do you know this result from somewhere?
  – Sambo
  Jul 17 at 13:34
  

  
  
  
  

add a comment | 

up vote
2
down vote

Hint: To start, try to prove it for $a=1,$ $b$ arbitrary. If false in this case, then for some $epsilon> 0,$ there is a sequence $b_n$ such that

$$||1+b_n|^p - 1 | > epsilon + n |b_n|^p,, n=1,2,dots.$$

That should lead to a contradiction. Having this special case should lead to the full result.

share|cite|improve this answer

edited Jul 17 at 1:06

answered Jul 16 at 21:26

zhw.

65.8k42870

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Define $k=dfracba$. This theorem is equivalent to show that $$left||1+k|^p-1right|leepsilon+C_epsilon|k|^p$$First let $|1+k|ge 1$, then we must have$$|1+k|^p-1leepsilon+C_epsilon|k|^p$$or equivalently$$dfracle C_epsilon$$if $|1+k|^ple1+epsilon$ the theorem holds with any $C_epsilon>0$ then assume that $|1+k|^p>1+epsilon$. By triangle inequality we have $$(1+|k|)^pge|1+k|^p>1+epsilon$$which yields to $$|k|>(1+epsilon)^frac1p-1$$or$$|dfrac1k|<dfrac1(1+epsilon)^frac1p-1$$therefore $$dfrac<(1+|dfrac1k|)^p<dfrac1+epsilon((1+epsilon)^frac1p-1)^p$$then by choosing $C_epsilon=dfrac1+epsilon((1+epsilon)^frac1p-1)^p$ we complete our proof in this case.

The other case is where $|1+k|<1$. Here we need to show that $$dfrac1+kle C_epsilon$$ for $|1+k|^pge 1-epsilon$ this automatically holds and for $|1+k|^p<1-epsilon$ we have $$|1+k|<(1-epsilon)^dfrac1p$$this is inside a circle of radius $(1-epsilon)^dfrac1p$ centered at $-1$ which geometrically means that $$|k|>1-(1-epsilon)^dfrac1p$$. Here taking $C_epsilon=dfrac1left(1-(1-epsilon)^dfrac1pright)^p$ fulfills our condition. Therefore the general $C_epsilon$ would be $$Large C_epsilon=maxleftdfrac1left(1-(1-epsilon)^frac1pright)^p,dfrac1+epsilon((1+epsilon)^frac1p-1)^pright$$

share|cite|improve this answer

answered Jul 16 at 22:19

Mostafa Ayaz

8,6023630

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for the detailed explanation! Did you come up with this on your own, or do you know this result from somewhere?
  – Sambo
  Jul 17 at 13:34
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

Define $k=dfracba$. This theorem is equivalent to show that $$left||1+k|^p-1right|leepsilon+C_epsilon|k|^p$$First let $|1+k|ge 1$, then we must have$$|1+k|^p-1leepsilon+C_epsilon|k|^p$$or equivalently$$dfracle C_epsilon$$if $|1+k|^ple1+epsilon$ the theorem holds with any $C_epsilon>0$ then assume that $|1+k|^p>1+epsilon$. By triangle inequality we have $$(1+|k|)^pge|1+k|^p>1+epsilon$$which yields to $$|k|>(1+epsilon)^frac1p-1$$or$$|dfrac1k|<dfrac1(1+epsilon)^frac1p-1$$therefore $$dfrac<(1+|dfrac1k|)^p<dfrac1+epsilon((1+epsilon)^frac1p-1)^p$$then by choosing $C_epsilon=dfrac1+epsilon((1+epsilon)^frac1p-1)^p$ we complete our proof in this case.

The other case is where $|1+k|<1$. Here we need to show that $$dfrac1+kle C_epsilon$$ for $|1+k|^pge 1-epsilon$ this automatically holds and for $|1+k|^p<1-epsilon$ we have $$|1+k|<(1-epsilon)^dfrac1p$$this is inside a circle of radius $(1-epsilon)^dfrac1p$ centered at $-1$ which geometrically means that $$|k|>1-(1-epsilon)^dfrac1p$$. Here taking $C_epsilon=dfrac1left(1-(1-epsilon)^dfrac1pright)^p$ fulfills our condition. Therefore the general $C_epsilon$ would be $$Large C_epsilon=maxleftdfrac1left(1-(1-epsilon)^frac1pright)^p,dfrac1+epsilon((1+epsilon)^frac1p-1)^pright$$

share|cite|improve this answer

answered Jul 16 at 22:19

Mostafa Ayaz

8,6023630

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for the detailed explanation! Did you come up with this on your own, or do you know this result from somewhere?
  – Sambo
  Jul 17 at 13:34
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Define $k=dfracba$. This theorem is equivalent to show that $$left||1+k|^p-1right|leepsilon+C_epsilon|k|^p$$First let $|1+k|ge 1$, then we must have$$|1+k|^p-1leepsilon+C_epsilon|k|^p$$or equivalently$$dfracle C_epsilon$$if $|1+k|^ple1+epsilon$ the theorem holds with any $C_epsilon>0$ then assume that $|1+k|^p>1+epsilon$. By triangle inequality we have $$(1+|k|)^pge|1+k|^p>1+epsilon$$which yields to $$|k|>(1+epsilon)^frac1p-1$$or$$|dfrac1k|<dfrac1(1+epsilon)^frac1p-1$$therefore $$dfrac<(1+|dfrac1k|)^p<dfrac1+epsilon((1+epsilon)^frac1p-1)^p$$then by choosing $C_epsilon=dfrac1+epsilon((1+epsilon)^frac1p-1)^p$ we complete our proof in this case.

The other case is where $|1+k|<1$. Here we need to show that $$dfrac1+kle C_epsilon$$ for $|1+k|^pge 1-epsilon$ this automatically holds and for $|1+k|^p<1-epsilon$ we have $$|1+k|<(1-epsilon)^dfrac1p$$this is inside a circle of radius $(1-epsilon)^dfrac1p$ centered at $-1$ which geometrically means that $$|k|>1-(1-epsilon)^dfrac1p$$. Here taking $C_epsilon=dfrac1left(1-(1-epsilon)^dfrac1pright)^p$ fulfills our condition. Therefore the general $C_epsilon$ would be $$Large C_epsilon=maxleftdfrac1left(1-(1-epsilon)^frac1pright)^p,dfrac1+epsilon((1+epsilon)^frac1p-1)^pright$$

share|cite|improve this answer

answered Jul 16 at 22:19

Mostafa Ayaz

8,6023630

Define $k=dfracba$. This theorem is equivalent to show that $$left||1+k|^p-1right|leepsilon+C_epsilon|k|^p$$First let $|1+k|ge 1$, then we must have$$|1+k|^p-1leepsilon+C_epsilon|k|^p$$or equivalently$$dfracle C_epsilon$$if $|1+k|^ple1+epsilon$ the theorem holds with any $C_epsilon>0$ then assume that $|1+k|^p>1+epsilon$. By triangle inequality we have $$(1+|k|)^pge|1+k|^p>1+epsilon$$which yields to $$|k|>(1+epsilon)^frac1p-1$$or$$|dfrac1k|<dfrac1(1+epsilon)^frac1p-1$$therefore $$dfrac<(1+|dfrac1k|)^p<dfrac1+epsilon((1+epsilon)^frac1p-1)^p$$then by choosing $C_epsilon=dfrac1+epsilon((1+epsilon)^frac1p-1)^p$ we complete our proof in this case.

The other case is where $|1+k|<1$. Here we need to show that $$dfrac1+kle C_epsilon$$ for $|1+k|^pge 1-epsilon$ this automatically holds and for $|1+k|^p<1-epsilon$ we have $$|1+k|<(1-epsilon)^dfrac1p$$this is inside a circle of radius $(1-epsilon)^dfrac1p$ centered at $-1$ which geometrically means that $$|k|>1-(1-epsilon)^dfrac1p$$. Here taking $C_epsilon=dfrac1left(1-(1-epsilon)^dfrac1pright)^p$ fulfills our condition. Therefore the general $C_epsilon$ would be $$Large C_epsilon=maxleftdfrac1left(1-(1-epsilon)^frac1pright)^p,dfrac1+epsilon((1+epsilon)^frac1p-1)^pright$$

share|cite|improve this answer

answered Jul 16 at 22:19

Mostafa Ayaz

8,6023630

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 22:19

Mostafa Ayaz

8,6023630

answered Jul 16 at 22:19

Mostafa Ayaz

8,6023630

answered Jul 16 at 22:19

Mostafa Ayaz

8,6023630

8,6023630

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for the detailed explanation! Did you come up with this on your own, or do you know this result from somewhere?
  – Sambo
  Jul 17 at 13:34
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for the detailed explanation! Did you come up with this on your own, or do you know this result from somewhere?
  – Sambo
  Jul 17 at 13:34
  

  
  
  
  

Thank you for the detailed explanation! Did you come up with this on your own, or do you know this result from somewhere?
– Sambo
Jul 17 at 13:34

Thank you for the detailed explanation! Did you come up with this on your own, or do you know this result from somewhere?
– Sambo
Jul 17 at 13:34

add a comment | 

up vote
2
down vote

Hint: To start, try to prove it for $a=1,$ $b$ arbitrary. If false in this case, then for some $epsilon> 0,$ there is a sequence $b_n$ such that

$$||1+b_n|^p - 1 | > epsilon + n |b_n|^p,, n=1,2,dots.$$

That should lead to a contradiction. Having this special case should lead to the full result.

share|cite|improve this answer

edited Jul 17 at 1:06

answered Jul 16 at 21:26

zhw.

65.8k42870

add a comment | 

up vote
2
down vote

Hint: To start, try to prove it for $a=1,$ $b$ arbitrary. If false in this case, then for some $epsilon> 0,$ there is a sequence $b_n$ such that

$$||1+b_n|^p - 1 | > epsilon + n |b_n|^p,, n=1,2,dots.$$

That should lead to a contradiction. Having this special case should lead to the full result.

share|cite|improve this answer

edited Jul 17 at 1:06

answered Jul 16 at 21:26

zhw.

65.8k42870

add a comment | 

up vote
2
down vote

up vote
2
down vote

Hint: To start, try to prove it for $a=1,$ $b$ arbitrary. If false in this case, then for some $epsilon> 0,$ there is a sequence $b_n$ such that

$$||1+b_n|^p - 1 | > epsilon + n |b_n|^p,, n=1,2,dots.$$

That should lead to a contradiction. Having this special case should lead to the full result.

share|cite|improve this answer

edited Jul 17 at 1:06

answered Jul 16 at 21:26

zhw.

65.8k42870

Hint: To start, try to prove it for $a=1,$ $b$ arbitrary. If false in this case, then for some $epsilon> 0,$ there is a sequence $b_n$ such that

$$||1+b_n|^p - 1 | > epsilon + n |b_n|^p,, n=1,2,dots.$$

That should lead to a contradiction. Having this special case should lead to the full result.

share|cite|improve this answer

edited Jul 17 at 1:06

answered Jul 16 at 21:26

zhw.

65.8k42870

share|cite|improve this answer

share|cite|improve this answer

edited Jul 17 at 1:06

edited Jul 17 at 1:06

edited Jul 17 at 1:06

answered Jul 16 at 21:26

zhw.

65.8k42870

answered Jul 16 at 21:26

zhw.

65.8k42870

answered Jul 16 at 21:26

zhw.

65.8k42870

65.8k42870

add a comment | 

add a comment | 

 

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