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Injective mapping theorem proof discrepancy.

2

$begingroup$

I do not understand from where the 2 comes in the proof of letter(a) of the statement of the theorem:

And this is proposition 12.2.4

I think the author has used the mean value inequality, but I do not know how, could anyone explain this for me please?

real-analysis calculus analysis multivariable-calculus proof-explanation

share|cite|improve this question

asked Apr 2 at 1:15

hopefullyhopefully

192215

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My answer should clear some of this up for you. The lemma I refer to is a bit involved but I can give you a reference.
  $endgroup$
  – RRL
  Apr 2 at 3:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you I know this lemma @RRL
  $endgroup$
  – hopefully
  Apr 2 at 3:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That is impressive.
  $endgroup$
  – RRL
  Apr 2 at 3:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RRL I am sorry .... but it is just mentioned before the theorem ..... it seems like my concentration is weak today.
  $endgroup$
  – hopefully
  Apr 2 at 3:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes -- it should use the mean value inequality $|f(u) - f(v)| leqslant |Df(c)(u-v)|$ for some $c$ on the segment joining $u$ and $v$.
  $endgroup$
  – RRL
  Apr 2 at 3:17
  
  

  
  
  
  

add a comment | 

2

$begingroup$

I do not understand from where the 2 comes in the proof of letter(a) of the statement of the theorem:

And this is proposition 12.2.4

I think the author has used the mean value inequality, but I do not know how, could anyone explain this for me please?

real-analysis calculus analysis multivariable-calculus proof-explanation

share|cite|improve this question

asked Apr 2 at 1:15

hopefullyhopefully

192215

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My answer should clear some of this up for you. The lemma I refer to is a bit involved but I can give you a reference.
  $endgroup$
  – RRL
  Apr 2 at 3:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you I know this lemma @RRL
  $endgroup$
  – hopefully
  Apr 2 at 3:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That is impressive.
  $endgroup$
  – RRL
  Apr 2 at 3:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RRL I am sorry .... but it is just mentioned before the theorem ..... it seems like my concentration is weak today.
  $endgroup$
  – hopefully
  Apr 2 at 3:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes -- it should use the mean value inequality $|f(u) - f(v)| leqslant |Df(c)(u-v)|$ for some $c$ on the segment joining $u$ and $v$.
  $endgroup$
  – RRL
  Apr 2 at 3:17
  
  

  
  
  
  

add a comment | 

$begingroup$

I do not understand from where the 2 comes in the proof of letter(a) of the statement of the theorem:

And this is proposition 12.2.4

I think the author has used the mean value inequality, but I do not know how, could anyone explain this for me please?

real-analysis calculus analysis multivariable-calculus proof-explanation

share|cite|improve this question

asked Apr 2 at 1:15

hopefullyhopefully

192215

$endgroup$

share|cite|improve this question

asked Apr 2 at 1:15

hopefullyhopefully

192215

share|cite|improve this question

asked Apr 2 at 1:15

hopefullyhopefully

192215

asked Apr 2 at 1:15

hopefullyhopefully

192215

asked Apr 2 at 1:15

hopefullyhopefully

192215

1 Answer
1

active

oldest

votes

2

$begingroup$

From Proposition 12.2.4, since $Df(c)$ is injective the exists a constant $gamma'$ such that $|Df(c)(u)| geqslant gamma'|u|$. This is easily proved by using the fact that a linear operator is continuous and attains a maximum and minimum on the (compact) unit sphere.

We can define $gamma = frac12gamma'$ so that $|Df(c)(u)| geqslant gamma'|u| geqslant 2gamma|u|$.

There is a lemma that states if $f$ is continuously differentiable, then given $gamma > 0$, for all points $u$ and $v$ sufficiently close to $c$, i.e. in the ball $B$ centered at $c$, we have

$$|f(u) - f(v) - Df(c)(u-v)| leqslant gamma |u-v|$$

The proof does, in fact, use the mean value theorem.

By the reverse triangle inequality,

$$|Df(c)(u-v)| - |f(u) - f(v)|leqslant|f(u) - f(v) - Df(c)(u-v)| leqslant gamma |u-v| $$

Hence,

$$|f(u) - f(v)| geqslant |Df(c)(u-v)| - gamma |u-v| geqslant 2gamma|u-v| - gamma |u-v| = gamma |u-v|$$

share|cite|improve this answer

answered Apr 2 at 3:09

RRLRRL

53.8k52675

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  math.stackexchange.com/questions/3170545/… could you please look at this question for me if you have time?
  $endgroup$
  – hopefully
  Apr 2 at 3:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @hopefully: Ok I'll take a look shortly.
  $endgroup$
  – RRL
  Apr 2 at 3:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much :) I appreciate it
  $endgroup$
  – hopefully
  Apr 2 at 3:26
  

  
  
  
  

add a comment | 

Your Answer

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2

$begingroup$

From Proposition 12.2.4, since $Df(c)$ is injective the exists a constant $gamma'$ such that $|Df(c)(u)| geqslant gamma'|u|$. This is easily proved by using the fact that a linear operator is continuous and attains a maximum and minimum on the (compact) unit sphere.

We can define $gamma = frac12gamma'$ so that $|Df(c)(u)| geqslant gamma'|u| geqslant 2gamma|u|$.

There is a lemma that states if $f$ is continuously differentiable, then given $gamma > 0$, for all points $u$ and $v$ sufficiently close to $c$, i.e. in the ball $B$ centered at $c$, we have

$$|f(u) - f(v) - Df(c)(u-v)| leqslant gamma |u-v|$$

The proof does, in fact, use the mean value theorem.

By the reverse triangle inequality,

$$|Df(c)(u-v)| - |f(u) - f(v)|leqslant|f(u) - f(v) - Df(c)(u-v)| leqslant gamma |u-v| $$

Hence,

$$|f(u) - f(v)| geqslant |Df(c)(u-v)| - gamma |u-v| geqslant 2gamma|u-v| - gamma |u-v| = gamma |u-v|$$

share|cite|improve this answer

answered Apr 2 at 3:09

RRLRRL

53.8k52675

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  math.stackexchange.com/questions/3170545/… could you please look at this question for me if you have time?
  $endgroup$
  – hopefully
  Apr 2 at 3:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @hopefully: Ok I'll take a look shortly.
  $endgroup$
  – RRL
  Apr 2 at 3:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much :) I appreciate it
  $endgroup$
  – hopefully
  Apr 2 at 3:26
  

  
  
  
  

add a comment | 

2

$begingroup$

From Proposition 12.2.4, since $Df(c)$ is injective the exists a constant $gamma'$ such that $|Df(c)(u)| geqslant gamma'|u|$. This is easily proved by using the fact that a linear operator is continuous and attains a maximum and minimum on the (compact) unit sphere.

We can define $gamma = frac12gamma'$ so that $|Df(c)(u)| geqslant gamma'|u| geqslant 2gamma|u|$.

There is a lemma that states if $f$ is continuously differentiable, then given $gamma > 0$, for all points $u$ and $v$ sufficiently close to $c$, i.e. in the ball $B$ centered at $c$, we have

$$|f(u) - f(v) - Df(c)(u-v)| leqslant gamma |u-v|$$

The proof does, in fact, use the mean value theorem.

By the reverse triangle inequality,

$$|Df(c)(u-v)| - |f(u) - f(v)|leqslant|f(u) - f(v) - Df(c)(u-v)| leqslant gamma |u-v| $$

Hence,

$$|f(u) - f(v)| geqslant |Df(c)(u-v)| - gamma |u-v| geqslant 2gamma|u-v| - gamma |u-v| = gamma |u-v|$$

share|cite|improve this answer

answered Apr 2 at 3:09

RRLRRL

53.8k52675

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  math.stackexchange.com/questions/3170545/… could you please look at this question for me if you have time?
  $endgroup$
  – hopefully
  Apr 2 at 3:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @hopefully: Ok I'll take a look shortly.
  $endgroup$
  – RRL
  Apr 2 at 3:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much :) I appreciate it
  $endgroup$
  – hopefully
  Apr 2 at 3:26
  

  
  
  
  

add a comment | 

$begingroup$

From Proposition 12.2.4, since $Df(c)$ is injective the exists a constant $gamma'$ such that $|Df(c)(u)| geqslant gamma'|u|$. This is easily proved by using the fact that a linear operator is continuous and attains a maximum and minimum on the (compact) unit sphere.

We can define $gamma = frac12gamma'$ so that $|Df(c)(u)| geqslant gamma'|u| geqslant 2gamma|u|$.

There is a lemma that states if $f$ is continuously differentiable, then given $gamma > 0$, for all points $u$ and $v$ sufficiently close to $c$, i.e. in the ball $B$ centered at $c$, we have

$$|f(u) - f(v) - Df(c)(u-v)| leqslant gamma |u-v|$$

The proof does, in fact, use the mean value theorem.

By the reverse triangle inequality,

$$|Df(c)(u-v)| - |f(u) - f(v)|leqslant|f(u) - f(v) - Df(c)(u-v)| leqslant gamma |u-v| $$

Hence,

$$|f(u) - f(v)| geqslant |Df(c)(u-v)| - gamma |u-v| geqslant 2gamma|u-v| - gamma |u-v| = gamma |u-v|$$

share|cite|improve this answer

answered Apr 2 at 3:09

RRLRRL

53.8k52675

$endgroup$

share|cite|improve this answer

answered Apr 2 at 3:09

RRLRRL

53.8k52675

answered Apr 2 at 3:09

RRLRRL

53.8k52675

answered Apr 2 at 3:09

RRLRRL

53.8k52675