Injective mapping theorem proof discrepancy. Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Prove or disprove: the Hilbert-Schmidt norm is independent of the choice of basis on $mathbbR^n$Inverse function theorem proof queryMultivariate mean value theorem in statistics contextA difficulty in understanding the proof of completeness of $l_2$.A difficulty in understanding theorem 4.2 in Israel Gohberg.Second difficulty in understanding the proof of theorem 1.14 in Hungerford.Need a help in understanding a solution of a forth problem in Israel Gohberg.A difficulty in obtaining a formula in the proof of a theorem.A difficulty in understanding a proof for L'Hospital's rule (in Petrovic)A difficulty in understanding a step in the proof of Thm. 11.5.6 in Petrovic.A difficulty in understanding multivariable Fermat theorem proof.
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Injective mapping theorem proof discrepancy.
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Prove or disprove: the Hilbert-Schmidt norm is independent of the choice of basis on $mathbbR^n$Inverse function theorem proof queryMultivariate mean value theorem in statistics contextA difficulty in understanding the proof of completeness of $l_2$.A difficulty in understanding theorem 4.2 in Israel Gohberg.Second difficulty in understanding the proof of theorem 1.14 in Hungerford.Need a help in understanding a solution of a forth problem in Israel Gohberg.A difficulty in obtaining a formula in the proof of a theorem.A difficulty in understanding a proof for L'Hospital's rule (in Petrovic)A difficulty in understanding a step in the proof of Thm. 11.5.6 in Petrovic.A difficulty in understanding multivariable Fermat theorem proof.
$begingroup$
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
asked Apr 2 at 1:15
hopefullyhopefully
192215
$endgroup$
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
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
asked Apr 2 at 1:15
hopefullyhopefully
192215
$endgroup$
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
real-analysis calculus analysis multivariable-calculus proof-explanation
asked Apr 2 at 1:15
hopefullyhopefully
192215
asked Apr 2 at 1:15
hopefullyhopefully
192215
asked Apr 2 at 1:15
hopefullyhopefully
192215
asked Apr 2 at 1:15
hopefullyhopefully
192215
asked Apr 2 at 1:15
hopefullyhopefully
192215
192215
$begingroup$
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 |
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
@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
$begingroup$
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 |
1 Answer
1
active
oldest
votes
$begingroup$
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,
$$|f(u) - f(v)| geqslant |Df(c)(u-v)| - gamma |u-v| geqslant 2gamma|u-v| - gamma |u-v| = gamma |u-v|$$
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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1 Answer
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1 Answer
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oldest
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$begingroup$
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,
$$|f(u) - f(v)| geqslant |Df(c)(u-v)| - gamma |u-v| geqslant 2gamma|u-v| - gamma |u-v| = gamma |u-v|$$
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,
$$|f(u) - f(v)| geqslant |Df(c)(u-v)| - gamma |u-v| geqslant 2gamma|u-v| - gamma |u-v| = gamma |u-v|$$
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,
$$|f(u) - f(v)| geqslant |Df(c)(u-v)| - gamma |u-v| geqslant 2gamma|u-v| - gamma |u-v| = gamma |u-v|$$
answered Apr 2 at 3:09
RRLRRL
53.8k52675
$endgroup$
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,
$$|f(u) - f(v)| geqslant |Df(c)(u-v)| - gamma |u-v| geqslant 2gamma|u-v| - gamma |u-v| = gamma |u-v|$$
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
answered Apr 2 at 3:09
RRLRRL
53.8k52675
53.8k52675
$begingroup$
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 |
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
@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
$begingroup$
Thank you so much :) I appreciate it
$endgroup$
– hopefully
Apr 2 at 3:26
add a comment |
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$begingroup$
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