Another form of the Sandwich theorem (for derivatives in dimension $1$)
Here is the theorem :
"Let $Isubseteq mathbb{R}$ an interval which contains $ain mathbb{R}$. Let $M$ and $m$ two functions defined on $I$, differentiable at $a$ and $f$ a function defined on $I$ which takes value in $mathbb{R}$.
If :
$i) forall x in I , m(x)le f(x)le M(x)$
$ii) m(a) = f(a) = M(a)$
$iii) m'(a)= M'(a)$
Then $f$ is differentiable at $a$ and we have $f'(a)=m'(a)=M'(a)$"
Here is my proof :
I consider the function : $xmapsto frac{f(x)-f(a)}{x-a}$ with $xne a$
Then $forall x in I setminus {a}$ by using $i)$ we have directly :
$frac{m(x)-f(a)}{x-a} le frac{f(x)-f(a)}{x-a} le frac{M(x)-f(a)}{x-a}$ if $x-a>0$ (for the other case we just have to change the sign of the inequality)
But by $ii)$ we have : $m(a)=f(a)=M(a)$ so we obtain :
$frac{m(x)-m(a)}{x-a} le frac{f(x)-f(a)}{x-a} le frac{M(x)-M(a)}{x-a}$
$Rightarrow$
$m'(a) le lim limits_{xto a} frac{f(x)-f(a)}{x-a} le M'(a)$
Then by using $iii)$ we have : $m'(a) le lim limits_{xto a} frac{f(x)-f(a)}{x-a} le m'(a)$ or $M'(a) le lim limits_{xto a} frac{f(x)-f(a)}{x-a} le M'(a)$
So we prove that $f$ is differentiable at $a$ (by the Squeeze theorem the limit gives $m'(a)$ or $M'(a)$) and we have $f'(a)=m'(a)=M'(a)$
The theorem is proved.
Am I right ? Thanks in advance.
real-analysis proof-verification
asked Dec 27 '14 at 14:43
Maman
1,134722
add a comment |
Here is the theorem :
"Let $Isubseteq mathbb{R}$ an interval which contains $ain mathbb{R}$. Let $M$ and $m$ two functions defined on $I$, differentiable at $a$ and $f$ a function defined on $I$ which takes value in $mathbb{R}$.
If :
$i) forall x in I , m(x)le f(x)le M(x)$
$ii) m(a) = f(a) = M(a)$
$iii) m'(a)= M'(a)$
Then $f$ is differentiable at $a$ and we have $f'(a)=m'(a)=M'(a)$"
Here is my proof :
I consider the function : $xmapsto frac{f(x)-f(a)}{x-a}$ with $xne a$
Then $forall x in I setminus {a}$ by using $i)$ we have directly :
$frac{m(x)-f(a)}{x-a} le frac{f(x)-f(a)}{x-a} le frac{M(x)-f(a)}{x-a}$ if $x-a>0$ (for the other case we just have to change the sign of the inequality)
But by $ii)$ we have : $m(a)=f(a)=M(a)$ so we obtain :
$frac{m(x)-m(a)}{x-a} le frac{f(x)-f(a)}{x-a} le frac{M(x)-M(a)}{x-a}$
$Rightarrow$
$m'(a) le lim limits_{xto a} frac{f(x)-f(a)}{x-a} le M'(a)$
Then by using $iii)$ we have : $m'(a) le lim limits_{xto a} frac{f(x)-f(a)}{x-a} le m'(a)$ or $M'(a) le lim limits_{xto a} frac{f(x)-f(a)}{x-a} le M'(a)$
So we prove that $f$ is differentiable at $a$ (by the Squeeze theorem the limit gives $m'(a)$ or $M'(a)$) and we have $f'(a)=m'(a)=M'(a)$
The theorem is proved.
Am I right ? Thanks in advance.
real-analysis proof-verification
asked Dec 27 '14 at 14:43
Maman
1,134722
Yes, your proof is (almost) correct, but in one of your first inequalities, you have to distinguish the cases $x<a$ and $x>a$, because this will change the sign of the denominator.
– PhoemueX
Dec 27 '14 at 15:07
@PhoemueX I was thinking about that fact and it can be more precise thanks... And do you think that the case $x=a$ is necessary for this proof ?
– Maman
Dec 27 '14 at 15:11
No, the case $x=a$ is not needed. The value around $x=a$ plays no role for $lim_{xto a}dots$. The value $f(a)$ only enters because of $frac{f(x)-color{red}{ f(a)}}{x-a}$.
– PhoemueX
Dec 27 '14 at 16:43
@PhoemueX ok perfect, thanks for confirmation !
– Maman
Dec 27 '14 at 16:48
add a comment |
Here is the theorem :
"Let $Isubseteq mathbb{R}$ an interval which contains $ain mathbb{R}$. Let $M$ and $m$ two functions defined on $I$, differentiable at $a$ and $f$ a function defined on $I$ which takes value in $mathbb{R}$.
If :
$i) forall x in I , m(x)le f(x)le M(x)$
$ii) m(a) = f(a) = M(a)$
$iii) m'(a)= M'(a)$
Then $f$ is differentiable at $a$ and we have $f'(a)=m'(a)=M'(a)$"
Here is my proof :
I consider the function : $xmapsto frac{f(x)-f(a)}{x-a}$ with $xne a$
Then $forall x in I setminus {a}$ by using $i)$ we have directly :
$frac{m(x)-f(a)}{x-a} le frac{f(x)-f(a)}{x-a} le frac{M(x)-f(a)}{x-a}$ if $x-a>0$ (for the other case we just have to change the sign of the inequality)
But by $ii)$ we have : $m(a)=f(a)=M(a)$ so we obtain :
$frac{m(x)-m(a)}{x-a} le frac{f(x)-f(a)}{x-a} le frac{M(x)-M(a)}{x-a}$
$Rightarrow$
$m'(a) le lim limits_{xto a} frac{f(x)-f(a)}{x-a} le M'(a)$
Then by using $iii)$ we have : $m'(a) le lim limits_{xto a} frac{f(x)-f(a)}{x-a} le m'(a)$ or $M'(a) le lim limits_{xto a} frac{f(x)-f(a)}{x-a} le M'(a)$
So we prove that $f$ is differentiable at $a$ (by the Squeeze theorem the limit gives $m'(a)$ or $M'(a)$) and we have $f'(a)=m'(a)=M'(a)$
The theorem is proved.
Am I right ? Thanks in advance.
real-analysis proof-verification
asked Dec 27 '14 at 14:43
Maman
1,134722
Here is the theorem :
"Let $Isubseteq mathbb{R}$ an interval which contains $ain mathbb{R}$. Let $M$ and $m$ two functions defined on $I$, differentiable at $a$ and $f$ a function defined on $I$ which takes value in $mathbb{R}$.
If :
$i) forall x in I , m(x)le f(x)le M(x)$
$ii) m(a) = f(a) = M(a)$
$iii) m'(a)= M'(a)$
Then $f$ is differentiable at $a$ and we have $f'(a)=m'(a)=M'(a)$"
Here is my proof :
I consider the function : $xmapsto frac{f(x)-f(a)}{x-a}$ with $xne a$
Then $forall x in I setminus {a}$ by using $i)$ we have directly :
$frac{m(x)-f(a)}{x-a} le frac{f(x)-f(a)}{x-a} le frac{M(x)-f(a)}{x-a}$ if $x-a>0$ (for the other case we just have to change the sign of the inequality)
But by $ii)$ we have : $m(a)=f(a)=M(a)$ so we obtain :
$frac{m(x)-m(a)}{x-a} le frac{f(x)-f(a)}{x-a} le frac{M(x)-M(a)}{x-a}$
$Rightarrow$
$m'(a) le lim limits_{xto a} frac{f(x)-f(a)}{x-a} le M'(a)$
Then by using $iii)$ we have : $m'(a) le lim limits_{xto a} frac{f(x)-f(a)}{x-a} le m'(a)$ or $M'(a) le lim limits_{xto a} frac{f(x)-f(a)}{x-a} le M'(a)$
So we prove that $f$ is differentiable at $a$ (by the Squeeze theorem the limit gives $m'(a)$ or $M'(a)$) and we have $f'(a)=m'(a)=M'(a)$
The theorem is proved.
Am I right ? Thanks in advance.
real-analysis proof-verification
real-analysis proof-verification
asked Dec 27 '14 at 14:43
Maman
1,134722
asked Dec 27 '14 at 14:43
Maman
1,134722
edited Dec 27 '14 at 16:49
asked Dec 27 '14 at 14:43
Maman
1,134722
asked Dec 27 '14 at 14:43
Maman
1,134722
asked Dec 27 '14 at 14:43
Maman
1,134722
1,134722
Yes, your proof is (almost) correct, but in one of your first inequalities, you have to distinguish the cases $x<a$ and $x>a$, because this will change the sign of the denominator.
– PhoemueX
Dec 27 '14 at 15:07
@PhoemueX I was thinking about that fact and it can be more precise thanks... And do you think that the case $x=a$ is necessary for this proof ?
– Maman
Dec 27 '14 at 15:11
No, the case $x=a$ is not needed. The value around $x=a$ plays no role for $lim_{xto a}dots$. The value $f(a)$ only enters because of $frac{f(x)-color{red}{ f(a)}}{x-a}$.
– PhoemueX
Dec 27 '14 at 16:43
@PhoemueX ok perfect, thanks for confirmation !
– Maman
Dec 27 '14 at 16:48
add a comment |
Yes, your proof is (almost) correct, but in one of your first inequalities, you have to distinguish the cases $x<a$ and $x>a$, because this will change the sign of the denominator.
– PhoemueX
Dec 27 '14 at 15:07
@PhoemueX I was thinking about that fact and it can be more precise thanks... And do you think that the case $x=a$ is necessary for this proof ?
– Maman
Dec 27 '14 at 15:11
No, the case $x=a$ is not needed. The value around $x=a$ plays no role for $lim_{xto a}dots$. The value $f(a)$ only enters because of $frac{f(x)-color{red}{ f(a)}}{x-a}$.
– PhoemueX
Dec 27 '14 at 16:43
@PhoemueX ok perfect, thanks for confirmation !
– Maman
Dec 27 '14 at 16:48
Yes, your proof is (almost) correct, but in one of your first inequalities, you have to distinguish the cases $x<a$ and $x>a$, because this will change the sign of the denominator.
– PhoemueX
Dec 27 '14 at 15:07
Yes, your proof is (almost) correct, but in one of your first inequalities, you have to distinguish the cases $x<a$ and $x>a$, because this will change the sign of the denominator.
– PhoemueX
Dec 27 '14 at 15:07
@PhoemueX I was thinking about that fact and it can be more precise thanks... And do you think that the case $x=a$ is necessary for this proof ?
– Maman
Dec 27 '14 at 15:11
@PhoemueX I was thinking about that fact and it can be more precise thanks... And do you think that the case $x=a$ is necessary for this proof ?
– Maman
Dec 27 '14 at 15:11
No, the case $x=a$ is not needed. The value around $x=a$ plays no role for $lim_{xto a}dots$. The value $f(a)$ only enters because of $frac{f(x)-color{red}{ f(a)}}{x-a}$.
– PhoemueX
Dec 27 '14 at 16:43
No, the case $x=a$ is not needed. The value around $x=a$ plays no role for $lim_{xto a}dots$. The value $f(a)$ only enters because of $frac{f(x)-color{red}{ f(a)}}{x-a}$.
– PhoemueX
Dec 27 '14 at 16:43
@PhoemueX ok perfect, thanks for confirmation !
– Maman
Dec 27 '14 at 16:48
@PhoemueX ok perfect, thanks for confirmation !
– Maman
Dec 27 '14 at 16:48
add a comment |
1 Answer
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It's correct. just a formal thing: you can't move to the inequality
$$
m^prime(a)leqlim_{xrightarrow a}frac{f(x)-f(a)}{x-a}leq M^prime(a)
$$
Because you haven't proved that the limit exists. you need to move from the first inequality straight to the conclusion.
answered Dec 27 '14 at 16:54
tzoorp
57637
You mean that it is better to write $lim_{xrightarrow a}frac{m(x)-m(a)}{x-a} = m'(a)$ before this inequality ?
– Maman
Dec 27 '14 at 17:11
no, i mean that you should write $frac{m(x)-m(a)}{x-a}leq frac{f(x)-f(a)}{x-a}leq frac{M(x)-M(a)}{x-a}$ and from there move to the conclusion that $f$ is derivable at $a$ and that $m^prime(a)=f^prime(a)=M^prime(a)$
– tzoorp
Dec 27 '14 at 17:15
But you used the squeeze theorem to get $m'(a)$ and $M'(a)$ no ?
– Maman
Dec 27 '14 at 17:22
Yes, the step in my comment is due to the sandwich theorem.
– tzoorp
Dec 27 '14 at 17:26
ok so it's better if I write : $frac{m(x)-m(a)}{x-a} le frac{f(x)-f(a)}{x-a} le frac{M(x)-M(a)}{x-a}$ then by applying the squeeze theorem and the third hypothesis we have : $m'(a) le lim limits_{xto a} frac{f(x)-f(a)}{x-a} le m'(a)$ which means that $f$ is differentiable at $a$ ?
– Maman
Dec 27 '14 at 17:35
|
show 1 more comment
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It's correct. just a formal thing: you can't move to the inequality
$$
m^prime(a)leqlim_{xrightarrow a}frac{f(x)-f(a)}{x-a}leq M^prime(a)
$$
Because you haven't proved that the limit exists. you need to move from the first inequality straight to the conclusion.
answered Dec 27 '14 at 16:54
tzoorp
57637
You mean that it is better to write $lim_{xrightarrow a}frac{m(x)-m(a)}{x-a} = m'(a)$ before this inequality ?
– Maman
Dec 27 '14 at 17:11
no, i mean that you should write $frac{m(x)-m(a)}{x-a}leq frac{f(x)-f(a)}{x-a}leq frac{M(x)-M(a)}{x-a}$ and from there move to the conclusion that $f$ is derivable at $a$ and that $m^prime(a)=f^prime(a)=M^prime(a)$
– tzoorp
Dec 27 '14 at 17:15
But you used the squeeze theorem to get $m'(a)$ and $M'(a)$ no ?
– Maman
Dec 27 '14 at 17:22
Yes, the step in my comment is due to the sandwich theorem.
– tzoorp
Dec 27 '14 at 17:26
ok so it's better if I write : $frac{m(x)-m(a)}{x-a} le frac{f(x)-f(a)}{x-a} le frac{M(x)-M(a)}{x-a}$ then by applying the squeeze theorem and the third hypothesis we have : $m'(a) le lim limits_{xto a} frac{f(x)-f(a)}{x-a} le m'(a)$ which means that $f$ is differentiable at $a$ ?
– Maman
Dec 27 '14 at 17:35
|
show 1 more comment
It's correct. just a formal thing: you can't move to the inequality
$$
m^prime(a)leqlim_{xrightarrow a}frac{f(x)-f(a)}{x-a}leq M^prime(a)
$$
Because you haven't proved that the limit exists. you need to move from the first inequality straight to the conclusion.
answered Dec 27 '14 at 16:54
tzoorp
57637
You mean that it is better to write $lim_{xrightarrow a}frac{m(x)-m(a)}{x-a} = m'(a)$ before this inequality ?
– Maman
Dec 27 '14 at 17:11
no, i mean that you should write $frac{m(x)-m(a)}{x-a}leq frac{f(x)-f(a)}{x-a}leq frac{M(x)-M(a)}{x-a}$ and from there move to the conclusion that $f$ is derivable at $a$ and that $m^prime(a)=f^prime(a)=M^prime(a)$
– tzoorp
Dec 27 '14 at 17:15
But you used the squeeze theorem to get $m'(a)$ and $M'(a)$ no ?
– Maman
Dec 27 '14 at 17:22
Yes, the step in my comment is due to the sandwich theorem.
– tzoorp
Dec 27 '14 at 17:26
ok so it's better if I write : $frac{m(x)-m(a)}{x-a} le frac{f(x)-f(a)}{x-a} le frac{M(x)-M(a)}{x-a}$ then by applying the squeeze theorem and the third hypothesis we have : $m'(a) le lim limits_{xto a} frac{f(x)-f(a)}{x-a} le m'(a)$ which means that $f$ is differentiable at $a$ ?
– Maman
Dec 27 '14 at 17:35
|
show 1 more comment
It's correct. just a formal thing: you can't move to the inequality
$$
m^prime(a)leqlim_{xrightarrow a}frac{f(x)-f(a)}{x-a}leq M^prime(a)
$$
Because you haven't proved that the limit exists. you need to move from the first inequality straight to the conclusion.
answered Dec 27 '14 at 16:54
tzoorp
57637
It's correct. just a formal thing: you can't move to the inequality
$$
m^prime(a)leqlim_{xrightarrow a}frac{f(x)-f(a)}{x-a}leq M^prime(a)
$$
Because you haven't proved that the limit exists. you need to move from the first inequality straight to the conclusion.
answered Dec 27 '14 at 16:54
tzoorp
57637
answered Dec 27 '14 at 16:54
tzoorp
57637
answered Dec 27 '14 at 16:54
tzoorp
57637
answered Dec 27 '14 at 16:54
tzoorp
57637
57637
You mean that it is better to write $lim_{xrightarrow a}frac{m(x)-m(a)}{x-a} = m'(a)$ before this inequality ?
– Maman
Dec 27 '14 at 17:11
no, i mean that you should write $frac{m(x)-m(a)}{x-a}leq frac{f(x)-f(a)}{x-a}leq frac{M(x)-M(a)}{x-a}$ and from there move to the conclusion that $f$ is derivable at $a$ and that $m^prime(a)=f^prime(a)=M^prime(a)$
– tzoorp
Dec 27 '14 at 17:15
But you used the squeeze theorem to get $m'(a)$ and $M'(a)$ no ?
– Maman
Dec 27 '14 at 17:22
Yes, the step in my comment is due to the sandwich theorem.
– tzoorp
Dec 27 '14 at 17:26
ok so it's better if I write : $frac{m(x)-m(a)}{x-a} le frac{f(x)-f(a)}{x-a} le frac{M(x)-M(a)}{x-a}$ then by applying the squeeze theorem and the third hypothesis we have : $m'(a) le lim limits_{xto a} frac{f(x)-f(a)}{x-a} le m'(a)$ which means that $f$ is differentiable at $a$ ?
– Maman
Dec 27 '14 at 17:35
|
show 1 more comment
You mean that it is better to write $lim_{xrightarrow a}frac{m(x)-m(a)}{x-a} = m'(a)$ before this inequality ?
– Maman
Dec 27 '14 at 17:11
no, i mean that you should write $frac{m(x)-m(a)}{x-a}leq frac{f(x)-f(a)}{x-a}leq frac{M(x)-M(a)}{x-a}$ and from there move to the conclusion that $f$ is derivable at $a$ and that $m^prime(a)=f^prime(a)=M^prime(a)$
– tzoorp
Dec 27 '14 at 17:15
But you used the squeeze theorem to get $m'(a)$ and $M'(a)$ no ?
– Maman
Dec 27 '14 at 17:22
Yes, the step in my comment is due to the sandwich theorem.
– tzoorp
Dec 27 '14 at 17:26
ok so it's better if I write : $frac{m(x)-m(a)}{x-a} le frac{f(x)-f(a)}{x-a} le frac{M(x)-M(a)}{x-a}$ then by applying the squeeze theorem and the third hypothesis we have : $m'(a) le lim limits_{xto a} frac{f(x)-f(a)}{x-a} le m'(a)$ which means that $f$ is differentiable at $a$ ?
– Maman
Dec 27 '14 at 17:35
You mean that it is better to write $lim_{xrightarrow a}frac{m(x)-m(a)}{x-a} = m'(a)$ before this inequality ?
– Maman
Dec 27 '14 at 17:11
You mean that it is better to write $lim_{xrightarrow a}frac{m(x)-m(a)}{x-a} = m'(a)$ before this inequality ?
– Maman
Dec 27 '14 at 17:11
no, i mean that you should write $frac{m(x)-m(a)}{x-a}leq frac{f(x)-f(a)}{x-a}leq frac{M(x)-M(a)}{x-a}$ and from there move to the conclusion that $f$ is derivable at $a$ and that $m^prime(a)=f^prime(a)=M^prime(a)$
– tzoorp
Dec 27 '14 at 17:15
no, i mean that you should write $frac{m(x)-m(a)}{x-a}leq frac{f(x)-f(a)}{x-a}leq frac{M(x)-M(a)}{x-a}$ and from there move to the conclusion that $f$ is derivable at $a$ and that $m^prime(a)=f^prime(a)=M^prime(a)$
– tzoorp
Dec 27 '14 at 17:15
But you used the squeeze theorem to get $m'(a)$ and $M'(a)$ no ?
– Maman
Dec 27 '14 at 17:22
But you used the squeeze theorem to get $m'(a)$ and $M'(a)$ no ?
– Maman
Dec 27 '14 at 17:22
Yes, the step in my comment is due to the sandwich theorem.
– tzoorp
Dec 27 '14 at 17:26
Yes, the step in my comment is due to the sandwich theorem.
– tzoorp
Dec 27 '14 at 17:26
ok so it's better if I write : $frac{m(x)-m(a)}{x-a} le frac{f(x)-f(a)}{x-a} le frac{M(x)-M(a)}{x-a}$ then by applying the squeeze theorem and the third hypothesis we have : $m'(a) le lim limits_{xto a} frac{f(x)-f(a)}{x-a} le m'(a)$ which means that $f$ is differentiable at $a$ ?
– Maman
Dec 27 '14 at 17:35
ok so it's better if I write : $frac{m(x)-m(a)}{x-a} le frac{f(x)-f(a)}{x-a} le frac{M(x)-M(a)}{x-a}$ then by applying the squeeze theorem and the third hypothesis we have : $m'(a) le lim limits_{xto a} frac{f(x)-f(a)}{x-a} le m'(a)$ which means that $f$ is differentiable at $a$ ?
– Maman
Dec 27 '14 at 17:35
|
show 1 more comment
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Yes, your proof is (almost) correct, but in one of your first inequalities, you have to distinguish the cases $x<a$ and $x>a$, because this will change the sign of the denominator.
– PhoemueX
Dec 27 '14 at 15:07
@PhoemueX I was thinking about that fact and it can be more precise thanks... And do you think that the case $x=a$ is necessary for this proof ?
– Maman
Dec 27 '14 at 15:11
No, the case $x=a$ is not needed. The value around $x=a$ plays no role for $lim_{xto a}dots$. The value $f(a)$ only enters because of $frac{f(x)-color{red}{ f(a)}}{x-a}$.
– PhoemueX
Dec 27 '14 at 16:43
@PhoemueX ok perfect, thanks for confirmation !
– Maman
Dec 27 '14 at 16:48