Question related to Darboux's theorem
$begingroup$
Darboux's theorem says that $f'$ has intermediate value property. More precisely,
( Darboux's theorem) If $f$ is differentiable on $[a,b]$, and if r is any number for which $f'(a)<r<f'(b)$ then $exists$ c in (a,b) such that $f'(r)=c$.
Thus Darboux's theorem implies that $f'$ cannot have any simple discontinuities on $[a,b].$
I have a question as follows:
Is there a function $f$ satisfying $displaystyle lim_{xto d} f'(x)=infty$ for some $d in (a,b)$ under the assumptions in Darboux's theorem?
If so, I can conclude that for each $x in (a,b),$ either $(i)$ $f'$ is continuous at $x$ or $(ii)$ $f'$ oscillates near $x.$
It seems that there is no differentiable function $f$ on $[a,b]$ satisfying function $displaystyle lim_{xto d} f'(x)=infty$ and $exists f'(d)$ for some $d in (a,b)$.
Please let me know if you have any idea or comment for my question. Thanks in advance!
real-analysis calculus analysis
asked Dec 30 '18 at 1:36
07060706
460311
$endgroup$
add a comment |
$begingroup$
Darboux's theorem says that $f'$ has intermediate value property. More precisely,
( Darboux's theorem) If $f$ is differentiable on $[a,b]$, and if r is any number for which $f'(a)<r<f'(b)$ then $exists$ c in (a,b) such that $f'(r)=c$.
Thus Darboux's theorem implies that $f'$ cannot have any simple discontinuities on $[a,b].$
I have a question as follows:
Is there a function $f$ satisfying $displaystyle lim_{xto d} f'(x)=infty$ for some $d in (a,b)$ under the assumptions in Darboux's theorem?
If so, I can conclude that for each $x in (a,b),$ either $(i)$ $f'$ is continuous at $x$ or $(ii)$ $f'$ oscillates near $x.$
It seems that there is no differentiable function $f$ on $[a,b]$ satisfying function $displaystyle lim_{xto d} f'(x)=infty$ and $exists f'(d)$ for some $d in (a,b)$.
Please let me know if you have any idea or comment for my question. Thanks in advance!
real-analysis calculus analysis
asked Dec 30 '18 at 1:36
07060706
460311
$endgroup$
1
$begingroup$
It is possible that $f'(d)$ exists but $lim_{x to d} f'(x)$ fails to exist (oscillating). It can even be unbounded. An example is $f(x) =x^2 sin x^{-2} $ for $x neq 0$ and $f(0) = 0$. You just can't have the derivative tend to $infty$.
$endgroup$
– RRL
Dec 30 '18 at 3:35
$begingroup$
@RRL Thanks for your comment. Now I get to know the information a lot from Darboux theorem.
$endgroup$
– 0706
Dec 30 '18 at 3:43
add a comment |
$begingroup$
Darboux's theorem says that $f'$ has intermediate value property. More precisely,
( Darboux's theorem) If $f$ is differentiable on $[a,b]$, and if r is any number for which $f'(a)<r<f'(b)$ then $exists$ c in (a,b) such that $f'(r)=c$.
Thus Darboux's theorem implies that $f'$ cannot have any simple discontinuities on $[a,b].$
I have a question as follows:
Is there a function $f$ satisfying $displaystyle lim_{xto d} f'(x)=infty$ for some $d in (a,b)$ under the assumptions in Darboux's theorem?
If so, I can conclude that for each $x in (a,b),$ either $(i)$ $f'$ is continuous at $x$ or $(ii)$ $f'$ oscillates near $x.$
It seems that there is no differentiable function $f$ on $[a,b]$ satisfying function $displaystyle lim_{xto d} f'(x)=infty$ and $exists f'(d)$ for some $d in (a,b)$.
Please let me know if you have any idea or comment for my question. Thanks in advance!
real-analysis calculus analysis
asked Dec 30 '18 at 1:36
07060706
460311
$endgroup$
Darboux's theorem says that $f'$ has intermediate value property. More precisely,
( Darboux's theorem) If $f$ is differentiable on $[a,b]$, and if r is any number for which $f'(a)<r<f'(b)$ then $exists$ c in (a,b) such that $f'(r)=c$.
Thus Darboux's theorem implies that $f'$ cannot have any simple discontinuities on $[a,b].$
I have a question as follows:
Is there a function $f$ satisfying $displaystyle lim_{xto d} f'(x)=infty$ for some $d in (a,b)$ under the assumptions in Darboux's theorem?
If so, I can conclude that for each $x in (a,b),$ either $(i)$ $f'$ is continuous at $x$ or $(ii)$ $f'$ oscillates near $x.$
It seems that there is no differentiable function $f$ on $[a,b]$ satisfying function $displaystyle lim_{xto d} f'(x)=infty$ and $exists f'(d)$ for some $d in (a,b)$.
Please let me know if you have any idea or comment for my question. Thanks in advance!
real-analysis calculus analysis
real-analysis calculus analysis
asked Dec 30 '18 at 1:36
07060706
460311
asked Dec 30 '18 at 1:36
07060706
460311
asked Dec 30 '18 at 1:36
07060706
460311
asked Dec 30 '18 at 1:36
07060706
460311
asked Dec 30 '18 at 1:36
07060706
460311
460311
1
$begingroup$
It is possible that $f'(d)$ exists but $lim_{x to d} f'(x)$ fails to exist (oscillating). It can even be unbounded. An example is $f(x) =x^2 sin x^{-2} $ for $x neq 0$ and $f(0) = 0$. You just can't have the derivative tend to $infty$.
$endgroup$
– RRL
Dec 30 '18 at 3:35
$begingroup$
@RRL Thanks for your comment. Now I get to know the information a lot from Darboux theorem.
$endgroup$
– 0706
Dec 30 '18 at 3:43
add a comment |
1
$begingroup$
It is possible that $f'(d)$ exists but $lim_{x to d} f'(x)$ fails to exist (oscillating). It can even be unbounded. An example is $f(x) =x^2 sin x^{-2} $ for $x neq 0$ and $f(0) = 0$. You just can't have the derivative tend to $infty$.
$endgroup$
– RRL
Dec 30 '18 at 3:35
$begingroup$
@RRL Thanks for your comment. Now I get to know the information a lot from Darboux theorem.
$endgroup$
– 0706
Dec 30 '18 at 3:43
1
1
$begingroup$
It is possible that $f'(d)$ exists but $lim_{x to d} f'(x)$ fails to exist (oscillating). It can even be unbounded. An example is $f(x) =x^2 sin x^{-2} $ for $x neq 0$ and $f(0) = 0$. You just can't have the derivative tend to $infty$.
$endgroup$
– RRL
Dec 30 '18 at 3:35
$begingroup$
It is possible that $f'(d)$ exists but $lim_{x to d} f'(x)$ fails to exist (oscillating). It can even be unbounded. An example is $f(x) =x^2 sin x^{-2} $ for $x neq 0$ and $f(0) = 0$. You just can't have the derivative tend to $infty$.
$endgroup$
– RRL
Dec 30 '18 at 3:35
$begingroup$
@RRL Thanks for your comment. Now I get to know the information a lot from Darboux theorem.
$endgroup$
– 0706
Dec 30 '18 at 3:43
$begingroup$
@RRL Thanks for your comment. Now I get to know the information a lot from Darboux theorem.
$endgroup$
– 0706
Dec 30 '18 at 3:43
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No, there can't be.
If $f'(x)to +infty$ for $xto d$, then there will be $delta$ such that $f'(x) > f'(d)+2$ on $(d, d+delta)$. And this contradicts Darboux's theorem on $[d,d+delta/2]$ -- there won't be any place with $f'(x) = f'(d)+1$.
Be a bit careful about the precise meaning of "oscillates" in your conclusion (ii), though. It is possible for $f'$ to have a strict global minimum at $d$, for example.
answered Dec 30 '18 at 2:26
Henning MakholmHenning Makholm
244k17312556
$endgroup$
$begingroup$
Thanks for your answer. In the last paragraph in your answer, you are right. The definition of oscillation is not clear. By the way, is it possible for $f’$ not to be oscillate? I think, at least one side, $f’$ should oscillate. Is it wrong?
$endgroup$
– 0706
Dec 30 '18 at 2:50
$begingroup$
@0706: That depends on exactly what "oscillate" means here. (I'm not sure if there is a standard definition and if so what it is).
$endgroup$
– Henning Makholm
Dec 30 '18 at 2:52
add a comment |
$begingroup$
If $f$ is differentiable on $(a,b)$ and $x_n to d in (a,b)$ from the right, we have
$$f'(d) = lim_{n to infty} frac{f(x_n) -f(d)}{x_n - d}$$
By the MVT there is a sequence $xi_n in (d,x_n)$ such that
$$f'(d) = lim_{n to infty} frac{f(x_n) -f(d)}{x_n - d} = lim_{n to infty} f'(xi_n)$$
However, if $f'(x) to infty$ as $x to d$, then we must have $f'(xi_n) to infty$ since $xi_n to d$, a contradiction.
Hence, your function does not exist.
answered Dec 30 '18 at 2:15
RRLRRL
53.8k52675
$endgroup$
add a comment |
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2 Answers
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$begingroup$
No, there can't be.
If $f'(x)to +infty$ for $xto d$, then there will be $delta$ such that $f'(x) > f'(d)+2$ on $(d, d+delta)$. And this contradicts Darboux's theorem on $[d,d+delta/2]$ -- there won't be any place with $f'(x) = f'(d)+1$.
Be a bit careful about the precise meaning of "oscillates" in your conclusion (ii), though. It is possible for $f'$ to have a strict global minimum at $d$, for example.
answered Dec 30 '18 at 2:26
Henning MakholmHenning Makholm
244k17312556
$endgroup$
$begingroup$
Thanks for your answer. In the last paragraph in your answer, you are right. The definition of oscillation is not clear. By the way, is it possible for $f’$ not to be oscillate? I think, at least one side, $f’$ should oscillate. Is it wrong?
$endgroup$
– 0706
Dec 30 '18 at 2:50
$begingroup$
@0706: That depends on exactly what "oscillate" means here. (I'm not sure if there is a standard definition and if so what it is).
$endgroup$
– Henning Makholm
Dec 30 '18 at 2:52
add a comment |
$begingroup$
No, there can't be.
If $f'(x)to +infty$ for $xto d$, then there will be $delta$ such that $f'(x) > f'(d)+2$ on $(d, d+delta)$. And this contradicts Darboux's theorem on $[d,d+delta/2]$ -- there won't be any place with $f'(x) = f'(d)+1$.
Be a bit careful about the precise meaning of "oscillates" in your conclusion (ii), though. It is possible for $f'$ to have a strict global minimum at $d$, for example.
answered Dec 30 '18 at 2:26
Henning MakholmHenning Makholm
244k17312556
$endgroup$
$begingroup$
Thanks for your answer. In the last paragraph in your answer, you are right. The definition of oscillation is not clear. By the way, is it possible for $f’$ not to be oscillate? I think, at least one side, $f’$ should oscillate. Is it wrong?
$endgroup$
– 0706
Dec 30 '18 at 2:50
$begingroup$
@0706: That depends on exactly what "oscillate" means here. (I'm not sure if there is a standard definition and if so what it is).
$endgroup$
– Henning Makholm
Dec 30 '18 at 2:52
add a comment |
$begingroup$
No, there can't be.
If $f'(x)to +infty$ for $xto d$, then there will be $delta$ such that $f'(x) > f'(d)+2$ on $(d, d+delta)$. And this contradicts Darboux's theorem on $[d,d+delta/2]$ -- there won't be any place with $f'(x) = f'(d)+1$.
Be a bit careful about the precise meaning of "oscillates" in your conclusion (ii), though. It is possible for $f'$ to have a strict global minimum at $d$, for example.
answered Dec 30 '18 at 2:26
Henning MakholmHenning Makholm
244k17312556
$endgroup$
No, there can't be.
If $f'(x)to +infty$ for $xto d$, then there will be $delta$ such that $f'(x) > f'(d)+2$ on $(d, d+delta)$. And this contradicts Darboux's theorem on $[d,d+delta/2]$ -- there won't be any place with $f'(x) = f'(d)+1$.
Be a bit careful about the precise meaning of "oscillates" in your conclusion (ii), though. It is possible for $f'$ to have a strict global minimum at $d$, for example.
answered Dec 30 '18 at 2:26
Henning MakholmHenning Makholm
244k17312556
edited Dec 30 '18 at 2:33
answered Dec 30 '18 at 2:26
Henning MakholmHenning Makholm
244k17312556
answered Dec 30 '18 at 2:26
Henning MakholmHenning Makholm
244k17312556
answered Dec 30 '18 at 2:26
Henning MakholmHenning Makholm
244k17312556
244k17312556
$begingroup$
Thanks for your answer. In the last paragraph in your answer, you are right. The definition of oscillation is not clear. By the way, is it possible for $f’$ not to be oscillate? I think, at least one side, $f’$ should oscillate. Is it wrong?
$endgroup$
– 0706
Dec 30 '18 at 2:50
$begingroup$
@0706: That depends on exactly what "oscillate" means here. (I'm not sure if there is a standard definition and if so what it is).
$endgroup$
– Henning Makholm
Dec 30 '18 at 2:52
add a comment |
$begingroup$
Thanks for your answer. In the last paragraph in your answer, you are right. The definition of oscillation is not clear. By the way, is it possible for $f’$ not to be oscillate? I think, at least one side, $f’$ should oscillate. Is it wrong?
$endgroup$
– 0706
Dec 30 '18 at 2:50
$begingroup$
@0706: That depends on exactly what "oscillate" means here. (I'm not sure if there is a standard definition and if so what it is).
$endgroup$
– Henning Makholm
Dec 30 '18 at 2:52
$begingroup$
Thanks for your answer. In the last paragraph in your answer, you are right. The definition of oscillation is not clear. By the way, is it possible for $f’$ not to be oscillate? I think, at least one side, $f’$ should oscillate. Is it wrong?
$endgroup$
– 0706
Dec 30 '18 at 2:50
$begingroup$
Thanks for your answer. In the last paragraph in your answer, you are right. The definition of oscillation is not clear. By the way, is it possible for $f’$ not to be oscillate? I think, at least one side, $f’$ should oscillate. Is it wrong?
$endgroup$
– 0706
Dec 30 '18 at 2:50
$begingroup$
@0706: That depends on exactly what "oscillate" means here. (I'm not sure if there is a standard definition and if so what it is).
$endgroup$
– Henning Makholm
Dec 30 '18 at 2:52
$begingroup$
@0706: That depends on exactly what "oscillate" means here. (I'm not sure if there is a standard definition and if so what it is).
$endgroup$
– Henning Makholm
Dec 30 '18 at 2:52
add a comment |
$begingroup$
If $f$ is differentiable on $(a,b)$ and $x_n to d in (a,b)$ from the right, we have
$$f'(d) = lim_{n to infty} frac{f(x_n) -f(d)}{x_n - d}$$
By the MVT there is a sequence $xi_n in (d,x_n)$ such that
$$f'(d) = lim_{n to infty} frac{f(x_n) -f(d)}{x_n - d} = lim_{n to infty} f'(xi_n)$$
However, if $f'(x) to infty$ as $x to d$, then we must have $f'(xi_n) to infty$ since $xi_n to d$, a contradiction.
Hence, your function does not exist.
answered Dec 30 '18 at 2:15
RRLRRL
53.8k52675
$endgroup$
add a comment |
$begingroup$
If $f$ is differentiable on $(a,b)$ and $x_n to d in (a,b)$ from the right, we have
$$f'(d) = lim_{n to infty} frac{f(x_n) -f(d)}{x_n - d}$$
By the MVT there is a sequence $xi_n in (d,x_n)$ such that
$$f'(d) = lim_{n to infty} frac{f(x_n) -f(d)}{x_n - d} = lim_{n to infty} f'(xi_n)$$
However, if $f'(x) to infty$ as $x to d$, then we must have $f'(xi_n) to infty$ since $xi_n to d$, a contradiction.
Hence, your function does not exist.
answered Dec 30 '18 at 2:15
RRLRRL
53.8k52675
$endgroup$
add a comment |
$begingroup$
If $f$ is differentiable on $(a,b)$ and $x_n to d in (a,b)$ from the right, we have
$$f'(d) = lim_{n to infty} frac{f(x_n) -f(d)}{x_n - d}$$
By the MVT there is a sequence $xi_n in (d,x_n)$ such that
$$f'(d) = lim_{n to infty} frac{f(x_n) -f(d)}{x_n - d} = lim_{n to infty} f'(xi_n)$$
However, if $f'(x) to infty$ as $x to d$, then we must have $f'(xi_n) to infty$ since $xi_n to d$, a contradiction.
Hence, your function does not exist.
answered Dec 30 '18 at 2:15
RRLRRL
53.8k52675
$endgroup$
If $f$ is differentiable on $(a,b)$ and $x_n to d in (a,b)$ from the right, we have
$$f'(d) = lim_{n to infty} frac{f(x_n) -f(d)}{x_n - d}$$
By the MVT there is a sequence $xi_n in (d,x_n)$ such that
$$f'(d) = lim_{n to infty} frac{f(x_n) -f(d)}{x_n - d} = lim_{n to infty} f'(xi_n)$$
However, if $f'(x) to infty$ as $x to d$, then we must have $f'(xi_n) to infty$ since $xi_n to d$, a contradiction.
Hence, your function does not exist.
answered Dec 30 '18 at 2:15
RRLRRL
53.8k52675
edited Dec 30 '18 at 2:38
answered Dec 30 '18 at 2:15
RRLRRL
53.8k52675
answered Dec 30 '18 at 2:15
RRLRRL
53.8k52675
answered Dec 30 '18 at 2:15
RRLRRL
53.8k52675
53.8k52675
add a comment |
add a comment |
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$begingroup$
It is possible that $f'(d)$ exists but $lim_{x to d} f'(x)$ fails to exist (oscillating). It can even be unbounded. An example is $f(x) =x^2 sin x^{-2} $ for $x neq 0$ and $f(0) = 0$. You just can't have the derivative tend to $infty$.
$endgroup$
– RRL
Dec 30 '18 at 3:35
$begingroup$
@RRL Thanks for your comment. Now I get to know the information a lot from Darboux theorem.
$endgroup$
– 0706
Dec 30 '18 at 3:43