differentiable function with unbounded, integrable derivation
up vote
0
down vote
favorite
(1) Is there an exmaple of function $fcolon [-1,1]tomathbb{R}$ which is differentiable, with unbounded derivation $f'$ and such that $f'$ is Lebesgue-integrable (i.e. $L^1$-.intebrale)?
I thought about $f(x)=x^2sin(frac{1}{x^2})$ if $xneq 0$ and $f(x)=0$ for $x=0$. This is differentiable, with unbounded derivation $f'(x)=2xsin(frac{1}{x^2})-frac{2}{x}cos(frac{1}{x^2})$ on $[-1,1]$. However, is $f'$ integrable? It think not, because it contains 1/x Showing that $1/x$ is NOT Lebesgue Integrable on $(0,1]$? I extracted this question here function is not Lebesgue integrable.
integration analysis derivatives lebesgue-integral
asked Nov 21 at 14:33
Gero
23319
add a comment |
up vote
0
down vote
favorite
(1) Is there an exmaple of function $fcolon [-1,1]tomathbb{R}$ which is differentiable, with unbounded derivation $f'$ and such that $f'$ is Lebesgue-integrable (i.e. $L^1$-.intebrale)?
I thought about $f(x)=x^2sin(frac{1}{x^2})$ if $xneq 0$ and $f(x)=0$ for $x=0$. This is differentiable, with unbounded derivation $f'(x)=2xsin(frac{1}{x^2})-frac{2}{x}cos(frac{1}{x^2})$ on $[-1,1]$. However, is $f'$ integrable? It think not, because it contains 1/x Showing that $1/x$ is NOT Lebesgue Integrable on $(0,1]$? I extracted this question here function is not Lebesgue integrable.
integration analysis derivatives lebesgue-integral
asked Nov 21 at 14:33
Gero
23319
1
It seems you are looking for a function that has a derivative at every point in [-1,1] and where the derivative is unbounded. That is true for $f(x) = x^2 sin(1/x^2)$ since $f'(0) = lim_{h to 0} frac{h^2 sin(1/h^2)}{h} = 0$. yet $f'(x)$ is unbounded as $x$ approaches $0$. Of course the derivative in this case is not Lebesgue (absolutely) integrable as you suggest. However the answer you accepted shows a function where $f'(x)$ is unbounded near $1$ but $f'(1)$ does not exist. Are you OK with that?
– RRL
Nov 21 at 16:40
1
Because it is very easy to find an example of a function with an unbounded derivative that is Lebesgue integrable but where the derivative does not exist at one point. The obvious choice being $f(x) = sqrt{x}$ on $[0,1]$.
– RRL
Nov 21 at 16:52
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
(1) Is there an exmaple of function $fcolon [-1,1]tomathbb{R}$ which is differentiable, with unbounded derivation $f'$ and such that $f'$ is Lebesgue-integrable (i.e. $L^1$-.intebrale)?
I thought about $f(x)=x^2sin(frac{1}{x^2})$ if $xneq 0$ and $f(x)=0$ for $x=0$. This is differentiable, with unbounded derivation $f'(x)=2xsin(frac{1}{x^2})-frac{2}{x}cos(frac{1}{x^2})$ on $[-1,1]$. However, is $f'$ integrable? It think not, because it contains 1/x Showing that $1/x$ is NOT Lebesgue Integrable on $(0,1]$? I extracted this question here function is not Lebesgue integrable.
integration analysis derivatives lebesgue-integral
asked Nov 21 at 14:33
Gero
23319
(1) Is there an exmaple of function $fcolon [-1,1]tomathbb{R}$ which is differentiable, with unbounded derivation $f'$ and such that $f'$ is Lebesgue-integrable (i.e. $L^1$-.intebrale)?
I thought about $f(x)=x^2sin(frac{1}{x^2})$ if $xneq 0$ and $f(x)=0$ for $x=0$. This is differentiable, with unbounded derivation $f'(x)=2xsin(frac{1}{x^2})-frac{2}{x}cos(frac{1}{x^2})$ on $[-1,1]$. However, is $f'$ integrable? It think not, because it contains 1/x Showing that $1/x$ is NOT Lebesgue Integrable on $(0,1]$? I extracted this question here function is not Lebesgue integrable.
integration analysis derivatives lebesgue-integral
integration analysis derivatives lebesgue-integral
asked Nov 21 at 14:33
Gero
23319
asked Nov 21 at 14:33
Gero
23319
edited Nov 21 at 15:32
asked Nov 21 at 14:33
Gero
23319
asked Nov 21 at 14:33
Gero
23319
asked Nov 21 at 14:33
Gero
23319
23319
1
It seems you are looking for a function that has a derivative at every point in [-1,1] and where the derivative is unbounded. That is true for $f(x) = x^2 sin(1/x^2)$ since $f'(0) = lim_{h to 0} frac{h^2 sin(1/h^2)}{h} = 0$. yet $f'(x)$ is unbounded as $x$ approaches $0$. Of course the derivative in this case is not Lebesgue (absolutely) integrable as you suggest. However the answer you accepted shows a function where $f'(x)$ is unbounded near $1$ but $f'(1)$ does not exist. Are you OK with that?
– RRL
Nov 21 at 16:40
1
Because it is very easy to find an example of a function with an unbounded derivative that is Lebesgue integrable but where the derivative does not exist at one point. The obvious choice being $f(x) = sqrt{x}$ on $[0,1]$.
– RRL
Nov 21 at 16:52
add a comment |
1
It seems you are looking for a function that has a derivative at every point in [-1,1] and where the derivative is unbounded. That is true for $f(x) = x^2 sin(1/x^2)$ since $f'(0) = lim_{h to 0} frac{h^2 sin(1/h^2)}{h} = 0$. yet $f'(x)$ is unbounded as $x$ approaches $0$. Of course the derivative in this case is not Lebesgue (absolutely) integrable as you suggest. However the answer you accepted shows a function where $f'(x)$ is unbounded near $1$ but $f'(1)$ does not exist. Are you OK with that?
– RRL
Nov 21 at 16:40
1
Because it is very easy to find an example of a function with an unbounded derivative that is Lebesgue integrable but where the derivative does not exist at one point. The obvious choice being $f(x) = sqrt{x}$ on $[0,1]$.
– RRL
Nov 21 at 16:52
1
1
It seems you are looking for a function that has a derivative at every point in [-1,1] and where the derivative is unbounded. That is true for $f(x) = x^2 sin(1/x^2)$ since $f'(0) = lim_{h to 0} frac{h^2 sin(1/h^2)}{h} = 0$. yet $f'(x)$ is unbounded as $x$ approaches $0$. Of course the derivative in this case is not Lebesgue (absolutely) integrable as you suggest. However the answer you accepted shows a function where $f'(x)$ is unbounded near $1$ but $f'(1)$ does not exist. Are you OK with that?
– RRL
Nov 21 at 16:40
It seems you are looking for a function that has a derivative at every point in [-1,1] and where the derivative is unbounded. That is true for $f(x) = x^2 sin(1/x^2)$ since $f'(0) = lim_{h to 0} frac{h^2 sin(1/h^2)}{h} = 0$. yet $f'(x)$ is unbounded as $x$ approaches $0$. Of course the derivative in this case is not Lebesgue (absolutely) integrable as you suggest. However the answer you accepted shows a function where $f'(x)$ is unbounded near $1$ but $f'(1)$ does not exist. Are you OK with that?
– RRL
Nov 21 at 16:40
1
1
Because it is very easy to find an example of a function with an unbounded derivative that is Lebesgue integrable but where the derivative does not exist at one point. The obvious choice being $f(x) = sqrt{x}$ on $[0,1]$.
– RRL
Nov 21 at 16:52
Because it is very easy to find an example of a function with an unbounded derivative that is Lebesgue integrable but where the derivative does not exist at one point. The obvious choice being $f(x) = sqrt{x}$ on $[0,1]$.
– RRL
Nov 21 at 16:52
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
$f(x) = sin^{-1} x$ has $f'(x)=(1-x^2)^{-1/2}$ which is unbounded, but $L^1$ (well it's Riemann integrable anyway).
answered Nov 21 at 14:40
Richard Martin
1,63918
1
Does $f'(1) = lim_{x to 1-} frac{sin^{-1}(1) - sin^{-1}(x)}{1 - x}$ exist?
– RRL
Nov 21 at 16:46
No. It doesn't need to; the function is to be cts on the closed interval and diffble on the open one.
– Richard Martin
Nov 21 at 16:55
I know you have that. OP needs to clarify because as worded above a "differentiable" function is sought and only a closed interval is shown. The OP proposed example is differentiable everwhere in the closed interval. That is the difficult part, otherwise $sqrt{x}$, would suffice.
– RRL
Nov 21 at 17:31
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
$f(x) = sin^{-1} x$ has $f'(x)=(1-x^2)^{-1/2}$ which is unbounded, but $L^1$ (well it's Riemann integrable anyway).
answered Nov 21 at 14:40
Richard Martin
1,63918
1
Does $f'(1) = lim_{x to 1-} frac{sin^{-1}(1) - sin^{-1}(x)}{1 - x}$ exist?
– RRL
Nov 21 at 16:46
No. It doesn't need to; the function is to be cts on the closed interval and diffble on the open one.
– Richard Martin
Nov 21 at 16:55
I know you have that. OP needs to clarify because as worded above a "differentiable" function is sought and only a closed interval is shown. The OP proposed example is differentiable everwhere in the closed interval. That is the difficult part, otherwise $sqrt{x}$, would suffice.
– RRL
Nov 21 at 17:31
add a comment |
up vote
2
down vote
accepted
$f(x) = sin^{-1} x$ has $f'(x)=(1-x^2)^{-1/2}$ which is unbounded, but $L^1$ (well it's Riemann integrable anyway).
answered Nov 21 at 14:40
Richard Martin
1,63918
1
Does $f'(1) = lim_{x to 1-} frac{sin^{-1}(1) - sin^{-1}(x)}{1 - x}$ exist?
– RRL
Nov 21 at 16:46
No. It doesn't need to; the function is to be cts on the closed interval and diffble on the open one.
– Richard Martin
Nov 21 at 16:55
I know you have that. OP needs to clarify because as worded above a "differentiable" function is sought and only a closed interval is shown. The OP proposed example is differentiable everwhere in the closed interval. That is the difficult part, otherwise $sqrt{x}$, would suffice.
– RRL
Nov 21 at 17:31
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
$f(x) = sin^{-1} x$ has $f'(x)=(1-x^2)^{-1/2}$ which is unbounded, but $L^1$ (well it's Riemann integrable anyway).
answered Nov 21 at 14:40
Richard Martin
1,63918
$f(x) = sin^{-1} x$ has $f'(x)=(1-x^2)^{-1/2}$ which is unbounded, but $L^1$ (well it's Riemann integrable anyway).
answered Nov 21 at 14:40
Richard Martin
1,63918
answered Nov 21 at 14:40
Richard Martin
1,63918
answered Nov 21 at 14:40
Richard Martin
1,63918
answered Nov 21 at 14:40
Richard Martin
1,63918
1,63918
1
Does $f'(1) = lim_{x to 1-} frac{sin^{-1}(1) - sin^{-1}(x)}{1 - x}$ exist?
– RRL
Nov 21 at 16:46
No. It doesn't need to; the function is to be cts on the closed interval and diffble on the open one.
– Richard Martin
Nov 21 at 16:55
I know you have that. OP needs to clarify because as worded above a "differentiable" function is sought and only a closed interval is shown. The OP proposed example is differentiable everwhere in the closed interval. That is the difficult part, otherwise $sqrt{x}$, would suffice.
– RRL
Nov 21 at 17:31
add a comment |
1
Does $f'(1) = lim_{x to 1-} frac{sin^{-1}(1) - sin^{-1}(x)}{1 - x}$ exist?
– RRL
Nov 21 at 16:46
No. It doesn't need to; the function is to be cts on the closed interval and diffble on the open one.
– Richard Martin
Nov 21 at 16:55
I know you have that. OP needs to clarify because as worded above a "differentiable" function is sought and only a closed interval is shown. The OP proposed example is differentiable everwhere in the closed interval. That is the difficult part, otherwise $sqrt{x}$, would suffice.
– RRL
Nov 21 at 17:31
1
1
Does $f'(1) = lim_{x to 1-} frac{sin^{-1}(1) - sin^{-1}(x)}{1 - x}$ exist?
– RRL
Nov 21 at 16:46
Does $f'(1) = lim_{x to 1-} frac{sin^{-1}(1) - sin^{-1}(x)}{1 - x}$ exist?
– RRL
Nov 21 at 16:46
No. It doesn't need to; the function is to be cts on the closed interval and diffble on the open one.
– Richard Martin
Nov 21 at 16:55
No. It doesn't need to; the function is to be cts on the closed interval and diffble on the open one.
– Richard Martin
Nov 21 at 16:55
I know you have that. OP needs to clarify because as worded above a "differentiable" function is sought and only a closed interval is shown. The OP proposed example is differentiable everwhere in the closed interval. That is the difficult part, otherwise $sqrt{x}$, would suffice.
– RRL
Nov 21 at 17:31
I know you have that. OP needs to clarify because as worded above a "differentiable" function is sought and only a closed interval is shown. The OP proposed example is differentiable everwhere in the closed interval. That is the difficult part, otherwise $sqrt{x}$, would suffice.
– RRL
Nov 21 at 17:31
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3007792%2fdifferentiable-function-with-unbounded-integrable-derivation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
It seems you are looking for a function that has a derivative at every point in [-1,1] and where the derivative is unbounded. That is true for $f(x) = x^2 sin(1/x^2)$ since $f'(0) = lim_{h to 0} frac{h^2 sin(1/h^2)}{h} = 0$. yet $f'(x)$ is unbounded as $x$ approaches $0$. Of course the derivative in this case is not Lebesgue (absolutely) integrable as you suggest. However the answer you accepted shows a function where $f'(x)$ is unbounded near $1$ but $f'(1)$ does not exist. Are you OK with that?
– RRL
Nov 21 at 16:40
1
Because it is very easy to find an example of a function with an unbounded derivative that is Lebesgue integrable but where the derivative does not exist at one point. The obvious choice being $f(x) = sqrt{x}$ on $[0,1]$.
– RRL
Nov 21 at 16:52