Is continuous increasing function in $H^1([0,1])$
Consider a function $f(x):[0,1]rightarrow mathbb{R}$. If $f(x)$ is continuous and increasing, is $f(x)$ in $H^1(Omega)$, the Sobolev space with norm $sqrt{int_0^1 (|f(x)|^2 + |Df(x)|^2) dx}$?
continuity sobolev-spaces
edited Dec 12 '18 at 22:18
Key Flex
7,74541232
asked Dec 12 '18 at 22:18
KKKKKK
1658
add a comment |
Consider a function $f(x):[0,1]rightarrow mathbb{R}$. If $f(x)$ is continuous and increasing, is $f(x)$ in $H^1(Omega)$, the Sobolev space with norm $sqrt{int_0^1 (|f(x)|^2 + |Df(x)|^2) dx}$?
continuity sobolev-spaces
edited Dec 12 '18 at 22:18
Key Flex
7,74541232
asked Dec 12 '18 at 22:18
KKKKKK
1658
If $f$ is the "Devil's staircase" then it doesn't have a weak derivative (however it has a distributional derivative)
– Calvin Khor
Dec 12 '18 at 22:30
@CalvinKhor Thank you. But this function is only nondecreasing right? By ``increasing'' I mean strictly increasing.
– KKK
Dec 12 '18 at 22:39
add a comment |
Consider a function $f(x):[0,1]rightarrow mathbb{R}$. If $f(x)$ is continuous and increasing, is $f(x)$ in $H^1(Omega)$, the Sobolev space with norm $sqrt{int_0^1 (|f(x)|^2 + |Df(x)|^2) dx}$?
continuity sobolev-spaces
edited Dec 12 '18 at 22:18
Key Flex
7,74541232
asked Dec 12 '18 at 22:18
KKKKKK
1658
Consider a function $f(x):[0,1]rightarrow mathbb{R}$. If $f(x)$ is continuous and increasing, is $f(x)$ in $H^1(Omega)$, the Sobolev space with norm $sqrt{int_0^1 (|f(x)|^2 + |Df(x)|^2) dx}$?
continuity sobolev-spaces
continuity sobolev-spaces
edited Dec 12 '18 at 22:18
Key Flex
7,74541232
asked Dec 12 '18 at 22:18
KKKKKK
1658
edited Dec 12 '18 at 22:18
Key Flex
7,74541232
asked Dec 12 '18 at 22:18
KKKKKK
1658
edited Dec 12 '18 at 22:18
Key Flex
7,74541232
edited Dec 12 '18 at 22:18
Key Flex
7,74541232
edited Dec 12 '18 at 22:18
Key Flex
7,74541232
7,74541232
asked Dec 12 '18 at 22:18
KKKKKK
1658
asked Dec 12 '18 at 22:18
KKKKKK
1658
asked Dec 12 '18 at 22:18
KKKKKK
1658
1658
If $f$ is the "Devil's staircase" then it doesn't have a weak derivative (however it has a distributional derivative)
– Calvin Khor
Dec 12 '18 at 22:30
@CalvinKhor Thank you. But this function is only nondecreasing right? By ``increasing'' I mean strictly increasing.
– KKK
Dec 12 '18 at 22:39
add a comment |
If $f$ is the "Devil's staircase" then it doesn't have a weak derivative (however it has a distributional derivative)
– Calvin Khor
Dec 12 '18 at 22:30
@CalvinKhor Thank you. But this function is only nondecreasing right? By ``increasing'' I mean strictly increasing.
– KKK
Dec 12 '18 at 22:39
If $f$ is the "Devil's staircase" then it doesn't have a weak derivative (however it has a distributional derivative)
– Calvin Khor
Dec 12 '18 at 22:30
If $f$ is the "Devil's staircase" then it doesn't have a weak derivative (however it has a distributional derivative)
– Calvin Khor
Dec 12 '18 at 22:30
@CalvinKhor Thank you. But this function is only nondecreasing right? By ``increasing'' I mean strictly increasing.
– KKK
Dec 12 '18 at 22:39
@CalvinKhor Thank you. But this function is only nondecreasing right? By ``increasing'' I mean strictly increasing.
– KKK
Dec 12 '18 at 22:39
add a comment |
1 Answer
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As in the comments, the Devil's staircase, $mathcal D=mathcal D(x)$ is a non-decreasing function with no weak derivative; therefore it cannot lie in any Sobolev space. Since the OP wants a function that is increasing, one can consider $f(x) := x + mathcal D(x)$ instead.
answered Dec 12 '18 at 23:12
Calvin KhorCalvin Khor
11.2k21438
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
votes
As in the comments, the Devil's staircase, $mathcal D=mathcal D(x)$ is a non-decreasing function with no weak derivative; therefore it cannot lie in any Sobolev space. Since the OP wants a function that is increasing, one can consider $f(x) := x + mathcal D(x)$ instead.
answered Dec 12 '18 at 23:12
Calvin KhorCalvin Khor
11.2k21438
add a comment |
As in the comments, the Devil's staircase, $mathcal D=mathcal D(x)$ is a non-decreasing function with no weak derivative; therefore it cannot lie in any Sobolev space. Since the OP wants a function that is increasing, one can consider $f(x) := x + mathcal D(x)$ instead.
answered Dec 12 '18 at 23:12
Calvin KhorCalvin Khor
11.2k21438
add a comment |
As in the comments, the Devil's staircase, $mathcal D=mathcal D(x)$ is a non-decreasing function with no weak derivative; therefore it cannot lie in any Sobolev space. Since the OP wants a function that is increasing, one can consider $f(x) := x + mathcal D(x)$ instead.
answered Dec 12 '18 at 23:12
Calvin KhorCalvin Khor
11.2k21438
As in the comments, the Devil's staircase, $mathcal D=mathcal D(x)$ is a non-decreasing function with no weak derivative; therefore it cannot lie in any Sobolev space. Since the OP wants a function that is increasing, one can consider $f(x) := x + mathcal D(x)$ instead.
answered Dec 12 '18 at 23:12
Calvin KhorCalvin Khor
11.2k21438
answered Dec 12 '18 at 23:12
Calvin KhorCalvin Khor
11.2k21438
answered Dec 12 '18 at 23:12
Calvin KhorCalvin Khor
11.2k21438
answered Dec 12 '18 at 23:12
Calvin KhorCalvin Khor
11.2k21438
11.2k21438
add a comment |
add a comment |
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If $f$ is the "Devil's staircase" then it doesn't have a weak derivative (however it has a distributional derivative)
– Calvin Khor
Dec 12 '18 at 22:30
@CalvinKhor Thank you. But this function is only nondecreasing right? By ``increasing'' I mean strictly increasing.
– KKK
Dec 12 '18 at 22:39